even number solutions

Answers toeven-numbered problems in

Mathematics forEconomic Analysis

Knut SydsæterPeter Hammond

Preface

Mathematics for Economic Analysis, Prentice Hall, 1995 has been out for a long time, and over the years wehave had many request for supplying solutions to the even-numbered problems. (Answers to the odd-numberedproblems are given in the main text.)

This manual provides answers to all the even-numbered problems in the text. These answers are takenfrom the old Instructors Manual which is no longer available. (In fact, some of the old answers have beenextended.) For many of the more interesting and/or difficult problems, detailed solutions are provided. Forsome of the simpler problems, only the final answer is presented.

Appendix A in the main text reviews elementary algebra. This manual includes a Test I, designed for thestudents themselves to see if they need to review particular sections of Appendix A.

Many students using our text will probably have some background in calculus. The accompanying Test IIis designed to give information to the students about what they actually know about single variable calculus,and about what needs to be studied more closely, perhaps in Chapters 6 to 9 of the text.

Oslo and Coventry, November 2010

Knut Sydsæter and Peter Hammond

Contact addresses:[email protected]@warwick.ac.uk

Version 1.0

07122010 1113

© Knut Sydsæter and Peter Hammond 2010

C H A P T E R 1 I N T R O D U C T I O N 1

Chapter 1 Introduction

1.3

2. (a) Put p/100 = x. Then the given expression becomes a + ax − (a + ax)x = a(1 − x2), as required.(b) $2000 · 1.05 · 0.95 = $1995. (c) The result is precisely the formula in (a). (d) With the notationused in the answer to (a), a − ax + (a − ax)x = a(1 − x2), which is the same expression as in (a).

4. (a) F = 32 yields C = 0; C = 100 yields F = 212. (b) F = 95C + 32

(c) F = 40 for C ≈ 4.4, F = 80 for C ≈ 26.7. The assertion is meaningless.

6. x = R(1 − p) − S(1 − q)

(p − q)(1 − p − q), y = pS − qR

(p − q)(1 − p − q)if (p − q)(1 − p − q) �= 0.

(Use, for example, (A.39) in Section A.9.)

1.4

2. (a) Correct. (b) Incorrect. (c) Incorrect. (d) Correct. (e) Incorrect.(f) Incorrect. (“Usually” the sum of two irrationals is irrational, but not always. For example, π and −π

are both irrationals, but π + (−π) = 0, which is rational.)

4. (a) y ≤ 3 − 34x (b) y > 3

2z (c) y ≤ (m − px)/q

6. |2 · 0 − 3| = | − 3| = 3, |2 · 12 − 3| = | − 2| = 2, |2 · 7

2 − 3| = |4| = 4

8. (a) 3 − 2x = 5 or 3 − 2x = −5, so −2x = 2 or −8. Hence x = −1 or x = 4.(b) −2 ≤ x ≤ 2 (c) 1 ≤ x ≤ 3 (d) −1/4 ≤ x ≤ 1 (e) x >

√2 or x < −√

2(f) 1 ≤ x2 ≤ 3, and so 1 ≤ x ≤ √

3 or −√3 ≤ x ≤ −1

1.5

2. (a) ⇒ right, ⇐ wrong (b) ⇒ wrong, ⇐ right (c) ⇒ right, ⇐ wrong(d) ⇒ and ⇐ both right (e) ⇒ wrong (0 · 5 = 0 · 4, but 5 �= 4), ⇐ right (f) ⇒ right, ⇐ wrong

4. x = 2. (x = −1, 0 or 1 make the equation meaningless. Multiplying each term by the common de-nominator x(x − 1)(x + 1) yields (x + 1)3 + (x − 1)3 − 2x(3x + 1) = 0. Expanding and simplifying,2x3 − 6x2 + 4x = 0, or 2x(x2 − 3x + 2) = 0, or 2x(x − 1)(x − 2) = 0. Hence, x = 2 is the onlysolution.)

6. (a) No solutions. (Squaring each side yields x−4 = x+5−18√

x + 5+81, which reduces to√

x + 5 = 5,with solution x = 20. But x = 20 does not satisfy the given equation.) (b) Just as in part (a) we findthat x must be a solution of

√x + 5 = 5, which has the solution x = 20. Inserting this value of x into

the given equation we find that x = 20 is the solution.

8. (a) x+√x + 4 = 2 ⇒ √

x + 4 = 2−x ⇒ x+4 = 4−4x+x2 ⇒ x2−5x = 0(1)⇒ x−5 = 0

(2)⇐ x = 5.Here implication (1) is incorrect (x2 − 5x = 0 ⇒ x − 5 = 0 or x = 0.) Implication (2) is correct, but itbreaks the chain of implications. (b) x = 0. (By modifying the argument we see that the given equationimplies x = 5 or x = 0. But only x = 0 is a solution.)

10. No. (To study law it seems that you need a course in logic.)


2 C H A P T E R 1 I N T R O D U C T I O N

1.6

2. Logically the two statements are equivalent, but the second statement is still an expressive poetic rein-forcement.

1.7

2. (a) No, not necessarily. (b) Yes (if anybody is both a painter and a poet).

4. F ∩B ∩C is the set of all female biology students in the university choir; M ∩F the female mathematicsstudents;

((M ∩B)\C

)\T the students who study both mathematics and biology but neither play tennisnor belong to the university choir.

6. (b) and (c) are true, the others are wrong. (Counter example for (a), (d), and (f): A = {1, 2}, B = {1},C = {1, 3}. As for (e), note in particular that A ∪ B = A ∪ C = A whenever B and C are subsets of A,even if B �= C.)

8. 50 − 35 = 15 liked only coffee, 40 − 35 = 5 liked only tea, 35 liked both, and 10 did not like either. Inall there were 15 + 5 + 35 + 10 = 65 who responded.

10. (a) Consider Fig. 1.7.10(a). Both sets consist of the elements in (1) and in (3).(b) Consider Fig. 1.7.10(b): A B consists of (3), (4), (2), and (5). Then (A B) C consists of (4),(5), (6), and (7), which is described verbally in the problem.

(1) (2) (3) BA

(4)(1)

(7)

(3)(2)

(5)

(6)(8)

A B

C

Figure 1.7.10(a) Figure 1.7.10(b)

12. (a) Consider Fig. 1.7.10(b), and let ni denote the number of people in the set marked (i), for i = 1, 2, . . . , 8.The responses reported imply that: n1+n3+n4+n7 = 420; n1+n2+n5+n7 = 316; n2+n3+n6+n7 =160; n1 + n7 = 116; n3 + n7 = 100; n2 + n7 = 30; n7 = 16. From these equations we easily findn1 = 100, n2 = 14, n3 = 84, n4 = 220, n5 = 186, n6 = 46, n7 = 16. (i) n3 + n4 = 304 had read A

but not B; (ii) n6 = 46; (iii) 1000 − (n1 + n2 + n3 + n4 + n5 + n6 + n7) = 334.(b) (i) n(A \ B) (ii) n(C \ (A ∪ B)) (iii) n(� \ (A ∪ B ∪ C)) = n(�) − n(A ∪ B ∪ C))

(c) The equality is n1 + n2 + n3 + n4 + n5 + n6 + n7 = (n1 + n3 + n4 + n7) + (n1 + n2 + n5 + n7) +(n2 + n3 + n6 + n7)− (n1 + n7)− (n3 + n7)− (n2 + n7)+ n7, which is easily verified. The last equalityis a special case of n(� \ D) = n(�) − n(D): The number of persons who are in �, but not in D, is thenumber of persons in all of � minus the number of those who are in D.


C H A P T E R 2 F U N C T I O N S O F O N E V A R I A B L E : I N T R O D U C T I O N 3

Chapter 2 Functions of One Variable: Introduction

2.2

2. F(0) = F(−3) = 10, F(a + h) − F(a) = 10 − 10 = 0

4. (a) f (−1/10) = −10/101, f (0) = 0, f (1/√

2) = √2/3, f (

√π) = √

π/(1 + π),f (2) = 2/5 (b) f (−x) = −x/(1 + (−x)2) = −x/(1 + x2) = −f (x) and moreover, f (1/x) =(1/x)/[1 + (1/x)2] = (1/x) · x2/[1 + (1/x)2] · x2 = x/(1 + x2) = f (x)

6. F(0) = 2, F(−3) = √19, F(t + 1) = √

t2 + 3

8. (a) b(0) = 0, b(50) = 100/11, b(100) = 200(b) b(50 + h) − b(50) is the additional cost of removing h% more than 50% of the impurities.

10. (a) No. f (2 + 1) = f (3) = 2 · 32 = 18, whereas f (2) + f (1) = 2 · 22 + 2 · 12 = 8 + 2 = 10.(b)Yes. f (2+1) = f (2)+f (1) = −9 (c) No. f (2+1) = f (3) = √

3 ≈ 1.73, whereas f (2)+f (1) =√2 + 1 ≈ 2.41.

12. See Figs. 2.2.12(a) and 2.2.12(b).

x2

x · 1 1 · 1

1 · x

x 1

x

1

x

1

x

1

Figure 2.2.12(a). Area = (x +1)2 = x2 +2x +1 Figure 2.2.12(b) Area = x2 +1

14. (a) (−∞, 2) ∪ (2, ∞) (b) f (8) = 5

(c) f (x) = 3x + 6

x − 2= 3 ⇐⇒ 3x + 6 = 3(x − 2) ⇐⇒ 6 = −6, which is absurd.

16. (a) is invalid: |2 + (−2)| = 0, whereas |2| + | − 2| = 2 + 2 = 4. (b), called the triangle inequality,|x + y| ≤ |x| + |y|, is valid. In fact, if x and y have the same sign, then |x + y| = |x| + |y|. If x and y

have opposite signs, then we see that |x + y| < |x|+ |y|. (See Problem 12.4.8 for a generalization.) (c)is valid: |xy| = |x| · |y|. (Look at the 4 different sign combinations of x and y: If x is positive and y isnegative, then xy is negative, |xy| = −xy and |x|·|y| = (−x)·y = −xy, and so on.) (d) is easily seen tobe valid. (e) is not valid (put x = 1). (f) is easily seen to be valid. (Note that

√−4)2 = −(−2) = 2.)(g) is valid: | − 2x| = 2|x|. (h) is valid because

∣∣|x| − |y|∣∣ ≤ |x − y| ⇔ x2 − 2|x||y| + y2 ≤x2 − 2xy + y2 ⇔ −2|x||y| ≤ −2xy ⇔ |x||y| ≥ x · y, which is true because (c) is valid.


4 C H A P T E R 2 F U N C T I O N S O F O N E V A R I A B L E : I N T R O D U C T I O N

2.32. See Figs. 2.3.2(a)—2.3.2(f).

1

1

y

x

1

1

y

x

1

2

3

4

1 2 3

y

x

Figure 2.3.2(a) Figure 2.3.2(b) Figure 2.3.2(c)

1

�1

1�1�2 2

y

x

2

1

1

y

x

1

1

y

x

Figure 2.3.2(d) Figure 2.3.2(e) Figure 2.3.2(f)

4. See Figs. 2.3.4(a)–(c).y

−2

−1

1

2

x−3 −2 −1 1 2 3

y

−2

−1

1

2

x1 2 3 4 5

y

−2

−1

1

2

x−3 −2 −1 1 2 3


y

x5

(2, 4)

2

2

y

x

A = (3, 2)

B = (5, −4)

2

2

P

y

1

x1

Figure 2.3.6 Figure 2.3.10 Figure 2.3.12


C H A P T E R 2 F U N C T I O N S O F O N E V A R I A B L E : I N T R O D U C T I O N 5

6. (5 − 2)2 + (y − 4)2 = 13, or y2 − 8y + 12 = 0, with solutions y = 2 and y = 6. Geometric explanation:The circle with center at (2, 4) and radius

√13 intersects the line x = 5 at two points. If the radius were

2, the circle would not intersect the line x = 5. See Fig. 2.3.6.

8. (a) (x − 2)2 + (y − 3)2 = 16 (b) (x − 2)2 + (y − 5)2 = 13

10. (x − 3)2 + (y − 2)2 = (x − 5)2 + (y + 4)2, which reduces to x − 3y = 7. See Fig. 2.3.10

12. (x, y) = (1, 0), (0, 1), (1/4, 1/4) are all solutions. See Fig. 2.3.12.

14. The condition is: p + r√

x2 + y2 = p + s√

(x − a)2 + y2. Cancelling p, then squaring each side andrearranging yields (r2 − s2)(x2 + y2) = s2(a2 − 2ax). If r = s, then the markets are separated by thestraight line x = 1

2a. Otherwise, completing squares yields the equation [x + as2/(r2 − s2)]2 + y2 =[ars/(r2 − s2)]2. This is a circle with center at

(− as2/(r2 − s2), 0)

and radius |ars/(r2 − s2)|.

2.42. (a) f (−5) = 0, f (−3) = −3, f (−2) = 0, f (0) = 2, f (3) = 4, f (4) = 0

(b) Df = [−5, 4], Vf = [−3, 4]

4. See Figs. 2.4.4(a)–(d).

1�1 x

1

y

�3 �1 x

1

y

�1 x

1

y

�1 x

1

y

Figure 2.4.4(a) Figure 2.4.4(b) Figure 2.4.4(c) Figure 2.4.4.4(d)

2.52. 0.78 4. See Figs. 2.5.4(a)–(c).

y

1

2

3

4

x1 2 3 4

y

−5

−4

−3

−2

−1

1

x1 2 3 4 5 6 7 8 10

y

−1

1

2

3

4

x1 2 3 4 5


6. (a) Assume F = aC + b. Then 32 = a · 0 + b and 212 = a · 100 + b. Therefore a = 180/100 = 9/5and b = 32, so F = (9/5)C + 32. (b) If X = (9/5)X + 32, then X = −40.


6 C H A P T E R 3 P O L Y N O M I A L S , P O W E R S , A N D E X P O N E N T I A L S

8. L: y = 3x − 2; M: y = − 34x + 5

4 . P : (13/15, 3/5). N : y = − 34x − 7

4 . See Fig. 2.5.8.

10. (a) 75 − 3P e = 20 + 2P e, and so P e = 11. (b) P e = 90

12. C = 45y + 100. (The general equation is C = ay + b. Here 900 = a · 1000 + b, and a = 80/100 = 4/5,

so b = 100.)

14. The slope is

1

x0 + h− 1

x0

x0 + h − x0. Multiplying both numerator and denominator by the product (x0 +h)x0, then

simplifying, yields the given expression.

16. See Figs. 2.5.16(a)–(c).

x

L

MN

y

1

1

x

y

1

1

x

y

1

1

x

y

1

1

Figure 2.5.8 Figure 2.5.16(a) Figure 2.5.16(b) Figure 2.5.16(c)

Chapter 3 Polynomials, Powers, and Exponentials

3.1

2. (a)x −4 −3 −2 −1 0 1 2

f (x) −2.5 0 1.5 2 1.5 0 −2.5

(b) See Fig. 3.1.2(b). (c) f (x) = (−1/2)(x + 1)2 + 2. Maximum point (−1, 2).(d) x = −3 and x = 1 (e) f (x) > 0 in the interval (−3, 1), while f (x) < 0 for x < −3 and for x > 1.

y

−4−3

−2−1

1

2

3

x−4 −3 −2−1 1 2 3

f (x) = − 12 x2 − x + 3

2

Figure 3.1.2(b)


C H A P T E R 3 P O L Y N O M I A L S , P O W E R S , A N D E X P O N E N T I A L S 7

4. (a) x(x + 4). Zeros 0 and −4. (b) Factorization not possible. No zeros.(c) −3(x − x1)(x − x2), where the zeros are x1 = 5 + √

15 and x2 = 5 − √15.

(d) 9(x − x1)(x − x2), where the zeros are x1 = 1/3 + √5 and x2 = 1/3 − √

5.(e) −(x + 300)(x − 100), where the zeros are −300 and 100.(f) (x + 200)(x − 100), where the zeros are −200 and 100.

6. (a) If the other side is y, then 2x + 2y = L, and so y = L/2 − x. The area is then A(x) = x(L/2 − x) =Lx/2 − x2. The square with side L/4 gives the largest area.(b) Yes; the radius is L/2π so the area is L2/4π > L2/16.

8. (a) x = ± 1, x = ± 2 (b) (i) x2 = 9 (x2 < 0 is impossible), so x = ± 3(ii) x3 = 1 or x3 = 8, so x = 1 or x = 2

10. y = 2x2 + x − 6. ((1, −3) belongs to the graph if −3 = a + b + c, (0, −6) belongs to the graph if−6 = c, and (3, 15) belongs to the graph if 15 = 9a +3b+ c. It follows that a = 2, b = 1, and c = −6.)

12. (a) (i) 289 ≤ 290 (ii) 361 ≤ 377 (b) If B2 − 4AC were > 0, then according to formula [3.2] theequation f (x) = 0 would have two distinct solutions, which is impossible when f (x) ≥ 0 for all x. Wefind that A = a2

1 + a22 + · · · + a2

n, B = 2(a1b1 + a2b2 + · · · + anbn), and C = b21 + b2

2 + · · · + b2n, so

the conclusion follows.

3.3

2. (a) Integer roots must divide 6. Thus ±1, ±2, ±3, and ±6 are the only possible integer solutions. Wefind that −2, −1, 1, 3 all are roots, and since there can be no more than 4 roots in a polynomial equationof degree 4, we have found them all.(b) The same possible integer solutions. Only −6 and 1 are integer solutions. (The third root is −1/2.)(c) Neither 1 nor −1 satisfies the equation, so there are no integer roots.(d) First multiply the equation by 4 to have integer coefficients. Then ±1, ±2, and ±4 are seen to be theonly possible integer solutions. In fact, 1, 2, −2 are all solutions.

4. (a) The answer is 2x2 + 2x + 4 + 3/(x − 1), because

(2x3 + 2x − 1) ÷ (x − 1) = 2x2 + 2x + 42x3 − 2x2

2x2 + 2x − 12x2 − 2x

4x − 14x − 4

3 remainder

(b) The answer is x2 + 1, because

(x4 + x3 + x2 + x) ÷ (x2 + x) = x2 + 1x4 + x3

x2 + x

x2 + x

0 no remainder


8 C H A P T E R 3 P O L Y N O M I A L S , P O W E R S , A N D E X P O N E N T I A L S

(c) The answer is 3x5 + 6x3 − 3x2 + 12x − 12 + (28x2 − 36x + 13)/(x3 − 2x + 1):

(3x8 x2 + 1) ÷ (x3 − 2x + 1) = 3x5 + 6x3 − 3x2 + 12x − 123x8 − 6x6 + 3x5

6x6 − 3x5 + x2 + 16x6 − 12x4 + 6x3

− 3x5 + 12x4 − 6x3 + x2 + 1− 3x5 + 6x3 − 3x2

12x4 − 12x3 + 4x2 + 112x4 − 24x2 + 12x

− 12x3 + 28x2 − 12x + 1− 12x3 + 24x − 12

28x2 − 36x + 13 remainder

(d) The answer is x3 − 4x2 + 3x + 1 − 4x/(x2 + x + 1), because

(x5 − 3x4 + 1) ÷ (x2 + x + 1) = x3 − 4x2 + 3x + 1x5 + x4 + x3

− 4x4 − x3 + 1− 4x4 − 4x3 − 4x2

3x3 + 4x2 + 13x3 + 3x2 + 3x

x2 − 3x + 1x2 + x + 1

− 4x remainder

6. (a) p(x) = x(x2 + x − 12) = x(x − 3)(x + 4), because x2 + x − 12 = 0 for x = 3 and x = −4. (b)±1, ±2, ±4, ±8 are the only possible integer zeros. By trial and error we find that q(2) = q(−4) =0, so 2(x − 2)(x + 4) = 2x2 + 4x − 16 is a factor for q(x). By polynomial division we find thatq(x) ÷ (2x2 + 4x − 16) = x − 1/2, so q(x) = 2(x − 2)(x + 4)(x − 1/2).

3.4

2. (a) 2.511886 (b) 0.000098 (c) 0.530094

4. 500.16 =(

100√

50)16

(Find first the number that, when raised to the power of 100, yields 50. Then this

number is raised to the power of 16.)

6. (a) 23 = 8, so x = 3/2 (b) 181 = 3−4, so 3x + 1 = −4, and then x = −5/3.

(c) x2 − 2x + 2 = 2, so x2 − 2x = 0, implying that x = 0 or x = 2.

8. (a) Multiply by 5K1/2 to obtain K1/2 = 15L1/3. Squaring each side, K = 225L2/3. (b) abxb−10 = p,

so xb−10 = p/ab. Now raise each side to the power 1/(b − 1).

(c) x = − b2a

. (Multiply each side by (ax + b)2/3.) (d) b = λ1/ρ(c−ρ − (1 − λ)a−ρ

)−1/ρ

10. K = 57 315.86 for Y = 100, L = 6, t = 10.


C H A P T E R 3 P O L Y N O M I A L S , P O W E R S , A N D E X P O N E N T I A L S 9

3.5

2. (a) P = 1.22 · 1.034t million (b) The doubling time t∗ is given by the equation (1.034)t∗ = 2, and we

find t∗ ≈ 20.7 (years).

4. (a) 210 = 1024 (b) f (t) = 2t (c) f (20) = (210)2 = (1024)2 > 10002 = 1 million.

6. The graph is shown in Fig. 3.5.6.

x −2 −1 0 1 2

2x216 2 1 2 16

y

x1

1

y

−4

−3

−2

−1

1

2

3

4

5

6

x−2 −1 1 2 3

y

2

4

6

x−5 −4 −3 −2 −1 1 2

y = x22x

Figure 3.5.6 Figure 3.5.10 Figure 3.5.12

8. If the initial time is t , the doubling time t∗ is given by the equation Aat+t∗ = 2Aat , which impliesAatat∗ = 2Aat , so at∗ = 2, independent of t .

10. (a) The graph is shown in Fig. 3.5.10.

x −3 −2 −1 0 1 2 3 4

1 − 2−x −7 −3 −1 0 1/2 3/4 7/8 15/16

(b) 1 − 2−x gets close to 1 as x becomes very large, and becomes very large negative as x becomes largenegative. (Formally, limx→∞ f (x) = 1, limx→−∞ f (x) = −∞. See Section 6.1.)

12. The graph is drawn in Fig. 3.5.12.

x −10 −5 −4 −3 −2 −1 0 1 2

x22x 0.1 0.8 1.0 1.1 1.0 0.5 0 2.0 16

14. I (t) = I0(1/2)t/8 remains after t days.


10 C H A P T E R 4 S I N G L E V A R I A B L E D I F F E R E N T I A T I O N

3.6

2. The function in (b) is one-to-one and has an inverse: the rule mapping each youngest child alive today tohis/her mother. (Though the youngest child of a mother with several children will have been different atdifferent dates.) The function in (d) is one-to-one and has an inverse: the rule mapping the surface area tothe volume. The function in (e) is one-to-one and has an inverse: the rule that maps (u, v) to (u − 3, v).The other functions are many-to-one, in general, and so have no inverses.

Chapter 4 Single Variable Differentiation

4.2

2. f ′(x) = 6x + 2, f ′(0) = 2, f ′(−2) = −10, f ′(3) = 20. The tangent equation is y = 2x − 1.

4.f (x + h) − f (x)

h= 1/(x + h) − 1/x

h= x − (x + h)

hx(x + h)= −h

hx(x + h)= −1

x(x + h)→ − 1

x2as h → 0,

which proves the implication.

6. (a) f (x + h) − f (x) = a(x + h)2 + b(x + h) + c − (ax2 + bx + c) = 2ahx + bh + ah2, so[f (x + h) − f (x)]/h = 2ax + b + ah → 2ax + b as h → 0. Thus f ′(x) = 2ax + b. (b) f ′(x) = 0for x = −b/2a. The tangent is parallel to the x-axis at the minimum/maximum point.

8. (a) Use [A.12] in Section A.3. (b) and (c) [f (x +h)−f (x)]/h = (√x + h−√x)/h. Multiplying both

numerator and denominator by√

x + h + √x , then using the identity in (a), yields the result. Letting

h → 0, the formula follows. (d)√

x = x1/2 and 1/√

x = x−1/2.

10. (a) [f (x + h) − f (x)]/h = [(x + h)1/3 − x1/3]/h. Then use the hint.(b) Follows from (a) by letting h → 0.

4.3

2. I is the fixed cost, whereas k is the marginal cost, and also the incremental cost of producing one additionalunit.

4. T ′(y) = t , so the marginal tax rate is constant.

4.42. (a) The following table seems to indicate that the limit is 9.

x 0.9 0.99 0.999 1 1.001 1.01 1.1

x2+7x−8x−1 8.9 8.99 8.999 ∗ 9.001 9.01 9.1

*not defined

(b) x2 + 7x − 8 = (x − 1)(x + 8), so for x �= 1,x2 + 7x − 8

x − 1= x + 8 → 9 as x → 1.

4. (a) 22 + 3 · 2 − 5 = 5 (b) 1/5 (c) 1 (d) −2 (e) 3x2 (f) h2

6. (a) 4 (b) 5 (c) 6 (d) 2a + 2 (e) 2a + 2 (f) 4a + 4 8. (a) 1/6 (b) 1/27 (c) n


C H A P T E R 4 S I N G L E V A R I A B L E D I F F E R E N T I A T I O N 11

4.5

2. (a) 2g′(x) (b) (−1/6)g′(x) (c) (1/3)g′(x) 4. (a) 8πr (b) A(b + 1)yb (c) (−5/2)A−7/2

6. Add an arbitrary constant to these answers: (a) (1/3)x3 (b) x2 + 3x (c)xa+1

a + 1

4.6

2. (a) (6/5)x − 14x6 − (1/2)x−1/2 (b) 4x(3x4 − x2 − 1) (c) 10x9 + 5x4 + 4x3 − x−2

4. (a)3

2√

x(√

x + 1)2(b)

−x4 + 5x2 + 18x + 2

(x2 + 2)2(x + 3)2(c)−2(1+2x)x−3 (d)

4x

(x2 + 1)2(e)

−2x2 + 2

(x2 − x + 1)2

(f)2(x3 − x2 + x + 1)

3x3(x + 1)2

6. (a) x = 2 (b) x = −√3, x = 0, x = √

3 (c) x = −√2, x = √

2(d) x = 0, x = −1/2 − √

5/2, x = −1/2 + √5/2

8. (a)ad − bc

(ct + d)2(b) a(n + 1/2)tn−1/2 + nbtn−1 (c)

−(2at + b)

(at2 + bt + c)2

10. Differentiating f (x) · f (x) = x gives f ′(x) · f (x) + f (x) · f ′(x) = 1, so 2f ′(x) · f (x) = 1. Hence,

f ′(x) = 1

2f (x)= 1

2√

x.

12. The Newton quotient of F is

F(x + h) − F(x)

h= f (x + h)/g(x + h) − f (x)/g(x)

h

Multiplying the numerator and denominator by g(x)g(x+h) yields the expression for the Newton quotientgiven in the hint. Letting h → 0 yields the desired result.

4.7

2. dy/dx = axa−1 − ax−a−1 and d2y/dx2 = a(a − 1)xa−2 + a(a + 1)x−a−2

4. g′(t) = (t2 − 2t)/(t − 1)2, so g′′(t) = [(2t − 2)(t − 1)2 − (t2 − 2t)2(t − 1)]/(t − 1)4 = 2 for t = 2.

6. For n = 1, the formula is true. Suppose it is true for n = k, so that y = xk ⇒ y(k) = k! . Then(dk+1/dxk+1)xk+1 = (dk/dxk)(d/dx)(xk+1) = (dk/dxk)(k+1)xk = (k+1)(dk/dxk)xk = (k+1) ·k!,by the induction hypothesis. This is equal to (k + 1)!, so the given formula is also true for n = k + 1. Byinduction, the formula is true for all n.


12 C H A P T E R 5 M O R E O N D I F F E R E N T I A T I O N

Chapter 5 More on Differentiation

5.1

2. (a) 3(2x + 1)2 · 2 = 6(2x + 1)2 (b) −5(1 − x)4 (c) 2(x2 − 2x + 2)(2x − 2)

(d) (x + 1)4(4x − 1)/x2 (e) −21(3x − 4)−8 (f) −2(4x + 3)(2x2 + 3x − 4)−3

4. (a) −6at (at2 + 1)−4 (b) na(at + b)n−1 (c)−b(a + 1)

nt2

(at + b

nt

)a

6. dx/dt = Ar(Ap + B)r−1(2at + b)

8. Puttingy = 1/(g(x))n implies thaty ′ = [0·(g(x))n−1·n(g(x))n−1g′(x)]/(g(x))2n = −n(g(x))−n−1g′(x).

10. For a = 1, the formula is true. Suppose the formula is true for a = k. Consider y = [g(x)

]k+1 =[g(x)

]k · g(x). The product rule for differentiation and the induction hypothesis yield

y ′ = k[g(x)

]k−1 · g′(x) · g(x) + [g(x)]k · g′(x) = (k + 1)

[g(x)

]k · g′(x), which is the given formula fora = k + 1. By induction, the formula is true for all natural numbers a.

5.2

2. (a) dY/dt = (dY/dV )(dV/dt) = (−3)5(V + 1)4t2 = −15(t3/3 + 1)4t2

(b) dK/dt = (dK/dL)(dL/dt) = AaLa−1b = Aab(bt + c)a−1

4. dY/dt = (dY/dK) · (dK/dt) = Y ′(K(t0))K′(t0)

6. x = b − √ap − c = b − √

u, with u = ap − c. Thendx

dp= − 1

2√

uu′ = − a

2√

ap − c. (The restriction

should be p > c/a, for x to be differentiable.)

8. (a) b(t) is the total fuel consumption after t hours. (b) b′(t) = B ′(s(t))s ′(t). So the rate of fuelconsumption per hour is equal to the rate per kilometer multiplied by the speed in kph.

10. f (f (x)) = f (3x + 7) = 3(3x + 7) + 7 = 9x + 28. Then 9x + 28 = 100 for x = 8.

12. dC/dx = q(25 − 12x)−1/2 14. p′(x) = 2(x − a)q(x) + (x − a)2q ′(x), so p′(a) = 0.

16. F ′(x) = f ′(xng(x))(

nxn−1g(x) + xng′(x))

5.3

2. By implicit differentiation, 4x + 6y + 6xy ′ + 2yy ′ = 0. So y ′ = −2x + 3y

3x + y. In particular, y ′ = −8/5

at the point (1, 2).

4. (a) 2x + 2yy ′ = 0, and solve for y ′ to get y ′ = −x/y.(b) 1/2

√x + y ′/2

√y = 0, and solve for y ′ to get y ′ = −√

y/x.(c) 4x3 − 4y3y ′ = 2xy3 + x23y2y ′, and so y ′ = 2x(2x2 − y3)/y2(3x2 + 4y).

6. (a) No relationship. (b) f ′(x0) = g′(x0)

8. Differentiatingyn = xm w.r.t.x yieldsnyn−1y ′ = mxm−1, soy ′ = mxm−1/nyn−1 = mxm−1/n(xm/n)n−1 =(m/n)xm−1−(m/n)(n−1) = (m/n)x(m/n)−1.


C H A P T E R 5 M O R E O N D I F F E R E N T I A T I O N 13

5.42. f (x) ≈ f (0) + f ′(0)x = 1

9 − 1027x

4. F(1) = A, F ′(K) = αAKα−1, so F ′(1) = αA. Then F(K) ≈ F(1) + F ′(1)(K − 1)=A + αA(K − 1) = A(1 + αA(K − 1)).

6. (a) (i) �y = 0.61, dy = 0.6 (ii) �y = 0.0601, dy = 0.06(b) (i) �y = 0.011494, dy = 0.011111 (ii) �y = 0.001124, dy = 0.001111(c) (i) �y = 0.012461, dy = 0.0125 (ii) �y = 0.002498, dy = 0.0025

8. g(0) = A − 1 and g′(μ) = (Aa/(1 + b)

)(1 + μ)[a/(1+b)]−1, so g′(0) = Aa/(1 + b). Hence, g(μ) ≈

g(0) + g′(0)μ = A − 1 + aA

1 + bμ

5.52. Follows from formula [5.9] with f = U , a = y, x = y + M − s.

4. We find x(0) = 2[x(0)]2 = 2 ·1 = 2. Differentiating the expression for x(t) yields x(t) = x(t)+ t x(t)+4[x(t)]x(t), and so x(0) = x(0)+4[x(0)]x(0) = 1+4·1·2 = 9. Hence, x(t) ≈ x(0)+x(0)t+ 1

2 x(0)t2 =1 + 2t + 9

2 t2.

6. h′(x) = (pxp−1 − qxq−1)(xp + xq) − (xp − xq)(pxp−1 + qxq−1)

(xp + xq)2= 2(p − q)xp+q−1

(xp + xq)2, so h′(1) =

12 (p − q). Since h(1) = 0, h(x) ≈ h(1) + h′(1)(x − 1) = 1

2 (p − q)(x − 1).

5.62. ElKT = 1.06. A 1% increase in expenditure on road building leads to an increase in the traffic volume

of approx. 1.06 %.

4. (a) Elxf (x) = 0 (b) Elxf (x) = x/(x + 1) (c) Elxf (x) = −20x2/(1 − x2)

6. ElxAf (x) = Elxf (x), Elx(A + f (x)

) = f (x)Elxf (x)

A + f (x)

8. (a) −3 (b)1 + 2x

1 + x(c)

30x3

x3 + 1(d) Elx5x2 = 2, so ElxElx5x2 = 0 (e)

2x2

1 + x2

(f) Elx

(x − 1

x5 + 1

)= Elx(x − 1)− Elx(x

5 + 1) = x Elxx

x − 1− x5Elx x5

x5 + 1= x

x − 1− 5x5

x5 + 1(Some answers

are easier if the results of Problem 6 are used as well.)

10. (a) 6Elxy = 5, so Elxy = 5/6 (b) Using rules (c) and (b) in Problem 7, we obtain Elxy − Elxx =Elx(x + 1)a + Elx(y − 1)b. Here Elxx = 1, whereas

Elx(x + 1)a = [x/(x + 1)a]a(x + 1)a−1 = ax/(x + 1)

Elx(y − 1)b = [x/(y − 1)b]b(y − 1)b−1y ′ = [by/(y − 1)]Elxy

Thus, Elxy − 1 = ax/(x + 1) + [by/(y − 1)]Elxy. Solving for Elxy yields

Elxy = (ax + x + 1)(y − 1)

(y − by − 1)(x + 1)


14 C H A P T E R 6 L I M I T S , C O N T I N U I T Y , A N D S E R I E S

Chapter 6 Limits, Continuity, and Series

6.1

2. (a)x − 3

x2 + 1= (1/x) − (3/x2)

1 + (1/x2)−→x→∞ (b)

√2 + 3x

x − 1=√

3 + 2/x

1 − 1/x−→x→∞

√3 (c) a2

4. limx→∞ fi(x) = ∞ for i = 1, 2, 3; limx→∞ f4(x) = 0. Then:(a) ∞ (b) 0 (c) −∞ (d) 1 (e) 0 (f) ∞ (g) 1 (h) ∞

6. y = Ax + A(b − c) + d is an asymptote as x → ∞.

6.2

2. (a) None of the six functions is continuous at a. (b) Only in (i) does f have a limit: limx→a f (x) = A

(c) (i) limx→a− f (x) = limx→a+ f (x) = limx→a f (x) = A,(ii) limx→a− f (x) = f (a), limx→a+ f (x) = A; (iii) limx→a− f (x) = limx→a+ f (x) = ∞;(iv) limx→a− f (x) = −∞, limx→a+ f (x) = f (a); (v) limx→a− f (x) = ∞, limx→a+ f (x) = f (a);(vi) limx→a− f (x) = A, limx→a+ f (x) = B. (d) Only the function in (ii) is left-continuous at x = a.The functions in (iv), (v), and (vi) are right-continuous at x = a. (e) (v) limx→∞ f (x) = A;(vi) limx→∞ f (x) does not exist.

4. (a) Continuous for all x. (b) Continuous for all x �= 1. (c) Continuous for all x < 2. (d) Continuousfor all x. (e) Continuous for all x where x �= √

3 − 1 and x �= −√3 − 1. (f) Continuous for all x

where x ≤ −1 or x > 1. (g) Continuous for all x > 0. (h) Continuous for all x �= 0. (i) Continuousfor all x > 0.

6. See Fig. 6.2.6; y is discontinuous at x = a. In the case where y is the nearest point on the ground, itwould be a continuous function.

y

xa

Figure 6.2.6

8. Because limx→0− f (x) = −2 and limx→2+ f (x) = 3, we need f (0) = −2 and f (2) = 3. Because f islinear on [0, 2], put f (x) = (5/2)x − 2 for x ∈ [0, 2].

6.3

2. f ′(0+) = 1, f ′(0−) = −1. At x = 0, f is continuous, but not differentiable.


C H A P T E R 6 L I M I T S , C O N T I N U I T Y , A N D S E R I E S 15

4. The tax function is

t (x) =⎧⎨⎩

0.15x for x ∈ [0, 20 250]0.28x − 2 632.50 for x ∈ (20 250, 49 300]0.31x − 4 111.50 for x ∈ (49 300, ∞)

In particular, t (22 000) = 3527.50 and t (50 000) = 11388.50.

6.4

2. (a) Converges to 5. (b) Diverges (to ∞). (c)3n√

2n2 − 1= 3√

2 − 1/n2−→n→∞

3√2

= 3√

2

2

6.52. (a) Geometric series with quotient 1/8. Its sum is 8/(1 − 1/8) = 64/7. (b) Geometric with quotient

−3. It diverges. (c) Geometric, with sum 21/3/(1 − 2−1/3). (d) Not geometric. (One can show thatthe series is convergent with sum ln 2.)

4. Geometric series with quotient (1 + p/100)−1. Its sum is b/[1 − (1 + p/100)−1] = b(1 + 100/p).

6. The general term does not approach 0 as n → ∞ in any of these three cases, so each of the series isdivergent.

8. sn = n/(n + 1) = 1/(1 + 1/n) → 1 as n → ∞. The infinite series converges to 1.

6.6

2. Offer (a) is best. The second offer has present value 46001 − (1.06)−5

1 − (1.06)−1≈ 20 540.

4. Schedule (b) has present value12 000 · 1.115

0.115[1 − (1.115)−8] ≈ 67 644.42.

The present value of the contract in (c) is 22, 000 + 7000(1.115)12 − 1

0.115(1.115)12= 66, 384.08. Thus the contract

in (c) is the best in any case. When the interest rate becomes 12.5 %, contracts (b) and (c) have presentvalues equal to 65907.61 and 64374.33, respectively.

6. This is a geometric series with first term a = D/(1 + r) and quotient k = (1 + g)/(1 + r). It converges

iff k < 1, i.e. iff 1 + g < 1 + r , or g < r . The sum isa

1 − k= D/(1 + r)

1 − (1 + g)/(1 + r)= D

r − g.

6.72. |(x +1)3 −1| = |x3 +3x2 +3x +1−1| ≤ |x|3 +3|x|2 +3|x| ≤ |x|+3+3|x|+3|x| = 7|x|. Let ε > 0.

Choose δ = ε/7. Then according to [6.24], limx→0(x + 1)3 = 1. Because f (0) = 1, f is continuous atx = 0.

4. (a) 4x2 − 100 = 4(x2 − 25) = 4(x + 5)(x − 5), so for x �= 5,4x2 − 100

x − 5= 4(x + 5). Hence, for x �= 5,∣∣∣4x2 − 100

x − 5− 40

∣∣∣ = |4(x + 5) − 40| = 4|x − 5|. Let ε > 0. Choose δ = ε/4. According to [6.24], the

conclusion follows. (b) |x2 − π2| = |(x + π)(x − π)|, so for x �= −π , |(x2 − π2)/(x + π) + 2π | =|x − π + 2π | = |x + π |. Let ε > 0. Choose δ = ε. According to [6.24], the conclusion follows.


16 C H A P T E R 7 I M P L I C A T I O N S O F C O N T I N U I T Y A N D D I F F E R E N T I A B I L I T Y

Chapter 7 Implications of Continuity and Differentiability

7.12. For x �= 0, we have f (x) = x3 + ax2 + bx + c = x3(1 + a/x + b/x2 + c/x3), which shows that f (x)

is (large) positive if x is sufficiently large, and that f (x) is (large) negative if x is sufficiently large andnegative. Hence Theorem 7.2 implies that f (x) = 0 has a solution. A similar argument applies to thegeneral case, when the order n of the polynomial is odd. If n is even, there may be no real roots. Forinstance, x2 + 1 = 0 has no real roots.

4. Recall that, arbitrarily close to any given real number, there are rational as well as irrational numbers.The function f is continuous at a = 0, because |f (x) − f (0)| = |f (x) − 0| = |f (x)| ≤ |x| for anyx, so f (x) → f (0) as x → 0. If a �= 0 is rational, then |f (x) − f (a)| = |f (x)|, which is equal to |x|when x is irrational. But if a �= 0 is irrational, then |f (x) − f (a)| = |f (a)| whenever x is rational. Ineither case, f (x) does not approach 0 as x approaches a. It follows that f is discontinuous for all x �= 0.

We prove that g is discontinuous by using the εδ definition of continuity. Let a be an arbitrary numberand choose ε = 1

2 . For each positive number δ, there is a rational number x1 and an irrational number x2 inthe interval (a−δ, a+δ). Then g(x1) = 1 and g(x2) = 0. If a is rational, then |g(x2)−g(a)| = |0−1| = 1,and if a is irrational, then |g(x1) − g(a)| = |1 − 0| = 1. In both cases there is a number x arbitrarilyclose to a for which |g(x) − g(a)| = 1 > ε. Hence g is discontinuous at a.

7.22. (a) No, because f (x) can get arbitrarily close to 1, by letting x be sufficiently close to 1. But there is no

value of x for which f (x) = 1. Similarly, there is no minimum, because f (x) can get arbitrarily closeto −1 by letting x be sufficiently close to −1, yet there is no value of x for which f (x) = −1.(b) No, not at x = −1 and x = 1.

�1 x1

y

1y � x2�3

y � �x � 3�

x1 2 3 4 5

y

3

2

1

1x � 1y �

1 2

y

x

1

�1


7.32. In (a), f is not differentiable at x = 0; in (b), f is not differentiable at x = 3; in (c), f is not even defined

at x = 1. See Fig. 7.3.2(a)–(c).

4. There is at least one point where you must be heading in the direction of the straight line joining A to B

(even if that straight line hits the shore).

7.42. (a) 3

√25 = 3(1 − 2/27)1/3 ≈ 3(1 − 1

3227 − 1

94

272 ) ≈ 2.924

(b) 5√

33 = 2(1 + 1/32)1/5 ≈ 2(1 + 15·32 − 2

251

322 ) ≈ 2.0125


C H A P T E R 7 I M P L I C A T I O N S O F C O N T I N U I T Y A N D D I F F E R E N T I A B I L I T Y 17

4. g(0) = g(x) = 0 follows immediately from [1] and [2]. Because f is thrice continuously differentiable,S(x) and g(x) are differentiable in (0, x). The formula for g′(t) follows easily, and the condition g′(c) = 0yields S(x) = f ′′′(c). The conclusion follows.

7.5

2. (a) limx→2

x4 − 4x3 + 6x2 − 8x + 8

x3 − 3x2 + 4= “0

0

” = limx→2

4x3 − 12x2 + 12x − 8

3x2 − 6x= “0

0

” =

limx→2

12x2 − 24x + 12

6x − 6= 12

6= 2

(b) limx→0

2(1 + x)1/2 − 2 − x

2(1 + x + x2)1/2 − 2 − x= “0

0

” = limx→0

(1 + x)−1/2 − 1

(1 + 2x)(1 + x + x2)−1/2 − 1= “0

0

” =

limx→0

− 12 (1 + x)−3/2

2(1 + x + x2)−1/2 + (1 + 2x)2(− 12 )(1 + x + x2)−3/2

= −1

3

4. G = limv→0+

1 − (1 + vβ)−γ

v= “0

0

” = limv→0+

γ (1 + vβ)−γ−1βvβ−1

1. If β = 1, then

G = γ . If β > 1, then G = 0, and if β < 1, then G = ∞.

6. limx→∞

f (x)

g(x)= lim

t→0+f (1/t)

g(1/t)= “0

0

” = limt→0+

f ′(1/t)(−1/t2)

g′(1/t)(−1/t2)= lim

t→0+f ′(1/t)

g′(1/t)= lim

x→∞f ′(x)

g′(x)

8.L = lim

x→a

f (x)

g(x)= lim

x→a

1/g(x)

1/f (x)= “0

0

” = limx→a

−1/(g(x))2

−1/(f (x))2· g′(x)

f ′(x)

= limx→a

(f (x))2

(g(x))2· g′(x)

f ′(x)= L2 lim

x→a

g′(x)

f ′(x)= L2 lim

x→a

1

f ′(x)/g′(x)

The conclusion follows. (Here, we have ignored problems with “division by 0”, when either f ′(x) or g′(x)

tends to 0 as x tends to a.)

7.6

2. p = (157.8/D)10/3

4. f (x) = x2 = (−x)2 is not one-to-one on (−∞, ∞), and therefore has no inverse. On [0, ∞), f isstrictly increasing and has the inverse g(x) = √

x.

6. f ′(x) = 7x6 + 25x4 + 2 > 0 for all x, so f has an inverse g. g′(−2) = 1/f ′(0) = 1/2.

8. (a) See Fig. 7.6.8(a). (b) Triangles OBA and OBC in Fig. 7.6.8(b) are congruent.

10. See Fig. 7.6.10. (This example shows that the commonly seen statement: “if the inverse function exists, theoriginal and the inverse function must both be monotonic” is wrong. This claim is correct for continuousfunctions, however.)

12. Differentiating f (g(x)) = x yields (∗) f ′((g(x))g′(x) = 1, so g′(x) = −1/f ′((g(x)

). Differentiating

(∗) yields f ′′(g(x))g′(x)g′(x) + f ′(g(x)

)g′′(x) = 0. Solving for g′′(x) gives

g′′(x) = −f ′′(g(x))(

g′(x))2

/f ′((g(x)) = −f ′′(g(x)

)/(f ′(g(x)

))3. The conclusions follow.


18 C H A P T E R 8 E X P O N E N T I A L A N D L O G A R I T H M I C F U N C T I O N S

y

x

(3, 1)

(5, 3)(1, 3)

(3, 5)

y = x

y

x

C = (b, a)

A = (a, b)

B

y = x

D

EO

y

x

Figure 7.6.8(a) Figure 7.6.8(b) Figure 7.6.10

14. f ′(x) = 12 (x+1)−1/2+ 1

2 (x−1)−1/2 > 0 for all x > 1. Hence, f is strictly increasing in [1, ∞), and has aninverse g. Because f (1) = √

2 and f (x) → ∞ as x → ∞, the range of f is [√

2, ∞). To find a formulafor g, let (1) y = √

x + 1 + √x − 1. Interchanging x and y, we have (2) x = √

y + 1 + √y − 1. The

next step is to solve (2) for y, with x ∈ [√

2, ∞) and y ∈ [1, ∞). We obtain x = √y + 1+√

y − 1 ⇐⇒x − √

y + 1 = √y − 1 �⇒ (x − √

y + 1 )2 = y − 1 ⇐⇒ x2 − 2x√

y + 1 + y + 1 = y − 1 ⇐⇒2x

√y + 1 = x2 + 2 �⇒ 4x2(y + 1) = x4 + 4x2 + 4 ⇐⇒ y = 1

4x2 + 1/x2.Thus, x = √

y + 1 + √y − 1 with y ≥ 1 implies y = x2/4 + 1/x2, x ≥ √

2. We prove the converse bysubstitution:√

y + 1 + √y − 1 =

√x2

4 + 1x2 + 1 +

√x2

4 + 1x2 − 1 =

√x4+4x2+4

4x2 +√

x4−4x2+44x2 = x2+2

2x+ x2−2

2x= x.

We conclude that g(x) = 14x2 + 1/x2, x ∈ [

√2, ∞) is the required inverse function.

Chapter 8 Exponential and Logarithmic Functions

8.1

2. (a) eex

ex = eex+x (b) 12 (et/2 − e−t/2) (c) − et − e−t

(et + e−t )2(d) z2ez3

(ez3 − 1)−2/3

4. p′(x) = kce−cx , p′′(x) = −kc2e−cx . The graph is shown in Fig. 8.1.4.

p

x

a + k

p(x) = a + k(1 − e−cx)

a

y

-3

-2

-1

1

2

3

x-3 -2 -1 1 2 3

y = 12 (ex − e−x)

y = 12 (ex + e−x)

Figure 8.1.4 Figure 8.1.6


C H A P T E R 8 E X P O N E N T I A L A N D L O G A R I T H M I C F U N C T I O N S 19

6. The graphs are in Fig. 8.1.6. Verifying the identities is straightforward. We prove only equality (a):cosh x cosh y + sinh x sinh y = 1

2 (ex + e−x) 12 (ey + e−y) + 1

2 (ex − e−x) 12 (ey − e−y) =

14 (ex+y + ex−y + e−x+y + e−x−y + ex+y − ex−y − e−x+y + e−x−y) = 1

2 (ex+y + e−x−y) = cosh(x + y)

8. f (z + x) = az+x = azax = f (z)f (x). Differentiating w.r.t. z while x is fixed gives f ′(z + x) =f ′(z)f (x). Putting z = 0 yields f ′(x) = f ′(0)f (x), which is [8.2].

8.2

2. (a) ln 3x = x ln 3 = ln 8, so x = ln 8/ ln 3. (b) x = e3 (c) x2 − 4x + 5 = 1, or x2 − 4x + 4 = 0,so (x − 2)2 = 0. Thus x = 2. (d) x(x − 2) = 1, so x = 1 − √

2 or x = 1 + √2. (e) x = 0 or

ln(x + 3) = 0, so x = 0 or x = −2. (f)√

x − 5 = 1, so x = 36.

4. (a) t = (ln x − b)/a (b) t = (ln 2)/a (c) t = ±√

ln(8/√

2π ) = ±√

52 ln 2 − 1

2 ln π

6. (a) True (πe ≈ 22.5, eπ ≈ 23.1) (b) True because (a) gives (πe)1/eπ < (eπ)1/eπ .

8. (a) Wrong. (Put A = B = C = 1.) (b) Correct by rule [8.7] (b). (c) Correct. (Use [8.7] (b) twice.)(d) Wrong. (If A = e and p = 2, then the equality becomes 0 = ln 2.) (e) Correct by [8.7](c).(f) Wrong. (Put A = 2, B = C = 1.)

10. (a) 1/(x + 1) (b) 1/x (c) ln x + 1 (d)ln x − 1

(ln x)2

12. (a) (i) y = x − 1 (ii) y = 2x − 1 − ln 2 (iii) y = x/e

(b) (i) y = x (ii) y = 2ex − e (iii) y = −e−2x − 4e−2

14. (a) Let f (x) = ex − (1 + x + x2/2). Then f (0) = 0 and f ′(x) = ex − (1 + x), which is positive forall x > 0, as shown in the problem. Hence f (x) > 0 for all x > 0, and the inequality follows. (b)Let f (x) = ln(1 + x) − 1

2x. Then f (0) = 0 and f ′(x) = 1/(x + 1) − 12 = (1 − x)/2(x + 1) > 0

in (0, 1), so ln(1 + x) > 12x in (0, 1). To prove the other inequality, let g(x) = x − ln(1 + x). Then

g(0) = 0 and g′(x) = 1 − 1/(x + 1) = x/(1 + x) > 0 in (0, 1), so x > ln(1 + x) in (0, 1). (c) Letf (t) = ln[(1 + t)/(1 − t)] − 2t . Then f (0) = 0 and f ′(t) = 2/[(1 + t)(1 − t)] − 2 = 2t2/(1 − t2) > 0in (0, 1), so the required inequality follows.

16. (a) x − x = 0 (b) 4 ln x − x (c) x2/y2 18. (a) x (b) 1/ ln x (c) x ln a

20. (a) ln y = √x ln x implies y ′/y = 1

2x−1/2 ln x + x−1/2 so y ′ = x√

x

(ln x + 2

2√

x

)(b) 1

2 (√

x )x(ln x + 1) (c) Differentiate ln ln y = x ln x + ln ln x to get

y ′/(y ln y) = ln x + 1 + 1/(x ln x) and so y ′ = xxx+x[(ln x)2 + ln x + 1

x

]22. First take logarithms to obtain the expression:

ln a + (v/α)[α ln N + α ln K − ln(Nα + bKα) = B − (v/α) ln(Nα + bKα)

for ln F(α), where B does not depend on α. Differentiation yields:

F ′(α) = vα−2F(α)

(ln(Nα + bKα) − α

Nα ln N + bKα ln K

Nα + bKα

)8.3

2. (a) x = 4 (b) x = e (c) x = 1/27 (d) x = −1050 and x = 1050


20 C H A P T E R 8 E X P O N E N T I A L A N D L O G A R I T H M I C F U N C T I O N S

4. (a) ex+1−4/x = e1, so x +1−4/x = 1, hence x = ±2. (b) Putting u = ln(x +e), the equation becomesu3 − 4u2 − u + 4 = 0, so (u − 4)(u2 − 1) = 0, implying that u = ±1 or 4. Because x = eu − e, onehas x = 0, x = −e + 1/e, or x = e4 − e.

6. (a) 1 (b) Does not exist. (c) 4/5 (d) 1 (x1/x = (eln x)1/x = eln x/x , and [8.19].) (e) 0. (x ln x =ln x/(1/x), and then l’Hôpital’s rule.) (f) 1. (xx = (eln x)x = ex ln x , and then use (e).)

8. (a) xex ≈ x + x2 + 12x3 (b) e2x ≈ 1 + 2x + 2x2 + 4

3x3 (c) x2 + e12 x ≈

1 + 12x + 9

8x2 + 148x3 (d)

√ex + 1 ≈ √

2 + 14

√2x + 3

32

√2x2 + 1

384 (7√

2)x3

10. For x �= 0, one has f ′(x) = x−32e−1/x2, f ′′(x) = x−6(−6x2 + 4)e−1/x2

, and therefore f ′′′(x) =x−9(24x4 − 36x2 + 8)e−1/x2

. Hence, f (k) = x−3kpk(x)e−1/x2is correct for k = 1, 2, 3. The general

formula is proved by induction. (Actually, all we need in the following is the fact thatpk(x) is a polynomial.Its degree is of no importance.) Now, f ′(0) = limx→0[f (x) − f (0)]/x = limx→0 e−1/x2

/x = 0because, in general, (∗) e−1/x2

/xp → 0 as x → 0 for any natural number p. (Substitute x = 1/√

t

and use [8.18].) The general case is proved by induction: Suppose that f (k)(0) = 0 for an arbitrarynatural number k. Then, by the definition of the derivative, f (k+1)(0) = limx→0[f (k)(x) − f (k)(0)]/x =limx→0 x−3kpk(x)e−1/x2

/x = limx→0 x−3k−1pk(x)e−1/x2 = 0, using (∗) again. (Note that pk(x), as apolynomial, approaches a constant as x → 0.)

8.4

2. (a) S(t) = S0e−at (b) t = (ln 2)/a is the time it takes for sales to halve.

4. k = 0.1 ln(705/641) ≈ 0.0095. P(15) ≈ 739, P(40) ≈ 938.

6. (a) In 1950 there were about 276 thousand. In the next 10 years the number increased by 155 thousand.(b) y → 479.36 as t → ∞. The graph is sketched in Fig. 8.4.6.

20151050(1970)(1960)(1950)

t

300

400

200

100

0

500

Tractors (in 1000)

8642 lnx

8

4

2

6

lny


8. h′ = −f ′K/f 2 = −rf (1 − f/K)K/f 2 = −r(K/f − 1) = −rh. Hence h = Ae−rt for some constantA. But then −1 + K/f = Ae−rt , and solving for f yields [8.24].


C H A P T E R 9 S I N G L E - V A R I A B L E O P T I M I Z A T I O N 21

10. (a) See Fig. 8.4.10 and the following table:

ln x 3.00 3.69 4.09 4.38 4.61 4.79 4.94 5.08 5.19 5.30

ln y 7.44 6.31 5.44 4.79 4.32 4.09 3.81 3.56 3.22 3.00

(b) ln y = ln A+a ln x. We see from the graph in Fig. 8.4.10 that a ≈ −2. Assuming that the graph passesthrough (say) (ln x, ln y) = (5.19, 3.22), we obtain 3.22 = ln A + (−2)5.19, implying that ln A = 13.6,so A = e13.6 ≈ 806 129. The required formula is y = 806 129x−2.

12. (a) y = 2.81x0.83 (b) a = ln y2 − ln y1

ln x2 − ln x1, A = y1x

−a1 14. Take ln of each side and solve for t0.

8.5

2. (a) A = 4, k = 0.25. The price after 5 years is f (5) = 4e1.25 ≈ 14. (b) Price controls are needed after≈ 6 years. The doubling time before price controls is 2.8 years, after price controls it is 7.3 years.

Chapter 9 Single-Variable Optimization

9.2

2. h′(x) = 8(2 − √3x)(2 + √

3x)

(3x2 + 4)2. Note that h(x) → 0 as x ± ∞. The function has a maximum at

x = 2√

3/3 and a minimum at x = −2√

3/3, with h(2√

3/3) = 2√

3/3 and h(−2√

3/3) = −2√

3/3.

4. (a) f ′(x) = [4x(x4 + 1) − 2x24x3]/(x4 + 1)2, then simplify and factor.(b) f has maximum 1 at x = 1, because f (x) increases in [0, 1] and decreases in [1, ∞). We see thatf (−x) = f (x) for all x, so f (x) is symmetric about the y-axis. On (−∞, ∞), f has maximum 1 atx = −1 and at x = 1.

6. (a) f ′(x) = −48x(3x2 + 4)−2, so f ′(x) > 0 for x < 0 and f ′(x) < 0 for x > 0. Hence, f has amaximum at x = 0. (b) g′(x) = −2(x − 2), so g′(x) > 0 for x < 2 and g′(x) < 0 for x > 2. Hence,g has a maximum at x = 2. (c) h′(x) = 20(x + 2)3, so h′(x) < 0 for x < −2 and h′(x) > 0 forx > −2. Hence, h has a minimum at x = −2. (d) F ′(x) = 4x/(2 +x2)−2, so F ′(x) < 0 for x < 0 andF ′(x) > 0 for x > 0. Hence, F has a minimum at x = 0. (e) G′(x) = 1/2

√1 − x > 0 for x < 1, so G

has a maximum at x = 1. (f) H ′(x) = −4x3/(1 + x4)2, so H ′(x) > 0 for x ∈ [−1, 0), and H ′(x) < 0for x ∈ (0, 1]. Hence, H has a maximum at x = 0, and minima at x = ±1.

8. Here d ′(x) = 2(x − a1) + 2(x − a2) + · · · + 2(x − an) = 2[nx − (a1 + a2 + · · · + an)]. So d ′(x) = 0for x = x, where x = 1

n(a1 + a2 + · · · + an), the arithmetic mean of a1, a2, . . . , an. If x > x, then

d ′(x) > 0, and if x < x, then d ′(x) < 0. We conclude that x minimizes d(x).

9.3

2. In all cases the maximum and minimum exist by the extreme value theorem. Follow the recipe in [9.5].(a) f (x) is strictly decreasing. Maximum −1 at x = 0, minimum −7 at x = 3.


22 C H A P T E R 9 S I N G L E - V A R I A B L E O P T I M I Z A T I O N

(b) Maximum 10 at x = −1 and x = 2. Minimum 6 at x = 1. (f ′(x) = 3x2 − 3 = 0 at x = ±1.)(c) Maximum 2.5 atx = 1/2 andx = 2. Minimum 2 atx = 1. (f (x) = x+1/x andf ′(x) = 1−1/x2 = 0at x = ±1.) (d) Maximum 4 at x = −1. Minimum −6

√3 at

√3. (f ′(x) = 5x2(x2 − 3).)

(e) Maximum 4.5 · 109 at x = 3000. Minimum 0 at x = 0.(f ′(x) = 3(x2 − 3000x + 2 · 106 = 3(x − 500)(x − 2000).)

4. (a) Total costs when there are 61, 70, and 80 passengers are: $ 48,190; $ 49,000; $ 48,000.(b) C(x) = (60 + x)(800 − 10x) = 48, 000 + 200x − 10x2, x ∈ [0, 20] (c) Maximum cost is with 70travelers (x = 10).

6. h′(t) = 1/2√

t − 12 = (1 − √

t)/2√

t . We see that h′(t) ≥ 0 in [0, 1] and h′(t) ≤ 0 in [1, ∞), so t = 1maximizes h(t). The plant is highest after 1 month, when it is 1/2 meter tall.

8. Because A′(Q) = [C ′(Q)Q − C(Q)]/Q2, the result follows from Example 4.20 in Section 4.6.

10. A′(Q) = a(b − 1)Qb−2 − c/Q2, so there is a minimum at Q = (c/a(b − 1))1/b

.

9.42. (a) No local extreme points.

(b) Local maximum 10 at x = −1. Local minimum 6 at x = 1.(c) Local maximum −2 at x = −1. Local minimum 2 at x = 1.(d) Local maximum 6

√3 at x = −√

3. Local minimum −6√

3 at x = √3.

(e) No local maximum point. Local minimum 0.5 at x = 3.(f) Local maximum 2 at x = −2. Local minimum −2 at x = 0.

4. (a) (a) We find f ′(t) = 0.05(t + 5)(35 − t)e−t . Obviously, f ′(t) > 0 for t < 35 and f ′(t) < 0 fort > 35, so t = 35 maximizes f (with f (35) ≈ 278). (b) f (t) → 0 as t → ∞. The graph is shown inFig. 9.4.4.

y

t30

100

200

60 90

y

x1 2 3�3

�5

�2

�√ 2

�1

5

10

√ 2


6. (a) a > 0, b = 0 (f ′(0) = 0 ⇒ b = 0. f ′′(0) ≥ 0 ⇒ a ≥ 0. If a = b = 0, then f (x) = x3 + c

has no loc. min. at x = 0.) (b) a = −6, b = 9. (f ′(1) = 0 and f ′(3) = 0 give 3 + 2a + b = 0 and27 + 6a + b = 0.)

8. (a) f ′(x) = 2 − x2

(x2 + 3x + 2)2; f ′′(x) = 2x3 − 12x − 12

(x2 + 3x + 2)3; x = √

2 is a local maximum point with

f (√

2) ≈ 0.17; x = −√2 is a local minimum point with f (−√

2) ≈ 5.83. (b) There are no globalextreme points. The graph is shown in Fig. 9.4.8.(c) ex = √

2 and so x = 12 ln 2 gives a global maximum, but g has no global minimum.


C H A P T E R 1 0 I N T E G R A T I O N 23

10. f (x) = x3 + ax + b → ∞ as x → ∞, and f (x) → −∞ as x → −∞. Thus f (x) has at least one realroot. We have f ′(x) = 3x2 + a. Thus, for a ≥ 0, f ′(x) > 0 for all x �= 0, so f is strictly increasing, andthere is only one real root. Note that for a ≥ 0, 4a3 +27b2 ≥ 0. Assume next that a < 0. Then f ′(x) = 0for x = ±√−a/3 = ±√

p, where p = −a/3 > 0. Then f has a local maximum at (−√p, b + 2p

√p)

and a local minimum at (√

p, b − 2p√

p). If one of the local extreme values is 0, the equation has adouble root, and this is the case iff 4p3 = b2, that is, iff 4a3 + 27b2 = 0. The equation has three realroots iff the local maximum value is positive and the local minimum value is negative. This occurs iff|b| < 2p

√p or iff b2 < 4p3 or iff 4a3 + 27b2 < 0.

9.52. (a) f ′(x) = 3(x − 1)(x + 2), f ′′(x) = 6x + 3 (b) Stationary points: x = −2 and x = 1. f increases

in (−∞, −2) and in (1, ∞). (c) x = −1/2 is the only inflection point.

4. x = −2 and x = 4 are minimum points, whereas x = 2 is a maximum point. Moreover, x = 0, x = 1,x = 3, and x = 5 are inflection points.

6. a = −2/5, b = 3/5 (f (−1) = 1 gives −a + b = 1. Moreover, f ′(x) = 3ax2 + 2bx and f ′′(x) =6ax + 2b, so f ′′(1/2) = 0 yields 3a + 2b = 0.)

8. π(Q) is stationary at Q = (P/ab)1/(b−1). Moreover, π ′′(Q) = −ab(b − 1)Qb−2 < 0 for all Q > 0, sothis is a maximum point.

9.62. a ≥ 0, b arbitrary.

4. (a) Since u(c) is concave, 1T

∑Tt=1 u(ct ) ≤ u

(1T

∑Tt=1 ct

)≤ u

(1T

∑Tt=1 yt

)= u(y), where y =

1T

∑Tt=1 yt . Thus y is optimal. (For ct = y, t = 1, 2, . . . , T , this holds with equality.)

(b) Put λt = (1 + r)−t /∑T

t=1(1 + r)−t . Then∑T

t=1 λt = 1 and

max∑T

t=1(1 + r)−t u(ct ) =∑Tt=1(1 + r)−t max

∑Tt=1 λtu(ct ). Also,

∑Tt=1 λtu(ct ) ≤ u

(∑Tt=1 λtct

)≤

u(∑T

t=1 λtyt

). So ct = c =∑T

t=1 λtyt (t = 1, 2, . . . , T ) solves the problem.

Chapter 10 Integration

10.12. See Figs. 10.1.2(a) to 10.1.2(d). (a)

∫ 20 3x2 dx = 2

0 x3 = 8 (b) 1/7 (c) e − 1/e (d) 9/10

x21

y2

1

x1

y1

x1�1

y

2

101 x

y

1

Figure 10.1.2(a) Figure 10.1.2(b) Figure 10.1.2(c) Figure 10.1.2(d)


24 C H A P T E R 1 0 I N T E G R A T I O N

4. A = 12

∫ 1−1(e

x + e−x) dx = 12

1−1(e

x − e−x) = e − e−1

10.22. (a)

∫(t3 + 2t − 3) dt =

∫t3 dt +

∫2t dt −

∫3 dt = 1

4 t4 + t2 − 3t + C

(b)∫

(x−1)2 dx =∫

(x2 −2x+1) dx = 13x3 −x2 +x+C. Alternative: Since

d

dx(x−1)3 = 3(x−1)2,

we have∫

(x − 1)2 dx = 13 (x − 1)3 + C1. This agrees with the first answer, with C1 = C + 1/3.

(c)∫

(x − 1)(x + 2) dx =∫

(x2 + x − 2) dx = 13x3 + 1

2x2 − 2x + C

(d) Either first evaluate (x + 2)3 = x3 + 6x2 + 12x + 8, to get∫(x + 2)3 dx = 1

4x4 + 2x3 + 6x2 + 8x + C, or:∫

(x + 2)3 = 14 (x + 2)4 + C1.

(e)∫

(e3x − e2x + ex) dx = 13e3x − 1

2e2x + ex + C

(f)∫

x3 − 3x + 4

xdx =

∫ (x2 − 3 + 4

x

)dx = 1

3x3 − 3x + 4 ln |x| + C

4. (a) Differentiate the right-hand side.(b) (i) 1

10 (2x + 1)5 + C (ii) 23 (x + 2)3/2 + C (iii) −2

√4 − x + C

6. (a) F(x) = ∫ ( 12 − 2x) dx = 1

2x − x2 + C. F(0) = 12 implies C = 1

2 .(b) F(x) = ∫ (x − x3) dx = 1

2x2 − 14x4 + C. F(1) = 5

12 implies C = 16 .

8. The general form for f ′ is f ′(x) = 13x3 + A, so that for f is f (x) = 1

12x4 + Ax + B. If we require thatf (0) = 1 and f ′(0) = −1, then B = 1 and A = −1, so f (x) = 1

12x4 − x + 1.

10.3

2. (a) −12/5 (b) 41/2 (c)∫ 3

2

( 1

t − 1+ t)

dt =3

2

[ln(t − 1) + 1

2 t2] = ln 2 + 5

2

4. (a) 6/5 (b) 26/3 (c) α(eβ − 1)/β (d) − ln 2 6. (a) 32/15 ≈ 2.133. See Fig. 10.3.6.

y

x

1

2

21 2

y

x

14/3

2

1 3 4x∗

y = √x


8. (a) 59/30 (b) A

(b − 1 + (b − c) ln

∣∣∣∣b + c

1 + c

∣∣∣∣)

+ d ln |b| (c) 2√

2 − 3/2

10. (a) Given F(x) = ∫ x

af (t) dt , according to the mean-value theorem, there exists a number x∗ ∈ (a, b)

such that [F(b) − F(a)]/(b − a) = F ′(x∗). Here F(b) = ∫ b

af (t) dt , F(a) = 0, and F ′(x∗) = f (x∗),

so the conclusion follows. (b) f (x∗) = 4/3 at x∗ = 16/9. See Fig. 10.3.10, in which the two shadedareas are equal.


C H A P T E R 1 1 F U R T H E R T O P I C S I N I N T E G R A T I O N 25

10.4

2. (a) Let n be the total number of individuals. The number of individuals with income in the interval [b, 2b]

is then N = n

∫ 2b

b

Br−2 dr = n2b

b

−Br−1 = nB

2b. Their total income is M = n

∫ 2b

b

Br−2r dr =

n

∫ 2b

b

Br−1 dr = n2b

b

B ln r = nB ln 2. Hence the mean income is m = M/N = 2b ln 2.

(b) Total demand is x(p) =∫ 2b

b

nD(p, r)f (r) dr =∫ 2b

b

nApγ rδBr−2 dr = nABpγ

∫ 2b

b

rδ−2 dr =

nABpγ2b

b

rδ−1

δ − 1= nABpγ bδ−1 2δ−1 − 1

δ − 1.

4. PDV =∫ 15

0500e−0.06t dt = 500

∣∣150

−1

0.06e−0.06t = 500

0.06

[1 − e−0.9] ≈ 4945.25.

FDV = e0.06·15PDV = e0.9PDV ≈ 2.4596 · 4945.25 ≈ 12163.3.

Chapter 11 Further Topics in Integration

11.1

2. (a)∫ 1−1 x ln(x + 2) dx = 1

−112x2 ln(x + 2) − ∫ 1

−112x2 1

x+2 dx

= 12 ln 3 − 1

2

∫ 1−1

(x − 2 + 4

x+2

)dx = 2 − 3

2 ln 3

(b) Recall that ddx

2x = 2x ln 2, and therefore 2x/ ln 2 is the indefinite integral of 2x . If follows that∫ 2

0x2x dx =

2

0x

2x

ln 2−∫ 2

0

2x

ln 2dx = 8

ln 2−

2

0

2x

(ln 2)2= 8

ln 2− 3

(ln 2)2

(c) First use integration by parts on the indefinite integral: withf (x) = x2 andg(x) = ex , (∗)∫

x2ex dx =x2ex−∫ 2xex dx.To evaluate the last integral we must use integration by parts once more: withf (x) = 2x

andg(x) = ex ,∫

2xex dx = 2xex−∫ 2ex dx = 2xex−(2ex+C). Inserted into (∗) this gives∫

x2ex dx =x2ex−2xex+2ex+C, and hence,

∫ 10 x2ex dx = 1

0(x2ex−2xex+2ex) = (e−2e+2e)−(0−0+2) = e−2.

Alternatively, more compactly using formula [11.2]:∫ 1

0x2ex dx =

1

0x2ex − 2

∫ 1

0xex dx = e − 2

[ 1

0xex −

∫ 1

0ex dx

]= e − 2[e − ∣∣10ex] = e − 2

4. Using integration by parts, we have

∫ t1

t0

F(t)μ(t) dt =∣∣∣t1t0F(t)μ(t) −

∫ t1

t0

F (t)μ(t) dt

= F(t1)μ(t1) − F(t0)μ(t0) −∫ t1

t0

F (t)μ(t) dt = −∫ t1

t0

F (t)μ(t) dt

6. Using [11.2], we have

∫ T

0U(C(t))e−rt dt = −

∣∣∣T0U(C(t))(1/r)e−rt +

∫ T

0(d/dt)U(C(t))(1/r)e−rt dt

which becomes the required expression when U(C(0)) = 0.


26 C H A P T E R 1 1 F U R T H E R T O P I C S I N I N T E G R A T I O N

11.22. (a) 1

9 (x2 + 1)9 + C. (Substitute u = x2 + 1, du = 2x dx.) (b) With u = x3 + 2, du = 3x2 dx and∫x2ex3+2 dx = ∫ 1

3eu du = 13eu + C = 1

3ex3+2 + C. (c) First attempt: u = x + 2, which gives

du = dx and∫

ln(x + 2)

2x + 4dx =

∫ln u

2udu. Not promising. A better idea: Substitute u = ln(x + 2).

Then du = dx

x + 2and

∫ln(x + 2)

2x + 4dx =

∫12u du = 1

4 (u)2 + C = 14 (ln(x + 2))2 + C.

(d) Attempt: u = 1 + x. Then, du = dx, and∫

x√

1 + x dx = ∫ (u − 1)√

u du = ∫ (u3/2 − u1/2) du =25u5/2 − 2

3u3/2 + C = 25 (1 + x)5/2 − 2

3 (1 + x)3/2 + C. Second attempt: u = √1 + x. Then u2 = 1 + x

and 2udu = dx. Then the integral is∫

x√

1 + x dx =∫

(u2 − 1)u2u du =∫

(2u4 − 2u3) du e.t.c.

Check that you get the same answer. Actually, even integration by parts works in this case. Put f (x) = x

and g′(x) = √1 + x, and choose g(x) = 2

3 (1 + x)3/2. (The answer looks different, but is not.)

(e) With u = 1 + x2, x2 = u − 1, and du = 2xdx, so∫

x3

(1 + x2)3dx =

∫x2 · x

(1 + x2)3dx =

12

∫u − 1

u3du = 1

2

∫(u−2 − u−3) du = − 1

2u−1 + 14u−2 + C = −1

2(1 + x2)+ 1

4(1 + x2)2+ C.

(f) With u = √4 − x3, u2 = 4−x3, and 2udu = −3x2dx, so

∫x5√

4 − x3 dx =∫

x3√

4 − x3 x2 dx =∫(4−u2) u (− 2

3 )u du =∫

(− 83u2 + 2

3u4) du = − 89u3 + 2

15u5 +C = − 89 (4−x3)3/2 + 2

15 (4−x3)5/2 +C

4. Substituting u = t2 − 2t and assuming x > 2 gives∫ x

3

2t − 2

t2 − 2tdt =

x2−2x

3ln u = ln 1

3 (x2 − 2x). Thus,

the equation implies that 13 (x2 −2x) = 2

3x −1. Hence, x2 −4x +3 = 0, with solutions x = 1 and x = 3.Here x = 1 is impossible, because the integral is not defined when x = 1. So the solution is x = 3.

6. Substitute z = x(t). Then dz = x ′(t)dt , and the result follows.

8. (a) Introduce z = x −λ as a new variable in the right-hand side integral. Then dz = dx, and∫ b+λ

a+λf (x −

λ) dx = ∫ b

af (z) dz. Then replace the dummy variable z by x in the last integral. (b) Introduce z = x/λ

as a new variable in the right-hand side integral.

10. Substitute u = x1/6. Then I = 6∫

u8

1 − u2du. Here u8 : (−u2 +1) = −u6 −u4 −u2 −1+1/(−u2 +1).

It follows that

I = − 67x7/6 − 6

5x5/6 − 2x1/2 − 6x1/6 − 3 ln |1 − x1/6| + 3 ln |1 + x1/6| + C

11.3

2. (a)∫ +∞

−∞f (x) dx =

∫ b

a

1

b − adx = 1

b − a

b

a

x = 1

b − a(b − a) = 1

(b)∫ +∞

−∞xf (x) dx = 1

b − a

∫ b

a

x dx = 1

2(b − a)

b

a

x2 = 1

2(b − a)(b2 − a2) = 1

2(a + b)

(c)1

3(b − a)

b

a

x3 = 1

3

b3 − a3

b − a= 1

3(a2 + ab + b2)

4. Divergence because∫ b

0 x/(1 + x2) dx = b

012 ln(1 + x2) = 1

2 ln(1 + b2) → ∞ as b → ∞. But∫ b

−bx/(1 + x2) dx = b

−b12 ln(1 + x2) = 0 for all b, so the limit is 0.


C H A P T E R 1 2 L I N E A R A L G E B R A : V E C T O R S A N D M A T R I C E S 27

6. 11+x2 ≤ 1

x2 for x ≥ 1, and∫ b

11x2 dx = b

1(−1/x) = 1 − 1/b → 1 as b → ∞, so by

Theorem 11.1 the given integral converges.

8. (a) z = 1rτ

(1 − e−rτ ) (b) z = 2rτ

(1 − 1

rτ(1 − e−rτ )

)10. Integrating by parts,

∫ln x/

√x dx = 2

√x ln x − 4

√x + C. Hence,∫ 1

h

ln x/√

x dx =1

h

(2√

x ln x − 4√

x) = −4 − (2√

h ln h − 4√

h) → −4 as h → 0+, so the given

integral converges to −4. (√

h ln h → 0 by l’Hôpital’s rule.)

12. See Fig. 11.3.12. If p > 1, then∑∞

n=1(1/np) = 1 +∑∞n=2(1/np) is finite because

∑∞n=2(1/np) is the

sum of the shaded rectangles, and this sum is certainly less than the area under the curve y = 1/xp over[1, ∞), which is equal to 1/(p − 1). If p ≤ 1, the sum

∑∞n=1(1/np) is the sum of the larger rectangles in

the figure, and this sum is larger than the area under the curve y = 1/xp over [1, ∞), which is unboundedwhen p ≤ 1. Hence,

∑∞n=1(1/np) diverges in this case.

y

1

x1 2 3 4

y = 1/xp

Figure 11.3.12

11.4

2. The Gini coefficients are approximately: 0.34 (US 1980), 0.37 (US 1990), 0.35 (Netherlands 1959), 0.27(Netherlands 1985), 0.69 (World 1989). (These numbers are obtained by approximating the Lorenz curvewith a broken line through the points corresponding to the observations, then computing the area betweenthis broken line and the diagonal of the square. Note that this underestimates the Gini coefficients.)

Chapter 12 Linear Algebra: Vectors and Matrices

12.1

2. (a) No sector delivers to itself. (b) The total amount of good i needed to produce one unit of each good.(c) This collection gives the number of units of each good which are needed to produce one unit of good j .(d) No economic interpretation. (The goods are usually measured in different units, so it is meaninglessto add them together.)

4. The Leontief model for this three sector model is as follows:

0.9x1 − 0.2x2 − 0.1x3 = 85−0.3x1 + 0.8x2 − 0.2x3 = 95−0.2x1 − 0.2x2 + 0.9x3 = 20

which does have the claimed solution.


28 C H A P T E R 1 2 L I N E A R A L G E B R A : V E C T O R S A N D M A T R I C E S

12.22. a + b + c = (−1, 6, −4), a − 2b + 2c = (−3, 10, 2), 3a + 2b − 3c = (9, −6, 9), −a − b − c =

−(a + b + c) = (1, −6, 4)

4. 3(x, y, z)+5(−1, 2, 3) = (3x −5, 3y +10, 3z+15) = (4, 1, 3) provided that 3x −5 = 4, 3y +10 = 1,and 3z + 15 = 3. So x = 3, y = −3, z = −4.

6. (a)3x − 2y = −1

−4x + 3y = 2, x = 1, y = 2 (b)

2x + 4y = 1

−3x − 6y = 0, which have no solution.

8. We need to find numbers t and s such that t (2, −1)+s(1, 4) = (4, −11).This vector equation is equivalentto (2t + s, −t + 4s) = (4, −11), which in turn is equivalent to the equation system (i) 2t + s = 4(ii) −t + 4s = −11. This system has the solution t = 3, s = −2, so (4, −11) = 3(2, −1) − 2(1, 4).

12.32. (a) λ = 0 gives x = (3, 1), λ = 1/4 gives x = (2, 5/4), λ = 1/2 gives x = (1, 3/2),

λ = 3/4 gives x = (0, 7/4), λ = 1 gives x = (−1, 2). See Fig. 12.3.2. (b) When λ runs through[0, 1], then x will run through all points on the line segment between a and b.

4. See Fig. 12.3.4. (The point R should be one unit further down.)

ab

λ�1/2λ�1/4

λ�0

λ�3/4λ�1

x

y

�1 1

1

2 3

S�(3,�2,4)

x

yQ

R

P

z


12.42. (a) a · b = b · a = −6, (a + b) · c = a · c + b · c = 9, a · (3b) = 3a · b = −18.

(b) ‖a‖ = 3, ‖b‖ = 3, ‖c‖ = √29 , (c) |ab| = 6 ≤ ‖a‖‖b‖ = 9

4. The scalar product of the two vectors is x2+(x−1)x+3·3x = x2+x2−x+9x = 2x2+8x = 2x(x+4),which is 0 for x = 0 and x = −4.

6. x = (5, 7, 12), u = (20, 18, 25), x · u = 526

8. ||a + b||2 = (a + b) · (a + b) = a · a + 2a · b + b · b ≤ ||a||2 + 2||a||||b|| + ||b||2 = (||a|| + ||b||)2.The conclusion follows because ||a + b|| and ||a|| + ||b|| are both nonnegative.

10. (a) Total output is

⎛⎝ x1

x2

x3

⎞⎠ = λ

⎛⎝ 8

44

⎞⎠+ (1 − λ)

⎛⎝ 2

610

⎞⎠ =

⎛⎝ 6λ + 2

−2λ + 6−6λ + 10

⎞⎠.


C H A P T E R 1 2 L I N E A R A L G E B R A : V E C T O R S A N D M A T R I C E S 29

To produce the output in (i) we can put λ = 1/2. To produce the output in (ii) would require a value of λ

such that 6λ+2 = 7, −2λ+6 = 5, and −6λ+10 = 5, and the second of these equations gives a differentvalue for λ from the other two. (b) In (a) we saw that (i) can be produced even without throwing awayoutputs. For (ii) to be possible if we were allowed to throw away output, there must exist a λ ∈ [0, 1]such that 6λ + 2 ≥ 7, −2λ + 6 ≥ 5, and −6λ + 10 ≥ 5. These inequalities reduce to λ ≥ 5/6, λ ≤ 1/2,λ ≤ 5/6, which are incompatible.(c) Revenue = R(λ) = p1x1 + p2x2 + p3x3 = p1(6λ + 2) + p2(−2λ + 6) + p3(−6λ + 10) =(6p1 − 2p2 − 6p3)λ + 2p1 + 6p2 + 10p3. If the constant slope 6p1 − 2p2 − 6p3 is > 0, then R(λ) ismaximized at λ = 1; if 6p1 − 2p2 − 6p3 is < 0, then R(λ) is maximized at λ = 0. Only in the specialcase where 6p1 − 2p2 − 6p3 = 0 can the two plants both remain in use.

12.52. (a) Direct verification. (To show that a lies on L, put t = 0.) (b) The direction of L is given by (−1, 2, 1).

(c) The equation of plane is (−1)(x1 − 2) + 2(x2 − (−1)) + 1 · (x3 − 3) = 0, or −x1 + 2x2 + x3 = −1.(d) We must have 3(−t + 2) + 5(2t − 1) − (t + 3) = 6, and so t = 4/3. Thus P = (2/3, 5/3, 13/3).

4. (a) Direct verification. (b) x1 = −2 − t , x2 = 1 + 2t , x3 = −1 + 3t .

12.6

2. (a) A =(

2 3 43 4 5

)(b) A =

(1 −1 1

−1 1 −1

)4. A + B =

(1 07 5

), 3A =

(0 36 9

)

12.7

2. A + B =⎛⎝ 4 1 −1

9 2 73 −1 4

⎞⎠, A − B =

⎛⎝−2 3 −5

1 −2 −3−1 −1 −2

⎞⎠, AB =

⎛⎝ 5 3 3

19 −5 161 −3 0

⎞⎠,

BA =⎛⎝ 0 4 −9

19 3 −35 1 −3

⎞⎠, (AB)C = A(BC) =

⎛⎝ 23 8 25

92 −28 764 − 8 −4

⎞⎠

4. T(Ts) =⎛⎝ 0.2875

0.22500.4875

⎞⎠

12.8

2. We start by performing the multiplication

⎛⎝ a d e

d b f

e f c

⎞⎠⎛⎝ x

y

z

⎞⎠ =

⎛⎝ ax + dy + ez

dx + by + f z

ex + fy + cz

⎞⎠. Next,

(x, y, z)

⎛⎝ ax + dy + ez

dx + by + f z

ex + fy + cz

⎞⎠ = (ax2 + by2 + cz2 + 2dxy + 2exz + 2fyz)

which is a 1 × 1 matrix.

4. Equality in both [1] and [2] ⇔ AB = BA. 6. (a) x0 = ±(1/√

3)

⎛⎝ 1

−11

⎞⎠ (b) Anx0 = x0 for all n.


30 C H A P T E R 1 3 D E T E R M I N A N T S A N D M A T R I X I N V E R S I O N

8. (a) Direct verification yields (1) A2 = (a + d)A − (ad − bc)I2 =(

a2 + bc ab + bd

ac + cd bc + d2

)(b) Multiplying (1) by A and using A3 = 0 yields (2) (a + d)A2 = (ad − bc)A, and further,(3) 0 = (a + d)A3 = (ad − bc)A2. If ad − bc �= 0, (3) yields A2 = 0. If ad − bc = 0,(2) yields (a + d)A2 = 0, and if a + d �= 0, again A2 = 0. Finally, if ad − bc = a + d = 0,

then (1) implies A2 = 0. (c) Let A =(

1 1−1 −1

). Then A2 = 0, and also (of course) A3 = 0.

12.9

2. A′ =(

3 −12 5

), B′ =

(0 22 2

), (A + B)′ =

(3 14 7

), (αA)′ =

(−6 2−4 −10

),

AB =(

4 1010 8

), (AB)′ =

(4 10

10 8

)= B′A′, A′B′ =

(−2 410 14

)

4. Symmetry if a2 − 1 = a + 1 and a2 + 4 = 4a, which reduces to a = 2.

6. (a) (A1A2A3)′ = (A1(A2A3))

′ = (A2A3)′A′

1 = (A′3A′

2)A′1 = A′

3A′2A′

1. (b) Easy induction proof.

8. (a) tr(A + B) =∑ni=1(aii + bii) =∑n

i=1 aii +∑ni=1 bii = tr(A) + tr(B)

(b) tr(cA) = ∑ni=1 caii = c

∑ni=1 aii = c tr(A) (c) tr(AB) = ∑n

i=1

∑nj=1 aij bji , whereas tr(BA) =∑n

i=1

∑nj=1 bij aji = ∑n

j=1

∑ni=1 bjiaij , because the indices i and j can be interchanged. The two are

equal because the order of multiplication and summation in each double sum is irrelevant. (If this is hardto understand, write out the sums in full for the cases when n = 2 and/or n = 3.)(d) tr(A′) =∑n

i=1 aii = tr(A)

Chapter 13 Determinants and Matrix Inversion

13.1

2. See Fig. 13.1.2. The shaded area is 18 =∣∣∣∣ 3 0

2 6

∣∣∣∣.

(3, 0)

(2, 6)

Figure 13.1.2

4. The matrix product is AB =(

a11b11 + a12b21 a11b12 + a12b22

a21b11 + a22b21 a21b12 + a22b22

), implying that

|AB| = (a11b11 + a12b21)(a21b12 + a22b22) − (a11b12 + a12b22)(a21b11 + a22b21). On the other hand,|A||B| = (a11a22 − a12a21)(b11b22 − b12b21). A tedious computation shows that the two expressionsare equal.


C H A P T E R 1 3 D E T E R M I N A N T S A N D M A T R I X I N V E R S I O N 31

6. We write the system as

{Y − C = I0 + G0

−bY + C = a. Then Cramer’s rule yields

Y =

∣∣∣∣ I0 + G0 −1a 1

∣∣∣∣∣∣∣∣ 1 −1−b 1

∣∣∣∣= a + I0 + G0

1 − b, C =

∣∣∣∣ 1 I0 + G0

−b a

∣∣∣∣∣∣∣∣ 1 −1−b 1

∣∣∣∣= a + b(I0 + G0)

1 − b

The expression for Y is most easily found by substituting the second equation into the first, and thensolving for Y . Then use C = a + bY to find C.

8. |A(t)| = 2t (1 − t), which is 0 for t = 0 and t = 1.

13.2

2. AB =⎛⎝−1 −1 −1

7 13 135 9 10

⎞⎠, |A| = −2, |B| = 3, |AB| = |A| · |B| = −6

4. By Sarrus’ rule the determinant is (1 + a)(1 + b)(1 + c) + 1 + 1 − (1 + b) − (1 + a) − (1 + c), whichreduces to the given expression.

6. Direct verification using [13.7] or Sarrus’ rule.

13.3

2. +a12a23a35a41a54 (Four lines between pairs of elements rise as one goes to the right.)

13.4

2. A′ =⎛⎝ 2 1 1

1 0 23 1 5

⎞⎠, |A| = |A′| = −2

4. (a) The first and the second columns are proportional. (b) Add the second column to the third. Then thefirst and the third columns are proportional. (c) The term x − y is a common factor for each entry inthe first row. If x = y, all elements in the first row are 0. If x − y �= 0, we divide this row by x − y, andthe resulting determinant has rows 1 and 2 identical, so it is 0.

6. |AB| = |A||B| = −12, 3|A| = 9, | − 2B| = (−2)3(−4) = 32, |A| + |B| = −1, whereas |A + B| is notdetermined.

8. (a) Because A2 = In it follows from part 8 of Theorem 13.1 that |A|2 = 1, and so |A| = ±1.(b) Direct verification. (c) (In − A)(In + A) = I2

n + InA − AIn − A2 = In − A2 = 0 ⇐⇒ A2 = In


32 C H A P T E R 1 3 D E T E R M I N A N T S A N D M A T R I X I N V E R S I O N

10. Start by adding the sum of the last n − 1 rows to the first row. Then na + b is a common factor in thefirst row, which we take out. Next, add the first row multiplied by −a to all the other n − 1 rows. Theupper triangular matrix resulting from these operations obviously has the determinant bn−1:

Dn =

∣∣∣∣∣∣∣∣a + b a · · · a

a a + b · · · a...

.... . .

...

a a · · · a

∣∣∣∣∣∣∣∣← · · · ←

1 · · ·. . .

1

=

∣∣∣∣∣∣∣∣na + b na + b · · · na + b

a a + b · · · a...

.... . .

...

a a · · · a + b

∣∣∣∣∣∣∣∣

= (na + b)

∣∣∣∣∣∣∣∣1 1 · · · 1a a + b · · · a...

.... . .

...

a a · · · a + b

∣∣∣∣∣∣∣∣−a · · · −a

← · · ·. . .

←= (na + b)

∣∣∣∣∣∣∣∣1 1 · · · 10 b · · · 0...

.... . .

...

0 0 · · · b

∣∣∣∣∣∣∣∣According to [13.14], the last determinant is bn−1. Thus Dn = (na + b)bn−1.

13.5

2. In each of these cases we keep expanding by the last (remaining) column. The answers are: (a) −abc

(b) abcd (c) 6 · 4 · 3 · 5 · 1 = 360

13.6

2. Multiply the two matrices to get I3.

4. Using

(2 −33 −4

)−1

=(−4 3

−3 2

), we get: (a) x = 3, y = 1 (b) x = 1, y = −2 (c) x = y = 0

6. (a) |A| = 1, A2 =⎛⎝ 0 1 1

1 1 21 1 1

⎞⎠, A3 =

⎛⎝ 1 1 2

2 2 31 2 2

⎞⎠, then direct verification of the equality.

(b) By (a), (A − I)2A = I. (c) P = (A − I)−1 =⎛⎝ 0 0 1

1 0 10 1 0

⎞⎠ does the job. So does −P.

8. (a) A2 = (PDP−1)(PDP−1) = PD(P−1P)DP−1 = PDIDP−1 = P(D)2P−1.(b) Suppose the formula is valid for m = k. Then Ak+1 = AAk = PDP−1(PDkP−1) = PD(P−1P)DkP−1

= PDIDkP−1 = PDDkP−1 = PDk+1P−1.

10. (a) If C2 + C = I, then C(C + I) = I, and so C−1 = C + I = I + C. (b) Because C2 = I − C, soC3 = C2C = (I − C)C = C − C2 = C − (I − C) = −I + 2C. Moreover, C4 = C3C = (−I + 2C)C =−C + 2C2 = −C + 2(I − C) = 2I − 3C.

12. (a) AT = 112

⎛⎝ s + 17 4t − 16 0

2s + 10 5t − 8 03s + 15 4t − 16 12

⎞⎠. We see that AT = I3 and so T = A−1 for s = −5, t = 4.

(b) BX = 2X + C ⇐⇒ BX − 2X = C ⇐⇒ (B − 2I)X = C. But it is easy to see that B − 2I = A,

so BX = 2X + C ⇐⇒ AX = C ⇐⇒ X = A−1C = TC. Hence, X =⎛⎝−1/2 0 0 1/6

1/2 3 −3 1/61/2 −1 2 1/6

⎞⎠ .


C H A P T E R 1 4 F U R T H E R T O P I C S I N L I N E A R A L G E B R A 33

13.7

2.1

|A|

⎛⎝C11 C21 C31

C12 C22 C32

C13 C23 C33

⎞⎠ = 1

72

⎛⎝−3 5 9

18 −6 186 14 −18

⎞⎠. 3. (I − A)−1 = 5

62

⎛⎝ 18 16 10

2 19 84 7 16

⎞⎠

4. Let B denote the n × p matrix whose ith column has the elements bi1, bi2, . . . , bin. The p systems of n

equations in n unknowns can be expressed as AX = B, where A is n × n and X is n × p. Following themethod in Example 13.20, exactly the same row operations that transforms the n× 2n matrix (A : I) into(I : A−1) will also transform the n × (n + p) matrix (A : B) into (I : B∗), where B∗ is the matrix withelements b∗

ij . Because these row operations are together equivalent to premultiplcation by A−1, it mustbe true that B∗ = A−1B. When i = k, the solution to the system is x1 = b∗

k1, x2 = b∗k2, . . . , xn = b∗

kn.

13.8

2. The determinant of the system is equal to −10, so the solution is unique. The determinants in (2) are

D1 =∣∣∣∣∣∣b1 1 0b2 −1 2b3 3 −1

∣∣∣∣∣∣ , D2 =∣∣∣∣∣∣

3 b1 01 b2 22 b3 −1

∣∣∣∣∣∣ , D3 =∣∣∣∣∣∣

3 1 b1

1 −1 b2

2 3 b3

∣∣∣∣∣∣Expanding each of these determinants by the column (b1, b2, b3), we find that D1 = −5b1 + b2 + 2b3,D2 = 5b1 −3b2 −6b3, D3 = 5b1 −7b2 −4b3. Hence, x1 = 1

2b1 − 110b2 − 1

5b3, x2 = − 12b1 + 3

10b2 + 35b3,

x3 = − 12b1 + 7

10b2 + 25b3.

Chapter 14 Further Topics in Linear Algebra

14.1

2. Only the vectors in (b) are linearly independent.

4. If x(1, 1, 1) + y(2, 1, 0) + z(3, 1, 4) + w(1, 2, −2) = (0, 0, 0), then x, y, z, and w satisfy the equationsx + 2y + 3z + w = 0, x + y + z + 2w = 0, x + 4z − 2w = 0. One solution is x = −2, y = −1, z = 1,w = 1, so the vectors are linearly dependent.

6. (a) Suppose αa+βb+γ c = 0. Taking the scalar product of each side with a yields a · (αa+βb+γ c) =a · 0 = 0. But a · b = a · c = 0, so α(a · a) = 0. Because a �= 0, α = 0. In a similar way we can provethat β = γ = 0, so a, b, and c are linearly independent. (b) Generalize the proof in (a).

14.2

2. (a) By cofactor expansion along the first row, the determinant of the matrix A is |A| = x · (−1) − 0 · 1 +(x2 − 2) · 1 = x2 − x − 2 = (x + 1)(x − 2). If x �= −1 and x �= 2, then |A| �= 0, so the rank of Aequals 3. If x = −1 or x = 2, then |A| = 0 and r(A) ≤ 2. On the other hand, the minor we get if we

strike out the first row and the third column in A is

∣∣∣∣ 0 1−1 x

∣∣∣∣ = 1 �= 0 for all x, so r(A) can never be

less than 2. Thus, r(A) = 2 if x = −1 or x = 2, r(A) = 3 otherwise.(b) A little calculation shows that the determinant is t3 +4t2 −4t −16, and if we note that this expressionhas t + 4 as a factor, it follows that the determinant is t3 + 4t2 − 4t − 16 = t2(t + 4) − 4(t + 4) =


34 C H A P T E R 1 4 F U R T H E R T O P I C S I N L I N E A R A L G E B R A

(t2 − 4)(t + 4) = (t + 2)(t − 2)(t + 4). Thus, if t is not any of the numbers −2, 2, and −4, the rankof the matrix is 3. If we strike out the second row and the first column of the matrix, we get the minor∣∣∣∣ 5 61 t + 4

∣∣∣∣ = 5t + 14, which is different from 0 for all the three special values of t that we found above,

and thus the rank of the matrix is 2 if t = −4, −2, or 2, and 3 otherwise.(c) The first and third rows are identical, as are the second and fourth. But the first two rows are alwayslinearly independent. So the rank is 2 for all values of x, y, z, and w.

14.3

2. Unique solution for c �= 3/2: x = (3 − kc)/(3 − 2c), y = (k − 2)/(3 − 2c). For c = 3/2, k = 2 thesolutions are x = (−3/2)a + 1, y = a, a arbitrary. For c = 3/2, k �= 2 there are no solutions.

4. A((1 − λ)x1 + λx2) = (1 − λ)Ax1 + λAx2 = (1 − λ)b + λb = b. This shows that if x1 and x2 aredifferent solutions, then so are all points on the straight line through x1 and x2.

6. (a) |At | = (t − 2)(t + 3), so At has an inverse for all t except t = 2 and t = −3.

(b) r(At ) = 3 for t �= 2 and for t �= −3. Because

(1 32 5

)is always nonsingular, r(A2) = 2, r(A−3) = 2.

(c) x1 = −46 + 19t , x2 = 19 − 7t , x3 = t , with t arbitrary.(d) Any z �= 0 satisfying zA2 = 0 has this property. In particular, z = (10a, −7a, a) for a �= 0.

14.4

2. |A − λI| = 0 iff |(A − λI)′| = 0 iff |A′ − λI| = 0 using [12.44] in Section 12.9 and Thm. 13.1 [2] inSection 13.4. The conclusion follows.

4. (a) X′AX = (ax2 + ay2 + bz2 + 2axy), A2 =⎛⎝ 2a2 2a2 0

2a2 2a2 00 0 b2

⎞⎠, A3 =

⎛⎝ 4a3 4a3 0

4a3 4a3 00 0 b3

⎞⎠

(b) λ = 0, λ = 2a, λ = b (c) p(λ) = −λ(b − λ)(2a − λ) = −λ3 + (2a + b)λ2 − 2abλ. Using A2 andA3 from part (a), it is easy to show that p(A) = −A3 + (2a + b)A2 − 2abA = 0.

6. Since A−1 exists, |A| �= 0. So λ is an eigenvalue for AB ⇐⇒ |AB − λI| = 0 ⇐⇒ |A(B − λA−1)| =0 ⇐⇒ |B − λA−1| = 0 ⇐⇒ |(B − λA−1)A| = 0 ⇐⇒ |BA − λI| = 0 ⇐⇒ λ is an eigenvalue forBA. (Using [14.11] in the next section, here is a simpler argument: AB and A−1(AB)A = BA have thesame eigenvalues.)

14.5

2. (a) No, because the eigenvectors are not linearly independent. (b) Yes, because the eigenvectors arelinearly independent.

14.6

2. (a) |A−λI| =∣∣∣∣ 1 − λ 2

2 −2 − λ

∣∣∣∣ = λ2 +λ−6 = 0, with the real eigenvalues λ1 = −3 and λ2 = 2. One

set of linearly independent eigenvectors is

(1

−2

)and

(21

). These two eigenvectors are orthogonal.


C H A P T E R 1 5 F U N C T I O N S O F S E V E R A L V A R I A B L E S 35

Thus, Theorem 14.7 (not 14.6!) is confirmed. Further, let

P =(

1 2−2 1

)for which P−1 =

(1/5 −2/52/5 1/5

)

It is easy to verify that P−1AP is the diagonal matrix diag (−3, 2), which confirms Theorem 14.8 (not14.6!).

(b) Here |A −λI| =∣∣∣∣∣∣−λ 0 1

0 −λ 11 1 −1 − λ

∣∣∣∣∣∣ = −λ(λ− 1)(λ+ 2) = 0, with the real eigenvalues λ1 = −2,

λ2 = 0 and λ3 = 1. One set of linearly independent eigenvectors is

⎛⎝ 1

1−2

⎞⎠,

⎛⎝ 1

−10

⎞⎠, and

⎛⎝ 1

11

⎞⎠.

Obviously, these eigenvectors are mutually orthogonal. Thus Theorem 14.7 is confirmed. Further, let

P =⎛⎝ 1 1 1

1 −1 1−2 0 1

⎞⎠ for which P−1 =

⎛⎝ 1/6 1/6 −1/3

1/2 −1/2 01/3 1/3 1/3

⎞⎠

It is easy to verify that P−1AP is the diagonal matrix diag (−2, 0, 2), which confirms Theorem 14.8.

Chapter 15 Functions of Several Variables

15.12. f (1, 1) = 2, f (−2, 3) = 51, f (1/x, 1/y) = 3/x2 − 2/xy + 1/y3, [f (x + h, y) − f (x, y)]/h =

6x − 2y + 3h, [f (x, y + k) − f (x, y)]/k = −2x + 3y2 + 3yk + k2

4. (a) F(1, 1) = 10, F(4, 27) = 60, F(9, 1/27) = 10, F(3,√

2 ) = 10√

3 · 21/6,F(100, 1000) = 1000, F(2K, 2L) = 25/6F(K, L) (b) a = 5/6

6. (a) y gets 21.053 ≈ 2.07 times larger.(b) ln y = ln 2.9 + 0.015 ln x1 + 0.25 ln x2 + 0.35 ln x3 + 0.408 ln x4 + 0.03 ln x5

8. (a) x + y > 0 (b) Must require 2 − (x2 + y2) ≥ 0, i.e. x2 + y2 ≤ 2.(c) Put a = x2 + y2. We must require (4 − a)(a − 1) ≥ 0, i.e. 1 ≤ a ≤ 4. (Use a sign diagram.)

10. Total output is∑n

i=1(T /ti) = T∑n

i=1(1/ti). If all n of the machines were equally efficient, and theirproduction per unit were tH , then each machine would produce T/tH units. Total output is the same asbefore if T

∑ni=1(1/ti) = nT/tH . So tH = n/

∑ni=1(1/ti), the harmonic mean of t1, . . . , tn.

15.22. x2 +y2 = 6 is a level curve of f at height c = √

6−4, because when x2 +y2 = 6, then f (x, y) = √6−

6+2 = √6−4. In the general case, a level curve is defined by f (x, y) = √x2 + y2 − (x2 +y2)+2 = c

for some constant c. Let x2 + y2 = u2, where u > 0. Then u − u2 + 2 = c. Solving for u yieldsu = 1

2 ± 12

√9 − 4c (provided that c ≤ 9/4) and thus x2 + y2 is a constant, which is the equation for a

circle centered at (0, 0). If 9 − 4c > 0 and 1 − √9 − 4c > 0, i.e. 2 < c < 9/4, there are two circles. If

c ≤ 2 or c = 9/4, there is only one positive value of u, so the level curve is one circle.


36 C H A P T E R 1 5 F U N C T I O N S O F S E V E R A L V A R I A B L E S

4. Generally, the graph of g(x, y) = f (x) in 3-space consists of a surface traced out by moving the graphof z = f (x) parallel to the y-axis in both directions. The graph of g(x, y) = x is the plane through they-axis at a 45◦ angle with the xy-plane. The graph of g(x, y) = −x3 is shown in Fig. 15.2.4. (Only aportion of the unbounded graph is indicated, of course.)

z

y

x

Figure 15.2.4

15.3

2. (a) f ′1(x, y) = 7x6, f ′

2(x, y) = −7y6, and f ′12(x, y) = 0

(b) f ′1(x, y) = 5x4 ln y, f ′

2(x, y) = x5/y, and f ′′12(x, y) = 5x4/y

(c) f ′1(x, y) = 5(x2 − 2y2)42x = 10x(x2 − 2y2)4, f ′

2(x, y) = 5(x2 − 2y2)4(−4y) = −20y(x2 − 2y2)4,and f ′

12(x, y) = 40x(x2 − 2y2)3(−4y) = −160xy(x2 − 2y2)3

4. (a) F ′S = 2.26·0.44S−0.56E0.48 = 0.9944S−0.56E0.48, F ′

E = 2.26·0.48S0.44E−0.52 = 1.0848S0.44E−0.52

(b) SF ′S + EF ′

E = S · 2.26 · 0.44S−0.56E0.48 + E · 2.26 · 0.48S0.44E−0.52 = 0.44 F + 0.48 F = 0.92 F ,so k = 0.92.

6. (a) z′x = 2x, z′

y = 2e2y , z′′xx = 2, z′′

yy = 4e2y , z′′xy = z′′

yx = 0 (b) z′x = y/x, z′

y = ln x, z′′xx = −y/x2,

z′′yy = 0, z′′

xy = z′′yx = 1/x (c) z′

x = y2−yexy , z′y = 2xy−xexy , z′′

xx = −y2exy , z′′yy = 2x−x2exy ,z′′

xy =z′′yx = 2y − xyexy − exy

8. Here ∂z/∂x = x/(x2 + y2), ∂z/∂y = y/(x2 + y2), ∂2z/∂x2 = (y2 − x2)/(x2 + y2)2, and ∂2z/∂y2 =(x2 − y2)/(x2 + y2)2. Thus, ∂2z/∂x2 + ∂2z/∂y2 = 0.

10. u′x = au/x and u′

y = bu/y, so u′′xy = au′

y/x = abu/xy. Hence, u′′xy/u

′xu

′y = 1/u (u �= 0). Then,

1

u′x

∂

∂x

(u′′

xy

u′xu

′y

)= 1

u′x

· −u′x

u2= −1

u2= 1

u′y

∂

∂y

(u′′

xy

u′xu

′y

)

15.4

2. F(1, 0) = F(0, 0) + ∫ 10 F ′

1(x, 0) dx ≥ 2, F(2, 0) = F(1, 0) + ∫ 21 F ′

1(x, 0) dx ≥ F(1, 0) + 2

F(0, 1) = F(0, 0) + ∫ 10 F ′

2(0, y) dy ≤ 1, F(1, 1) = F(0, 1) + ∫ 10 F ′

1(x, 1) dx ≥ F(0, 1) + 2

F(1, 1) = F(1, 0) + ∫ 10 F ′

2(1, y) dy ≤ F(1, 0) + 1

4. With u = y/x, z′x = f (u) − uf ′(u) and z′

y = f ′(u). The equation for the tangent plane at (x, y, z)

is Z − xf (u) = [f (u) − uf ′(u)](X − x) + f ′(u)(Y − y). This goes through the origin (0, 0, 0) iff


C H A P T E R 1 5 F U N C T I O N S O F S E V E R A L V A R I A B L E S 37

−xf (u) = [f (u) − uf ′(u)](−x) + f ′(u)(−y) iff 0 = xuf ′(u) + f ′(u)(−y), which is satisfied becausexu = y.

15.5

2. F ′1(x, y, z) = 2xexz + x2zexz + y4exy , so F ′

1(1, 1, 1) = 4e; F ′2(x, y, z) = 3y2exy + xy3exy , so

F ′2(1, 1, 1) = 4e; F ′

3(x, y, z) = x3exz, so F ′3(1, 1, 1) = e.

4. First-order partials: w′1 = 3yz + 2xy − z3, w′

2 = 3xz + x2, w′3 = 3xy − 3xz2. Second-order partials:

w′′11 = 2y, w′′

12 = w′′21 = 3z + 2x, w′′

13 = w′′31 = 3y − 3z2, w′′

22 = 0, w′′23 = w′′

32 = 3x, w′′33 = −6xz.

6. For (x, y) �= (0, 0), f ′1 = y(x4 + 4x2y2 − y4)(x2 + y2)−2. Thus, for y �= 0, f ′

1(0, y) = −y. This is alsocorrect for y = 0, because f ′

1(0, 0) = limh→0[f (h, 0) − f (0, 0)]/h = 0. Similarly, for (x, y) �= (0, 0),f ′

2 = x(x4 − 4x2y2 − y4)(x2 + y2)−2, and f ′2(x, 0) = x for all x. It follows that f ′′

12(0, 0) = −1 andf ′′

21(0, 0) = 1. Also, for (x, y) �= (0, 0), f ′′12 = f ′′

21 = (x6 + 9x4y2 − 9x2y4 − y6)(x2 + y2)−3. Thus,f ′′

12(0, y) = −1, f ′′12(x, 0) = 1, and f ′′

12(x, y) has no limit as (x, y) → (0, 0). Similarly for f ′′21.

15.6

2. (a) Y ′K = aAKa−1 and Y ′

K = aBLa−1, so KY ′K + LY ′

L = aAKa + aBLa = a(AKa + BLa) = aY

(b) KY ′K +LY ′

L = KaAKa−1Lb +LAKabLb−1 = aAKaLb +bAKaLa = (a +b)AKaLb = (a +b)Y

(c) Y ′K = 2aKL5 − bK4L2

(aL3 + bK3)2and Y ′

L = 2bK5L − aK2L4

(aL3 + bK3)2, so KY ′

K + LY ′L =

2aK2L5 − bK5L2 + 2bK5L2 − aK2L5

(aL3 + bK3)2= K2L2(aL3 + bK3)

(aL3 + bK3)2= K2L2

aL3 + bK3= Y . (According to

Section 16.5 these functions are homogeneous of degrees a, a + b, and 1, respectively, so the results areimmediate consequences of Euler’s Theorem.)

4. ∂D/∂p and ∂E/∂q are normally negative, because the demand for a commodity goes down when theprice of that commodity increases. ∂D/∂q and ∂E/∂p are (usually) positive, because the demand for acommodity increases when the price of a substitute increases.

6. Y ′K = maK−ρ−1Aeλt

[aK−ρ + bL−ρ

]−(m/ρ)−1, Y ′

L = mbL−ρ−1Aeλt[aK−ρ + bL−ρ

]−(m/ρ)−1.

Thus, KY ′K + LY ′

L = m(aK−ρ + bL−ρ)Aeλt[aK−ρ + bL−ρ

]−(m/ρ)−1 = mY . (This function is homo-geneous of degree m, so the result is an immediate consequences of Euler’s Theorem.)

15.7

2. If the monopolist is not allowed to discriminate, he must charge the same price P in both markets. Thus,from [1], P = a1 − b1Q1 = a2 − b2Q2. Substituting Q1 = (a1 − P)/b1 and Q2 = (a2 − P)/b2 in theexpression for profit gives the quadratic function π = −{(a1b2 +a2b1)α−[a1b2 +a2b1 +(b1 +b2)α]P +(b1 + b2)P

2}/b1b2. This is maximized when P = 1

2α + [a1b2 + a2b1)/2(b1 + b2)], giving maximumprofit of [a1b2 + a2b1 − α(b1 + b2)]2/4b1b2(b1 + b2). If this expression is subtracted from the (higher)maximum profit in the example, we eventually find that the loss of profit is (a1 − a2)

2/4(b1 + b2). Notethat the loss is 0 if a1 = a2.

4. L = σyy − σ 2xy/σxx


38 C H A P T E R 1 6 T O O L S F O R C O M P A R A T I V E S T A T I C S

15.8

2. (a) ξ = −5/4, η = −1/2, d = −7/8 (b) ξ = 3/5, η = −1/5, d = 26/5

4.(

a b

b c

)(x

y

)=(

ax + by

bx + cy

), so (x, y)

(a b

b c

)(x

y

)= x(ax + by) + y(bx + cy), which reduces

to ax2 + 2bxy + cy2.

15.9

2. (a) (x, y)

(1 11 1

)(x

y

)(b) (x1, x2, x3)

⎛⎝ 3 −1 3/2

−1 1 −23/2 −2 3

⎞⎠⎛⎝ x1

x2

x3

⎞⎠

4. Let λ1, . . . , λn be the eigenvalues of A. If A is positive semidefinite and symmetric, then according toTheorem 15.2(b), all the eigenvalues are nonnegative. Because |A| is not 0, 0 is not an eigenvalue, so allthe eigenvalues must be positive. Then according to Theorem 15.2(a), A is positive definite.

Chapter 16 Tools for Comparative Statics

16.1

2.dz

dt= F ′

1(t, y)+F ′2(t, y)

dy

dt= ∂z

∂t+ ∂z

∂y

dy

dt. If z = t2 +yey and y = t2, then

∂z

∂t= 2t ,

∂z

∂y= ey +yey ,

anddy

dt= 2t . Hence we get

dz

dt= 2t (1 + et2 + t2et2

).

4. The usual rules for differentiating (a) a sum; (b) a difference; (c) a product; (d) a quotient; (e) a compositefunction of one variable.

6. (a) F is defined for x > 0, y > 0. F(x, y) = 0 ⇐⇒ y = 1/x2. See Fig. 16.1.6.

(b)dz

dt= 2 + e−3t

1 + e−3t→ 2 as t → ∞.

y

1

2

x1 2

F(x, y) > 0

y = 1/x2

Figure 16.1.6

8. From [16.1], dz/dt = F ′1(x, y)dx/dt + F ′

2(x, y)dy/dt . Differentiating w.r.t. t yields d2z/dt2 =(d/dt)[F ′

1(x, y)dx/dt] + (d/dt)[F ′2(x, y)dy/dt]. Here (d/dt)[F ′

1(x, y)dx/dt] = [F ′′11(x, y)dx/dt +

F ′′12(x, y)dy/dt]dx/dt+F ′

1(x, y)d2x/dt2. Also, (d/dt)[F ′2(x, y)dy/dt] = [F ′′

21(x, y)dx/dt+F ′′22(x, y)dy/dt]dy/dt+

F ′2(x, y)d2y/dt2. Assuming F ′′

12 = F ′′21, the conclusion follows.


C H A P T E R 1 6 T O O L S F O R C O M P A R A T I V E S T A T I C S 39

16.22. ∂z/∂t1 = F ′(x)∂x/∂t1, ∂z/∂t2 = F ′(x)∂x/∂t2

4. (a) Let u = ln v, where v = x3 + y3 + z3 − 3xyz. Then ∂u/∂x = (1/v)(∂v/∂x) = (3x2 − 3yz)/v.Similarly, ∂u/∂y = (3y2 − 3xz)/v, and ∂u/∂z = (3z2 − 3xy)/v. Hence,

x∂u

∂x+ y

∂u

∂y+ z

∂u

∂z= x

v(3x2 − 3yz) + y

v(3y2 − 3xz) + z

v(3z2 − 3xy) = 3v

v= 3

which proves (i). Equation (ii) is then proved by elementary algebra. (b) Here ∂z/∂x = 2xyf ′(x2y)

and ∂z/∂y = x2f ′(x2y), so x∂z/∂x = 2x2yf ′(x2y) = 2y∂z/∂y.

6. F ′(α) = ∫ 10 (∂/∂α)(xeαx2

) dx = ∫ 10 x3eαx2

dx. Introduce u = αx2 as a new variable. Then the integral

is (1/2α2)∫ α

0 ueu du = (1/2α2)∣∣α0 (ueu − eu) = (αeα − eα + 1)/2α2. Also, F(α) = (1/2α)

∣∣10e

αx2 =(eα − 1)/2α, so F ′(α) = (αeα − eα + 1)/2α2 is correct.

8. With T = T (t), N(t) = n(t)e−δ(t−T ) −n(t −T )e−δ(t−T )(1− T )+∫ t

t−T

n(τ)(−δ)(1− T )e−δ(t−T ) dτ =[n(t) − (1 − T )n(t − T )]e−δ(t−T ) − δ(1 − T )N(t).

10. We have z(t) = ∫ 2t

tF (τ, t) dτ , where F(τ, t) = x(τ)e− ∫ τ

tr(s) ds . Leibniz’s formula gives z(t) =

2F(2t, t) − F(t, t) + ∫ 2t

t(∂F (τ, t)/∂t) dτ = 2x(2t)e− ∫ 2t

tr(s) ds − x(t) + ∫ 2t

tx(τ )e− ∫ τ

tr(s) dsr(t) dτ =

2x(2t)p(t) − x(t) + ∫ 2t

tF (τ, t)r(t) dτ = 2p(t)x(2t) − x(t) + r(t)z(t), and therefore z(t) − r(t)z(t) =

2p(t)x(2t) − x(t).

12. (a) g′(Q) = c + h∫ Q

0 f (D) dD − p∫ a

Qf (D) dD and g′′(Q) = (h + p)f (Q) ≥ 0 for all Q, so g

is convex. (b) g′(Q∗) = 0 yields c + h∫ Q∗

0 f (D) dD − p∫ a

Q∗ f (D) dD = 0. Here∫ a

Q∗ f (D) dD =∫ a

0 f (D) dD − ∫ Q∗0 f (D) dD = 1 − F(Q∗). It follows that F(Q∗) = (p − c)/(h + p).

16.32. (a) With F(x, y) = xy, F ′

1 = y, F ′2 = x, F ′′

11 = F ′′22 = 0, and F ′′

12 = 1. Hence, [16.9] and [16.12] yield

y ′ = −y

x, y ′′ = − 1

x3(−2 · 1 · yx) = 2yx

x3= 2

x3

(b) y ′ = −(1 + 3y)/(3x − 1), y ′′ = 6(1 + 3y)/(3x − 1)2

(c) y ′ = 5x4/6y5, y ′′ = −(10/3)x3y−5 − (125/36)x8y−11

4. (a) With F(x, y) = 2x2 + xy + y2, y ′ = −F ′1/F

′2 = −(4x + y)/(x + 2y) = −4 at (2, 0). Moreover,

y ′′ = −(28x2 + 14y2 + 14xy)/(x + 2y)3 = −14 at (2, 0). The tangent has the equation y = −4x + 8.(b) y ′ = 0 requires y = −4x. Inserting this into the original equation gives the two points.

6. For x = −1 and y = 2 both LHS and RHS are −1, so the curve passes through (−1, 2). Implicitdifferentiation yields 2yy ′ + 5 = ex(y−2) + xex(y−2)(y − 2 + xy ′). At (−1, 2) we find y ′ = −4/3, so theequation for the tangent is y = 1

3 (−4x + 2).

16.42. Let z = ug with u = a1x

d1 +a2x

d2 +a3x

d3 . Then, for i = 1, 2, 3, Eliz = EluugEliu = g(xi/u)aidxd−1

i =dgaix

di /u, so El1z + El2z + El3z = dga1x

d1 /u + dga2x

d2 /u + dga3x

d3 /u = dg. (This result follows


40 C H A P T E R 1 6 T O O L S F O R C O M P A R A T I V E S T A T I C S

easily from the fact that the function is homogeneous of degree dg (see Problem 2(b) in Section 16.6)and from the elasticity form [16.28] of the Euler equation.)

4. (a) Elxz = 20, Elyz = 30, Elt x = t/(t + 1), Elt y = 2t/(t + 1), so, according to [16.15], Elt z =Elxz Elt x + Elyz Elt y = 20t/(t + 1) + 30 · 2t/(t + 1) = 80t/(t + 1).

(b) Elt z = 2x2

x2 + y2· 1

ln t+ 2y2

x2 + y2· (−t + 2) = 2[ln t + t4e−2t (2 − t)]

(ln t)2 + t4e−2t

6. Taking logarithms yields ln z = ax + by and a ln x + b ln y + c ln z = 0. Differentiating w.r.t. x yieldsz′x/z = a+by ′

x and a/x+by ′x/y+cz′

x/z = 0. Solving for y ′x and z′

x , and then multiplying the expressions

obtained by x/y and x/z respectively, yields finally Elxy = −a(1 + cx)

b(1 + cy), Elxz = a(x − y)

1 + cy.

8. The easiest way to solve this problem is to note thatKL

(aL2/3 + bK2/3)3/2= KL(aL2/3 + bK2/3)−3/2 = (aK−2/3 + bL−2/3)−3/2

Comparing this with Example 16.20, we see that the production function is actually a CES-function withA = A0e

0.021t and ρ = 2/3. According to [2] in that example, σKL = 1/(2/3 + 1) = 3/5.

10. With F(K, L) = AKaLb, F ′K = aF/K , F ′

L = bF/L, F ′′KK = a(a − 1)F/K2, F ′′

KL = abF/KL, andF ′′

LL = b(b−1)F/L2. Also, −F ′KF ′

L(KF ′K+LF ′

L) = −(aF/K)(bF/L)(a+b)F = −ab(a+b)F 3/KL.Moreover, KL

[(F ′

L)2F ′′KK − 2F ′

KF ′2F

′′KL + (F ′

K)2F ′′LL

] = −ab(a + b)F 3/KL, so σKL = 1.

16.5

2. x(tp, tr) = A(tp)−1.5(tr)2.08 = At−1.5p−1.5 t2.08 r2.08 = t−1.5 t2.08Ap−1.5r2.08 = t0.58x(p, r), so thefunction is homogeneous of degree 0.58 .

4. By using [16.18] we find that f is homogeneous of degree 0, and using the results in Example 15.13(b)in Section 15.3 we confirm that xf ′

1(x, y) + yf ′2(x, y) = 0.

6. f (tx, ty) = a ln[g(tx, ty)/tx] = a ln[tg(x, y)/tx] = a ln[g(x, y)/x] = f (x, y), so f is homogeneousof degree 0.

8. Suppose F is homogeneous of degree k �= 0. Then F−1 is homogeneous of degree 1/k iff F−1(tx) =t1/kF−1(x) for all t > 0. But this equation is equivalent to F(F−1(tx)) = F(t1/kF−1(x)). Here theLHS is tx, and because F is homogeneous of degree 1/k, the RHS is (t1/k)kF (F−1(x)) = tx. Thefunction F(x) = √

x defined for x ≥ 0 with range [0, ∞) has an inverse F−1 = x2, defined for x ≥ 0.F is homogeneous of degree 1/2, while F−1 is homogeneous of degree 2.

16.6

2. (a) F(tx1, tx2, tx3) = (tx1tx2tx3)2

(tx1)4 + (tx2)4 + (tx3)4

(1

tx1+ 1

tx2+ 1

tx3

)

= t6(x1x2x3)2

t4(x41 + x4

2 + x43)

1

t

(1

x1+ 1

x2+ 1

x3

)= tF (x1, x2, x3)

The function is homogeneous of degree 1.

(b) G(tx1, tx2, tx3) = [a(tx1)d + b(tx2)

d + c(tx3)d]g = [td (axd

1 + bxd2 + cxd

3 )]g =

(td)gG(x1, x2, x3) = tdgG(x1, x2, x3), so G is homogeneous of degree dg.


C H A P T E R 1 6 T O O L S F O R C O M P A R A T I V E S T A T I C S 41

4. All are homogeneous of degree 1, as is easily checked by using [16.23].

6. When F(x, y) = a ln x +b ln y, xF ′1 +yF ′

2 = x(a/x)+y(b/y) = a +b, which is not equal to kF (x, y)

for any choice of k, so F is not homogeneous of any degree.

8. (a) f ((tx1)m, . . . , (txn)

m) = f (tmxm1 , . . . , tmxm

n ) = (tm)rf (xm1 , . . . , xm

n ) = tmrh(x1, . . . , xn), so h ishomo. of degree mr . (b) Homo. of degree sp. (c) Homo. of degree r for r = s, not homo. for r �= s.(d) Homo of degree r + s. (e) Homo. of degree r − s.

16.72. Differentiating partially w.r.t. x yields (∗) 3x2 + 3z2z′

x − 3z′x = 0, so z′

x = x2/(1 − z2). Similarly, or bysymmetry, z′

y = y2/(1 − z2). To find z′′xy , differentiate (∗) partially w.r.t. y to obtain 6zz′

yz′x + 3z2z′′

xy −3z′′

xy = 0, so z′′xy = 2zx2y2/(1 − z2)3.

4. Implicit differentiation gives f ′P (r, P )P ′

w = g′w(w, P ) + g′

P (w, P )P ′w. Hence

P ′w = −g′

w(w, P )

g′P (w, P ) − f ′

P (r, P )< 0, because g′

w > 0, g′P > 0, and f ′

P < 0.

6. Differentiating the equation w.r.t. x gives 1 − az′x = f ′(y − bz)(−bz′

x). Differentiating w.r.t. y gives−az′

y = f ′(y − bz)(1 − bz′y). Simple algebra yields az′

x + bz′y = 1.

16.82. Write the function on the form g∗(μ, ε) = (1 + μ)a(1 + ε)αa − 1, where a = 1/(1 + β). Then

∂g∗(μ, ε)/∂μ = a(1 + μ)a−1(1 + ε)αa, ∂g∗(μ, ε)/∂ε = (1 + μ)aαa(1 + ε)αa−1

Hence, g∗(0, 0) = 0, ∂g∗(0, 0)/∂μ = a, ∂g∗(0, 0)/∂ε = αa, and g∗(μ, ε)≈aμ + αaε.

4. f (0.98, −1.01) ≈ −4.97. The exact value is −4.970614, so the error is 0.000614.

6. (a) and (b): dz = (y2 + 3x2)dx + 2xy dy

8. (a) dU = 2a1u1du1 + · · · + 2anundun

(b) dU = A(δ1u−ρ1 + · · · + δnu

−ρn )−1−1/ρ(δ1u

−ρ−11 du1 + · · · + δnu

−ρ−1n dun)

10. Let T (x, y, z) = [x2 + y2 + z2]1/2 = u1/2, where u = x2 + y2 + z2. Then dT = 12u−1/2du =

12u−1/2(2xdx + 2ydy + 2zdz). For x = 2, y = 3, and z = 6, we have u = 49, T = 7 and dT =17 (xdx+ydy+zdz) = 1

7 (2dx+3dy+6dz). Accordingly, T (2+0.01, 3−0.01, 6+0.02) ≈ T (2, 3, 6)+17 [2 · 0.01 + 3(−0.01) + 6 · 0.02] = 7 + 1

7 · 0.11 ≈ 7.016. (The exact value is√

49.2206 ≈ 7.015739.)

12. (a) dX = AβNβ−1eρtdN + ANβρeρtdt

(b) dX1 = BEXE−1N1−EdX + B(1 − E)XEN−EdN

14. g(0) = f (x01 , . . . , x0

n). Using the result in part (a) of Problem 11 in Section 16.2 gives g′(0) =f ′

1(x01 , . . . , x0

n) dx1 + · · · + f ′n(x

01 , . . . , x0

n) dxn. The approximation g(1) ≈ g(0) + g′(0) now yields,using vector notation, f (x0 + dx) ≈ f (x0) + f ′

1(x0) dx1 + · · · + f ′

n(x0) dxn.

16.92. By differentiation, 3 dx1 + 2x2 dx2 − dy1 − 9y2

2 dy2 = 0, 3x21 dx1 − 2dx2 + 6y2

1 dy1 − dy2 = 0. Lettingdx2 = 0 and solving for dy1 and dy2 leads to dy1 = Adx1 and dy2 = Bdx1, where A = ∂y1/∂x1 =J−1(3 − 27x2

1y22 ) and B = ∂y2/∂x1 = J−1(3x2

1 + 18y21 ) with J = 1 + 54y2

1y22 .


42 C H A P T E R 1 7 M U L T I V A R I A B L E O P T I M I Z A T I O N

4. Differentiation w.r.t. x yields y + u′xv + uv′

x = 0, u + xu′x + yv′

x = 0. Solving this system for u′x and

v′x yields

u′x = u2 − y2

yv − xu= u2 − y2

2yv, v′

x = xy − uv

yv − xu= 2xy − 1

2yv

where we substituted xu = −yv and uv = 1 − xy. Now,

∂2u

∂x2= ∂

∂xu′

x = 2uu′x2yv − (u2 − y2)2yv′

x

4y2v2

= 2u(u2 − y2) − (u2 − y2)(2xy − 1)/v

4y2v2= (u2 − y2)(4uv − 1)

4y2v3

(The answer to this problem can be expressed in many different ways.)

6. From w = G(x, y, z) we have (∗) (∂w

∂x)y = G′

x + G′z(

∂z

∂x)y . Here, (

∂w

∂x)y,z = G′

x and (∂w

∂z)x,y = G′

z.

Substituting these expressions into (∗) yields the result.

16.102. Denote the left hand sides of the three equations by f1, f2, and f3, respectively. Then the Jacobian

determinant of these functions w.r.t. u, v, and w is∣∣∣∣∣∣∂f1/∂u ∂f1/∂v ∂f1/∂w

∂f2/∂u ∂f2/∂v ∂f2/∂w

∂f3/∂u ∂f3/∂v ∂f3/∂w

∣∣∣∣∣∣ =∣∣∣∣∣∣

1 −1 −3w2

1 3v2 −1−1 −1 3w2

∣∣∣∣∣∣ = 2(3w2 − 1)

At P the value of this determinant is 4 �= 0, and the functions f1, f2, and f3 are C1 functions, so accordingto the implicit function theorem the system defines u, v, and w as differentiable functions of x, y, and z

in a neighborhood of P . Differentiating the system w.r.t. x (holding y and z constant) yields

u′x − v′

x − 3w2w′x = 0

u′x + 3v2v′

x − w′x = 2

−u′x − v′

x + 3w2w′x = −2x

Substituting (x, y, z, u, v, w) = (1, 1, 0, −1, 0, 1) and solving the system yields u′x = 5/2, v′

x = 1, andw′

x = 1/2.

Chapter 17 Multivariable Optimization17.1

2. First-order conditions: f ′1(x, y) = −4x−2y+36 = 0, f ′

2(x, y) = −2x−4y+42 = 0, with the solutionx = 5, y = 8.

4. (a) P(10, 8) = P(12, 10) = 98 (b) First-order conditions: P ′x = −2x + 22 = 0, P ′

y = −2y + 18 = 0.It follows that x = 11 and y = 9, with P(11, 9) = 100.

6. Solving the constraint for z yields z = 4x + 2y − 5, so we have to minimize P(x, y) = x2 + y2 +(4x + 2y − 5)2 w.r.t. x and y. We find P(x, y) = 17x2 + 5y2 + 16xy − 40x − 20y + 25. The first-orderconditions are P ′

1 = 34x + 16y − 40 = 0, P ′2 = 16x + 10y − 20 = 0, with the solution x = 20/21,

y = 10/21. The minimum value is 525/441.


C H A P T E R 1 7 M U L T I V A R I A B L E O P T I M I Z A T I O N 43

8. F ′K = −2(K −3)−(L−6) and F ′

L = −4(L−6)−(K −3), so here [17.3] yields −2(K −3)−(L−6) =0.65, −4(L − 6) − (K − 3) = 1.2. The solution is K = 2.8, L = 5.75.

10. x = am/(a + b + c)p, y = bm/(a + b + c)q, z = cm/(a + b + c)r . (The constraint yieldsz = (m − px − qy)/r , so we maximize P = Axayb [(m − px − qy)/r]c w.r.t. x and y. Keeping inmind that z depends on x and y, the first-order conditions are P ′

x = Axa−1ybzc−1 [az − c(px/r)] = 0and P ′

y = Axayb−1zc−1 [bz − c(qy/r)] = 0. These give px = (a/c)rz and qy = (b/c)rz. Inserted intothe constraint and solved for z, these yield the given value for z, and the other two follow easily.)

17.2

2. Use Theorem 9.2 with F(u) = ln u.

4. By the chain rule, g′1(x, y) = F ′(f (x, y))f ′

1(x, y) and g′2(x, y) = F ′(f (x, y))f ′

2(x, y). Because F ′ > 0,the conclusion follows.

6. (a) S = {(x, y) : g(x, y) = c}, where g(x, y) = x2 + xy + y2, c = 3, with g continuous.(b) The given equation is symmetric in x and y, so −2 ≤ x ≤ 2. (c) The set is closed becauseg(x, y) = x2 + xy − y2 is continuous. The pair given in the hint does satisfy the equation, so theconclusion follows.

17.3

2. (a) f ′1(x, y) = 3x2 − 9y and f ′

2(x, y) = 3y2 − 9x. The only stationary point in the interior of S is (3, 3).((0, 0) is a stationary point on the boundary of S.) Along the four parts of the boundary we find thefollowing candidates for extreme points: (0, 0), (4, 0), (4,

√12), (4, 4), (

√12, 4), and (0, 4). Comparing

the values of f at all these points, we find that f has mimimum 91 at (0, 4) and at (4, 0), minimum 0 at(3, 3). (A maximum and a minimum exist by the extreme-value theorem.)(b) Maximum 9/4 at (−1/2,

√3/2) and at (−1/2, −√

3/2). Minimum −1/4 at (1/2, 0).(c) Maximum 3 at (0, 0) and at (1, 0). Minimum 2 at (0, −1) and at (0, 1).(d) Maximum 2e4 at (0, 0). Minimum 0 for all (x, 1/2) where x ∈ [0, 2], and for all (2, y) wherey ∈ [0, 1/2].

4. (a)The domainD is shown in Fig. 17.3.4. The first order partials aref ′1(x, y) = e− 1

4 (x+y2)(1 − 1

4 (x + y)),

f ′2(x, y) = e− 1

4 (x+y2)(1 − 1

2y(x + y)). (b) The point (7/2, 1/2) gives the maximum value 4e−15/16.

6. (a) f ′1 = 9x(x2 +y2)1/2 − 4x(x2 +y2)−1/2, f ′

2 = 9y(x2 +y2)1/2 − 4y(x2 +y2)−1/2 + 1. At a stationarypoint, f ′

1 = 0 implies that x = 0 or x2 + y2 = 4/9. But x2 + y2 = 4/9 implies f ′2 = 1, so x = 0 at the

stationary point. Then f ′2 = 9y2 − 3 if y > 0, whereas f ′

2 = −9y2 + 5 if y < 0. The stationary pointsare (0, 1

3

√3) and (0, − 1

3

√5).

(b) S is closed and bounded and f is continuous, so extreme points exist according to the extreme-valuetheorem. There are no stationary points in the interior of S, so the extreme points must be on the boundary.Along x = 0, y ∈ [−1, 1], f (0, y) = 3y3 − 3y if y ≥ 0, and f (0, y) = −3y3 + 5y if y ≤ 0. Along thispart of the boundary f has its largest value 0 at (0, 1) and at (0, 0), and its smallest value −10

√5/9 at

(0, − 13

√5). Along x2 + y2 = 1, y ∈ [−1, 1], f (x, y) = y − 1, which has its largest value 0 at (0, 1) and

its smallest value −2 at (0, −1). We conclude that the maximum value is 0 at (0, 1) and at (0, 0), andthat its smallest value is −10

√5/9 at (0, − 1

3

√5).



y

x

y = 1 − x

D

x

y

z

x

y

z

z = ln(1 + x2y)

y

1

2

x−1 1

y = 2x2

y = x2

y = kx

Figure 17.3.4 Figure 17.4.2(f) Figure 17.4.8

17.4

2. (a) (0, 0) saddle point, (1, 1) local minimum. (b) (−4/3, 1/3) local minimum.(c) (2, 3) local maximum, (a, 0) and (0, b) no decision (a and b arbitrary numbers).(d) (0, 0) saddle point, (a, 0) and (−a, 0) local maxima where a = √

3/2.(e) (0, 0) saddle point, (1/2, 1/4) and (−1/2, −1/4) local mininma.(f) (0, b) no decision (b an arbitrary number). (The graph of this function is shown in Fig. 17.4.2(f). f

is defined for x = 0 and for y > −1/x2. f ′1(x, y) = 2xy/(1 + x2y) and f ′

2(x, y) = x2/(1 + x2y).Here f ′

1 = f ′2 = 0 at all points (0, b) with b ∈ �. Because AC − B2 = 0 when (x, y) = (0, b), the

second-derivative test fails. But by studying the function directly, we see that (0, b) is a local maximumpoint if b < 0; a saddle point if b = 0; and a local minimum point if b > 0. See Fig. 17.4.2(f).)

4. (a) The level curves are x = 1 and y = 0. g(x, y) is positive if y > 0 and x �= 1.(b) g′

1 = y(x2 −1)ex+3y , g′2 = (x −1)2(1+3y)ex+3y . (−1, −1/3) is a local minimum. (1, y0) is a local

minimum for y0 > 0, a local maximum for y0 < 0, and a saddle point for y0 = 0. (Use the results in (a).)(c) g(x, 1) = (x−1)2ex+3 → ∞ as x → ∞, so g has no maximum; g(x, −1) = −(x−1)2ex−3 → −∞as x → ∞, so g has no minimum either.

6. We find easily in all three cases that (0, 0) is a stationary point where z = 0 and AC − B2 = 0. In case(a), z ≤ 0 for all (x, y), so the origin is a maximum point. In case (b), z ≥ 0 for all (x, y), so the originis a minimum point. In (c), z takes positive and negative values at points arbitrarily close to the origin,so it is a saddle point.

8. (a) Fig. 17.4.8. The domain where f (x, y) is negative is shaded. (b) (0, 0) is seen to be a stationarypoint at which z = 0. As Fig. 17.4.8 shows, this function takes positive and negative values for pointsarbitrary close to (0, 0), so the origin is a saddle point. (c) g(t) = f (th, tk) = (tk−t2h2)(tk−2t2h2) =2h4t4 − 3h2kt3 + k2t2, so g′(t) = 8h4t3 − 9h2kt2 + 2k2t and g′′(t) = 24h4t2 − 18h2kt + 2k2. Theng′(0) = 0 and g′′(0) = 2k2, so t = 0 is a minimum point for k �= 0. For k = 0, g(t) = 2t4h4, which hasa minimum at t = 0.

17.5

2. (a) The set is convex; see Fig. 17.5.2(a). (b) The set is convex; see Fig. 17.5.2(b).(c) The set is not convex; see Fig. 17.5.2(c). (d) The set is convex; see Fig. 17.5.2(d).(e) The set is not convex; see Fig. 17.5.2(e). (f) The set is not convex; see Fig. 17.5.2(f). ((4, 0) and(0, 4) belong to the set, but not 1

2 (4, 0) + 12 (0, 4) = (2, 2).)



y

-3

-2

-1

1

2

3

x-3 -2 -1 1 2 3

y

x

y

-4-3-2-1

1234

x-4 -3 -2 -1 1 2 3 4

Figure 17.5.2(a) Convex. Figure 17.5.2(b) Convex. Figure 17.5.2(c) Not convex.

y

-1

1

2

3

x-1 1 2 3

y

-3

-2

-1

1

2

3

x-3 -2 -1 1 2 3

y

1

2

3

4

x1 2 3 4

Figure 17.5.2(d) Convex. Figure 17.5.2(e) Not convex. Figure 17.5.2(f) Not convex.

4. To prove that aS + bT is convex, take x ∈ aS + bT , y ∈ aS + bT , and a number λ ∈ [0, 1]. Becausex ∈ aS + bT , there are points x1 ∈ S and y1 ∈ T such that x = ax1 + by1. Similarly, there are pointsx2 ∈ S and y2 ∈ T such that y = ax2 +by2. Then (1−λ)x+λy = (1−λ)(ax1 +by1)+λ(ax2 +by2) =a[(1 − λ)x1 + λx2] + b[(1 − λ)y1 + λy2]. This belongs to aS + bT because S and T are convex, so(1 − λ)x1 + λx2 ∈ S and (1 − λ)y1 + λy2 ∈ T . Therefore aS + bT is convex.

17.6

2. If f is linear, then f is both concave and convex according to Example 17.12. On the other hand, if f isboth concave and convex on the convex set S, then [17.15] is satisfied with equality for any λ ∈ (0, 1)

and for all x, x0 in S. Thus the graph of f contains any line segment between two points on its graph. Sothe graph of f is a hyperplane, and f is linear.

4. We prove the first statement in [17.16]. The second statement is proved in the same way. Suppose firstthat f is concave. Take any two points (x1, y1) and (x2, y2) in Mf . Then y1 ≤ f (x1) and y2 ≤ f (x2). Letλ ∈ [0, 1]. Then (1−λ)(x1, y1)+λ(x2, y2) = ((1−λ)x1+λx2, (1−λ)y1+λy2). Here (1−λ)x1+λx2 ∈ S,because S is convex. Moreover, by the concavity of f , f ((1 − λ)x1 + λx2) ≥ (1 − λ)f (x1) + λf (x2) ≥(1−λ)y1 +λy2. This inequality implies that (1−λ)(x1, y1)+λ(x2, y2) belongs to Mf , so Mf is convex.

Suppose on the other hand that Mf is convex. Take any x1, x2 in S and let λ ∈ [0, 1]. Of course,f (x1) ≤ f (x1) and f (x2) ≤ f (x2), so (x1, f (x1)) and (x2, f (x2)) both belong to Mf . Because Mf isconvex, (1 − λ)(x1, f (x1)) + λ(x2, f (x2)) = ((1 − λ)x1 + λx2, (1 − λ)f (x1) + λf (x2)) ∈ Mf . Thisimplies that (1 − λ)f (x1) + λf (x2) ≤ f ((1 − λ)x1 + λx2), and so f is concave.



17.7

2. Part (a) remains valid iff either a or b is strictly positive. This follows immediately from the argumentsin the proof of part (a) of Theorem 17.6. (Of course, if a = b = 0, af + bg is certainly not strictlyconcave.) It follows from the proof of part (c) that if F is strictly concave, then the inequalityF(f ((1 − λ)x0 + λx)

) ≥ F((1 − λ)f (x0) + λf (x)

)becomes strict for x0 �= x, so U is strictly concave.

There is no need to assume that f is strictly concave. For part (e), suppose f and g are strictly concaveover the convex set S. Let λ ∈ (0, 1) and let x �= y. Then, because f and g are strictly concave,f ((1 − λ)x + λy) > (1 − λ)f (x) + λf (y), and g((1 − λ)x + λy) > (1 − λ)g(x) + λg(y). It follows that

h((1 − λ)x + λy) = min{f ((1 − λ)x + λ(y)), g((1 − λ)x + λy

)}> min{(1 − λ)f (x) + λf (y), (1 − λ)g(x) + λg(y)}

But(1 − λ)f (x) + λf (y) ≥ (1 − λ) min{f (x), g(x)} + λ min{f (y), g(y)}

= (1 − λ)h(x) + λh(y)

Similarly, (1−λ)g(x)+λg(y) ≥ (1−λ)h(x)+λh(y). Therefore, h((1−λ)x+λy) > (1−λ)h(x)+λh(y).

17.8

2. (a) f ′′11 = 2a, f ′′

12 = 2b, f ′′22 = 2c, and f ′′

11f′′22 − (f ′′

12)2 = 2a2c− (2b)2 = 4(ac−b2). The result follows

from Theorem 17.10. (b) According to Theorem 17.9, f is concave iff a ≤ 0, b ≤ 0, and ac − b2 ≥ 0;f is convex iff a ≥ 0, b ≥ 0, and ac − b2 ≥ 0.

4. (a) f ′′11(x, y) = −ex − ex+y , f ′′

12(x, y) = −ex+y , and f ′′22(x, y) = −ex+y , so f ′′

11(x, y) < 0 andf ′′

11f′′22 − (f ′′

12)2 = e2x+y > 0. Hence, f is strictly concave.

(b) g is strictly convex. (g′′11 = ex+y + ex−y > 0 and g′′

11g′′22 − (g′′

12)2 = 4e2x > 0.)

6. f is concave where f ′′11 = 2 − 6x ≤ 0, f ′′

22 = −2 ≤ 0, and f ′′11f

′′22 − (f ′′

12)2 = −2(2 − 6x) − 1 ≥ 0.

Here 2 − 6x ≤ 0 iff x ≥ 1/3 and −2(2 − 6x) − 1 ≥ 0 iff x ≥ 5/12. We conclude that the largest convexdomain on which f is concave is S = {(x, y) : x ≥ 5/12}.

8. f ′1 = −4x + 4, f ′

2 = −2y + 4, f ′′11 = −4, f ′′

12 = 0, f ′′22 = −2. Clearly (1, 2) is a stationary point for f

and the conditions in Theorem 17.11(a) are satisfied, so (1, 2) is the maximum point.

10. (a) The firms choose p and q to maximize px+qy−(5+x)−(3+2y) = 26p+24q−5p2−6q2+8pq−69.This is a strictly concave function. The maximum is given by p = 9, q = 8, x = 16, y = 4. A’s profit is123, whereas B’s is 21. (b) Firm A’s profit is now πA(p) = px − 5 − x = p(29 − 5p + 4q)− 5 − 29 +5p − 4q = 34p − 5p2 + 4pq − 4q − 34, with q fixed. This quadratic polynomial is maximized at p =pA(q) = 1

5 (2q+17). Likewise, firm B’s profit is now πB(q) = qy−3−2y = 28q−6q2+4pq−8p−35,with p fixed. This quadratic polynomial is maximized at q = qB(p) = 1

3 (p + 7).(c) Equilibrium occurs where p = pA(q) and q = qB(p). So 5p = 2q + 17 and 3q = p + 7. These givep = 5, q = 4, x = 20, y = 12. A gets 75, B gets 21.(d) The process starts at (9, 8) and converges to (5, 4). See Fig. 17.8.10.



q

2

4

6

8

10

p2 4 6 8 10 12 14

(5, 4)

q = qB(p)

p = pA(q)

(9, 8)

Figure 17.8.10

17.9

2. (a) Stationary points must satisfy the three equations: f ′1 = 2x + 2y + 2z = 0, f ′

2 = 4y + 2x = 0, andf ′

3 = 6z + 2x = 0. The only solution is (0, 0, 0). The Hessian matrix is⎛⎜⎝

f ′′11 f ′′

12 f ′′13

f ′′21 f ′′

22 f ′′23

f ′′31 f ′′

32 f ′′33

⎞⎟⎠ =

⎛⎜⎝

2 2 2

2 4 0

2 0 6

⎞⎟⎠

The leading principal minors are 2,

∣∣∣∣ 2 22 4

∣∣∣∣ = 4,

∣∣∣∣∣∣2 2 22 4 02 0 6

∣∣∣∣∣∣ = 8. So (0, 0, 0) is a local minimum point.

(b) f has two stationary points, (3, 3, 3) and ( 13y2

0 , y0, y0), where y0 ≈ 1.63 is the other real root (apartfrom y = 3) of the equation y4 − 54y + 81 = 0. (There are two complex roots.) The Hessian matrix is

H =⎛⎜⎝

6x −9 −9

−9 6y 0

−9 0 6z

⎞⎟⎠

The leading principal minors areD1 = 6x, D2 =∣∣∣∣ 6x −9−9 6y

∣∣∣∣ = 36xy − 81, D3 = |H| =54(4xyz − 9y − 9z). At (3, 3, 3) the leading principal minors are all positive, so this is a local min-imum point. At ( 1

3y20 , y0, y0), D1 > 0, D2 < 0, and D3 < 0. According to [17.27], this is a saddle point.

(c) The stationary points are the solutions of the equation system

f ′1 = 8x2 − 8x1 = 0, f ′

2 = 20 + 8x1 − 12x22 = 0, f ′

3 = 48 − 24x3 = 0, f ′4 = 6 − 2x4

The first equation gives x2 = x1, and then the second equation gives 12x21 − 8x1 − 20 = 0 with the two

solutions x1 = 5/3 and x1 = −1. The last two equations determine x3 and x4. There are two stationarypoints, (5/3, 5/3, 2, 3) and (−1, −1, 2, 3). The Hessian matrix is

f ′′(x1, x2, x3, x4) =

⎛⎜⎜⎝

−8 8 0 08 −24x2 0 00 0 −24 00 0 0 −2

⎞⎟⎟⎠

and the leading principal minors of the Hessian are D1 = −8, D2 = 192x2−64 = 64(3x2−1), D3 = −24D2,and D4 = 48D2. At (−1, −1, 2, 3) we get D2 < 0, so this point is a saddle point. The other stationary point,(5/3, 5/3, 2, 3), we get D1 < 0, D2 > 0, D3 < 0, and D4 > 0, so this point is a local maximum point.



17.102. If f is increasing, or decreasing, then the upper level sets {x : f (x) ≥ a} are all intervals, which are

convex sets, so f is quasi-concave.

4. Let x, x0 be any pair of vectors in S, and λ ∈ [0, 1]. Let z = (1 − λ)x + λx0. Now, if f (z) ≥min{f (x), f (x0)}, then f (x) ≥ f (x0) implies f (z) ≥ f (x0), as in [17.31]. Hence, Theorem 17.15implies that f is quasi-concave.

Conversely, suppose f is quasi-concave. If f (x) ≥ f (x0), then [17.31] implies f (z) ≥ f (x0). Onthe other hand, if f (x) ≤ f (x0), then interchanging x and x0 in [17.31] and replacing λ by 1 − λ impliesthat f (z) ≥ f (x). In either case, f (z) ≥ min{f (x), f (x0)}.

6. (a) This is mainly an exercise in manipulating determinants. If you feel that the calculations below lookfrightening, try to write them out in full for the case k = 3 (or k = 2). Note that z′′

ij = aiaj z/xixj for

i �= j , and z′′ii = ai(ai − 1)z/x2

i . Rule 3 in Theorem 13.1 tells us that a common factor in any column(or row) in a determinant can be “moved outside”. Therefore,

Dk =

∣∣∣∣∣∣∣∣∣∣

z′′11 z′′

12 . . . z′′1k

z′′21 z′′

22 . . . z′′2k

......

. . ....

z′′k1 z′′

k2 . . . z′′kk

∣∣∣∣∣∣∣∣∣∣=

∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣

a1(a1 − 1)

x21

za1a2

x1x2z · · · a1ak

x1xk

z

a2a1

x2x1z

a2(a2 − 1)

x22

z · · · a2ak

x2xk

z

......

. . ....

aka1

xkx1z

aka2

xkx2z · · · ak(ak − 1)

x2k

z

∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣

(1)= a1a2 . . . ak

x1x2 . . . xk

zk

∣∣∣∣∣∣∣∣∣∣∣∣∣∣

a1 − 1

x1

a1

x1· · · a1

x1

a2

x2

a2 − 1

x2· · · a2

x2...

.... . .

...ak

xk

ak

xk

· · · ak − 1

xk

∣∣∣∣∣∣∣∣∣∣∣∣∣∣

(2)= a1a2 . . . ak

(x1x2 . . . xk)2zk

∣∣∣∣∣∣∣∣a1 − 1 a1 · · · a1

a2 a2 − 1 · · · a2...

.... . .

...

ak ak · · · ak − 1

∣∣∣∣∣∣∣∣

where equality (1) holds because aj z/xj is a common factor in column j for each j and equality (2) holdsbecause 1/xi is a common factor in row i for each i.

(b) Let sk = ∑ki=1 ai = a1 + · · · + ak , and Pk = a1a2 . . . ak

(x1x2 . . . xk)2zk . We use the expression for Dk that

we found in part (a), and add rows 2, 3, . . . , k to the first row. Then each entry in the first row becomesequal to sk − 1. Afterwards we take the common factor sk − 1 in row 1 and move it outside.

Dk = Pk

∣∣∣∣∣∣∣∣sk − 1 sk − 1 · · · sk − 1

a2 a2 − 1 · · · a2...

.... . .

...

ak ak · · · ak − 1

∣∣∣∣∣∣∣∣= (sk − 1)Pk

∣∣∣∣∣∣∣∣1 1 · · · 1a2 a2 − 1 · · · a2...

.... . .

...

ak ak · · · ak − 1

∣∣∣∣∣∣∣∣Now subtract column 1 from all the other columns. Rule 7 in Theorem 13.1 says that this does not change

the value of the determinant, so Dk = (sk − 1)Pk

∣∣∣∣∣∣∣∣1 0 · · · 0a2 −1 · · · 0...

.... . .

...

ak 0 · · · −1

∣∣∣∣∣∣∣∣= (−1)k−1(sk − 1)Pk .


C H A P T E R 1 8 C O N S T R A I N E D O P T I M I Z A T I O N 49

(c) By assumption, ai > 0 for all i, so if∑n

i=1 ai < 1, then sk =∑ki=1 ai < 1 for all k. Therefore Dk has

the same sign as (−1)k . Combining Theorem 15.3 and Theorem 17.14, it follows that f is strictly concave.

Chapter 18 Constrained Optimization

18.1

2. (a) When f (x, y) = 3xy, g(x, y) = x2 + y2, and c = 8, equation [18.3] reduces to 3y/3x = 2x/2y,and so x2 = y2. Inserted into the constraint this yields x2 = 4, and so x = ±2. The solution candidatesare therefore: (2, 2), (2, −2), (−2, 2), (−2, −2). Here f (2, 2) = f (−2, −2) = 12 and f (−2, 2) =f (2, −2) = −12. So (2, 2) and (−2, −2) are the only possible solutions of the maximization problem,and (−2, 2) and (2, −2) are the only possible solutions of the minimization problem. (The extreme-valuetheorem ensures that we have found the solutions, because f is continuous and the constraint curve is aclosed bounded set (a circle).)(b) With L = x + y − λ(x2 + 3xy + 3y2 − 3), the first-order conditions are 1 − 2λx − 3λy = 0and 1 − 3λx − 6λy = 0. From these equations we get 2λx + 3λy = 3λx + 6λy, or λ(3y + x) = 0.Here λ = 0 is impossible, so x = −3y. Inserted into the constraint we have (3, −1) and (−3, 1) as theonly possible solutions of the maximization and minimization problems, respectively. The extreme valuetheorem ensures that solutions exist. (The objective function is continuous and the constraint curve is aclosed bounded set (an ellipse).)

18.2

2. (a) x = a/5, y = 2a/5 solve the problem. Solving x + 2y = a for y yields y = 12a − 1

2x, and thenx2 + y2 = x2 + ( 1

2a − 12x)2 = 5

4x2 − 12ax + 1

4a2. This quadratic function certainly has a minimum atx = a/5. (b) L(x, y) = x2 + y2 − λ(x + 2y − a). The necessary conditions are L′

1 = 2x − λ = 0,L′

2 = 2y − 2λ = 0, implying that 2x = y. From the constraint, x = a/5 and then y = 2a/5, λ = 2a/5.(c) Find the point on the linex+2y = a which has the minimal distance from the origin. The correspondingmaximization problem has no solution. (See Fig. 18.2.2.)(d) f ∗(a) = (a/5)2 + (2a/5)2 = a2/5, so df ∗(a)/da = 2a/5, which is also the value of the Lagrangeanmultiplier. Equation [18.7] is confirmed.

y

x

(a/5, 2a/5)

a

a/2

Figure 18.2.2

4. L(2, 2) = 2 > L(1, 1) = 1. (In fact, L(x, y) has a saddle point at (1, 1).)


50 C H A P T E R 1 8 C O N S T R A I N E D O P T I M I Z A T I O N

6. (a) L(x, y) = 100−e−x−e−y−λ(px+qy−m), so the first-order conditions are: (1) L′x = e−x−λp = 0

and (2) L′y = e−y − λq = 0. These imply that −x = ln(λp) = ln λ + ln p, −y = ln λ + ln q. Inserting

the expressions for x and y into the constraint, and solving for ln λ, yields

ln λ = −(m + p ln p + q ln q)/(p + q). Therefore x(p, q, m) = m + q ln q − q ln p

p + qand

y(p, q, m) = m + p ln p − p ln q

p + q.

(b) x(tp, tq, tm) = tm + tq ln(tq) − tq ln(tp)

tp + tq= tm + tq(ln t + ln q) − tq(ln t + ln p)

t (p + q)= x(p, q, m), so x is homogeneous of degree 0. The constraint px + qy = m remains unchanged whenx, y, m are all multiplied by t , so the maximization of U is over the same constraint set in both cases.Hence, the optimal values of x and y have to be unchanged.

8. (a) g(x, y) = ∫ y

0 xt (y− t) dt = ∫ y

0 (xyt −xt2) dt = ∣∣y0( 12xyt2 − 1

3xt3) dt = 16xy3. Similar computations

yield f (x, y) = − 16αxy5 + 1

12xy4 + 16xy3. The solution of [∗] is x = 384α3M , y = 1/4α. (Note that,

because 16xy3 = M , the problem reduces to that of maximizing M + 1

2My − αMy2 for y ≥ 0.)(b) As α tends to 0, the cost of extraction tends to infinity. Because the sales price increases as t increases,it is reasonable to expect that the terminal time y tends to infinity. (c) The value function is f ∗(M) =M +M/16α. The Lagrange multiplier must satisfy the equation f ′

1 −λg′1 = 0, or − 1

6αy5 + 112y4 + 1

6y3 −λ 1

6y3 = 0. Inserting the optimal values of x and y and solving for λ yields λ = 1 + 1/16α. Clearly,∂f ∗(M)/∂M = λ.

18.32. (a) (i) The constraint implies that x = y2/8 ≥ 0, so the problem reduces to that of minimizing h(x) =

(x − 1)2 + 8x for x ≥ 0. Because h′(x) = 2x + 6 ≥ 6 for all x ≥ 0, the minimum value is 1 at x = 0.(ii) The problem reduces to minimizing k(y) = ( 1

8y2 − 1)2 + y2. Here k′(y) = 2( 18y2 − 1) 1

4y + 2y =y( 1

16y2 + 32 ). Thus k′(y) < 0 for y < 0, and k′(y) > 0 for y > 0. We conclude that k(y) is minimized

at y = 0, which gives the same solution as in (i).(b) For the Lagrangean L = (x−1)2+y2−λ(y2−8x), the first-order conditions are: (1) 2(x−1)+8λ = 0;(2) 2y − 2λy = 0. From (2), either λ = 1 or y = 0. But inserting λ = 1 into (1) yields x = −3, and thenno value of y satisfies the constraint. So y = 0, implying the solution x = y = 0.(c) The problem is to find which point on the parabola y2 = 8x is closest to (1, 0).

18.42. In this case, f (x, y) = x2 + y2, g(x, y) = x + 2y, so

D(x, y) =

∣∣∣∣∣∣∣0 g′

1 g′2

g′1 f ′′

11 − λg′′11 f ′′

12 − λg′′12

g′2 f ′′

21 − λg′′21 f ′′

22 − λg′′22

∣∣∣∣∣∣∣ =∣∣∣∣∣∣∣0 1 2

1 2 0

2 0 2

∣∣∣∣∣∣∣ = −10

By [18.14], there is a local minimum at the stationary point (a/5, 2a/5).

18.52. Here L = x + 4y + 3z − λ(x2 + 2y2 + 1

3z2 − b). So necessary conditions are:(1) L′

1 = 1 − 2λx = 0; (2) L′2 = 4 − 4λy = 0; (3) L′

3 = 3 − 23λz = 0. It follows that λ �= 0,



and so x = 1/2λ, y = 1/λ, z = 9/2λ. Inserting these values into the constraint yields λ2 = 9/b, soλ = ±3/

√b. The value of the criterion function is x + 4y + 3z = 18/λ, so λ = −3/

√b minimizes the

criterion function. The minimum point is (a, 2a, 9a), where a = −√b/6.

4. With x = 1615 + h, y = 1

3 + k, and z = − 1115 + �, constraints [1] and [2] reduce to h + 2k + l = 0 and

2h − k − 3� = 0, so k = −h and � = h. But then easy algebra shows that f (x, y, z) = (16/15)2 +(1/3)2 + (−11/15)2 + 3h2. Because 3h2 ≥ 0 for all h, (16/15)2 + (1/3)2 + (−11/15)2 must then be themaximum value of f (x, y, z) subject to the constraints. The maximum value is obviously attained whenh = k = � = 0, or when (x, y, z) = (16/15, 1/3, −11/15).

6. (a) Here L = x + y + z − λ(x2 + y2 + z2 − 1) − μ(x − y − z − 1). Necessary conditions are:(1) L′

1 = 1 − 2λx − μ = 0; (2) L′2 = 1 − 2λy + μ = 0; (3) L′

3 = 1 − 2λz + μ = 0. Adding (1) and(2) yields λ(x + y) = 1. Adding (1) and (3) yields λ(x + z) = 1. It follows that λ �= 0 and also thatz = y. Substituting z = y in the two constraints yields: (4) x2 + 2y2 = 1; (5) x − 2y = 1. From (5),x = 2y + 1, so (4) gives 6y2 + 4y = 0, with solutions y = 0 and y = −2/3. The only possible solutionsto the problem are therefore (1, 0, 0) and (−1/3, −2/3, −2/3). In fact, (1, 0, 0) solves the maximizationproblem, while (−1/3, −2/3, −2/3) solves the minimization problem.

(b) The constraints are represented geometrically by a closed bounded curve which is the intersectionbetween a sphere (x2 + y2 + z2 = 1) and a plane (x − y − z = 1). The continuous function f (x, y, z) =x + y + z must therefore attain maximum and minimum values subject to the constraints, and in (a) wehave found the extreme points.

8. Here L = a21x2

1 +· · ·+a2nx

2n −λ(x1 +· · ·+xn −1). Necessary conditions are that L′

j = 2a2j xj −λ = 0,

j = 1, . . . , n, and so xj = λ/2a2j . Inserted into the constraint, this implies that 1 = 1

2λ(1/a21 +· · ·+1/a2

n).Thus, for j = 1, . . . , n, we have

xj = 1

a2j (1/a2

1 + · · · + 1/a2n)

= 1

a2j

∑ni=1(1/a2

i )

(Implicitly, this solution assumes that each ai �= 0. If at least one ai is 0, the minimum value is 0, whichis attained by letting a corresponding xi be 1, with the other xj all equal to 0.)

10. (a) The Lagrangean is here L = Axa11 · · · xan

n − λ(p1x1 + · · · + pnxn − m), and soL′

i = aiU/xi − λpi = 0, i = 1, . . . , n. In particular, p1x1 = a1U/λ, and so λ = a1U/p1x1. Itfollows that pixi = (ai/a1)p1x1. Inserted into the budget constraint, we have p1x1 + (a2/a1)p1x1 +· · · + (an/a1)p1x1 = m, which implies that x1 = a1m

p1(a1 + · · · + an). The demand functions are given

by xi = aim

pi(a1 + · · · + an)for i = 1, . . . , n.

(b) The Lagrangean is L = xa1 + · · · + xa

n − λ(p1x1 + · · · + pnxn − m), and so L′i = axa−1

i − λpi = 0.

In particular, λ = axa−11 /p1. Algebraic manipulations yield pixi = p

−a/(1−a)

i p1/(1−a)1 x1, i = 1, . . . , n.

Using the budget constraint equation gives an expression for x1, and subsequently we find the other

demand quantities xi . The consumer’s demand functions are xi = mp−1/(1−a)

in∑

j=1

p−a/(1−a)

j

, for i = 1, . . . , n.



12. For n = 2, the Lagrangean is L = a11x21 +2a12x1x2 +a22x

22 −λ(x2

1 +x22 −1). The first-order conditions

for x1 = x01 , x2 = x0

2 , with λ = λ0 to solve the problem are:

(∗)2a11x

01 + 2a12x

02 − 2λ0x

01 = 0

2a12x01 + 2a22x

02 − 2λ0x

02 = 0

Cancelling the common factor 2 and writing (∗) as a matrix equation yields

(a11 a12

a21 a22

)(x0

1x0

2

)= λ0

(x0

1x0

2

)

or, in obvious matrix notation, (∗∗) Ax0 = λ0x0. Thus, if x0 solves the problem, then x0 is an eigenvectorof A, with λ0 as an eigenvalue. Multiplying (∗∗) from the left by x′

0 = (x01 , x0

2 ) then implies thatx′

0Ax0 = x′0(λ0x0) = λ0x′

0x0 = λ0, because x′0x0 = 1. We conclude that the maximum value of x′Ax

subject to x′x = 1 must be the largest eigenvalue of A, whereas the minimum value must be the smallesteigenvalue. Both a maximum and a minimum value do exist because of the extreme-value theorem.

The argument in the general case of n variables is entirely similar. In fact, equating the partials ofthe Lagrangean to 0 leads to a system of equations like (∗), which is then easily seen to reduce to (∗∗)

18.6

2. (a) The Lagrangean is L = 4z−x2 −y2 − z2 −λ(z−xy). The first-order conditions are −2x +λy = 0,−2y + λx = 0, 4 − 2z − λ = 0, and z = xy. The only triples (x, y, z) that satisfy these conditions are(0, 0, 0) with λ = 4, (1, 1, 1) with λ = 2, and (−1, −1, 1) with λ = 2. (b) �f ∗ ≈ λ·0.1 = 2·0.1 = 0.2

18.7

2. If L(x, c) = f (x) −∑mj=1 λj (gj (x) − cj ), then ∂L/∂cj = λj , so [18.24] follows from [18.31].

18.8

2. (a) Writing the constraints as g1(x, y) = x + e−x − y ≤ 0 and g2(x, y) = −x ≤ 0, the Lagrangean isL = 1

2x − y − λ1(x + e−x − y) − λ2(−x). The Kuhn–Tucker conditions are then:

∂L

∂x= 1

2− λ1(1 − e−x) + λ2 = 0(1)

∂L

∂y= −1 + λ1 = 0(2)

λ1 ≥ 0 (= 0 if x + e−x < y)(3)

λ2 ≥ 0 (= 0 if x > 0)(4)

From (2) and (3), we conclude that λ1 = 1 and so x + e−x = y.

Either x = 0 or x > 0. In the latter case, (4) implies λ2 = 0. Then (1) implies 12 − (1 − e−x) = 0,

or e−x = 12 . Hence x = ln 2, and so y = x + e−x = ln 2 + 1

2 .

If x = 0, then (1) implies λ2 = − 12 , which contradicts λ2 ≥ 0.



We conclude that (x, y) = (ln 2, ln 2 + 12 ) is the only point satisfying the Kuhn–Tucker conditions.

(By sketching the constraint set and studying the level curves 12x − y = c, it is easy to see that the point

we found solves the problem.)

(b) With the Lagrangean L = x2 + 2y − λ1(x2 + y2) − λ2(−y), the Kuhn–Tucker conditions are:

∂L

∂x= 2x − 2λ1x = 0(1)

∂L

∂y= 2 − 2λ1y + λ2 = 0(2)

λ1 ≥ 0 (= 0 if x2 + y2 < 5)(3)

λ2 ≥ 0 (= 0 if y > 0)(4)

Note that the second constraint cannot be active, because if y = 0, then (2) implies λ2 = −2, whichcontradicts (4). So y > 0, and from (4), λ2 = 0. Then from (2), λ1y = 1. Hence λ1 > 0, and so from(3), x2 + y2 = 5. Also, (1) evidently implies that x = 0 or λ1 = 1.

If x = 0, then x2 + y2 = 5 implies y = ±√5. With y > 0, only y = √

5 is possible. Then (2)implies that λ1 = 1/

√5. Thus (x, y, λ1, λ2) = (0,

√5, 1/

√5, 0) satisfies (1) to (4) and is a solution

candidate.

If λ1 = 1, then λ1y = 1 implies y = 1 and from x2 +y2 = 5 it follows that x = ±2. Thus (2, 1, 1, 0)

and (−2, 1, 1, 0) are both solution candidates.

Now, f (0,√

5) = 2√

5, while f (2, 1) = f (−2, 1) = 6. The constraint set is closed and bounded(as a half-disk), so the extreme-value theorem applies. Thus, (2, 1) and (−2, 1) both solve the problem,with λ1 = 1 and λ2 = 0.

4. (a) The Lagrangean is L = x + ay − λ(x2 + y2 − 1) + μ(x + y). There must exist numbers λ and μ

such that: (1) 1 − 2λx + μ = 0; (2) a − 2λy + μ = 0; (3) λ ≥ 0 (= 0 if x2 + y2 < 1);(4) μ ≥ 0 (= 0 if x + y > 0).

(b) For a ≥ −1, the point (1/√

1 + a2, a/√

1 + a2) solves the problem, with λ = 12

√1 + a2, μ = 0.

For a < −1, the point (1/√

2, −1/√

2) solves the problem, with λ = (1 − a)/2√

2, μ = − 12 (1 + a).

6. The Lagrangean is L = ln(x2 +2y)− 12x2 −y −λ1(2−xy)+λ2(x −1)+λ3(y −1). The Kuhn–Tucker

conditions are that there must exist numbers λ1 ≥ 0, λ2 ≥ 0, and λ3 ≥ 0 such that

L′1(x, y) = 2x

x2 + 2y− x + λ1y + λ2 = 0(1)

L′2(x, y) = 2

x2 + 2y− 1 + λ1x + λ3 = 0(2)

λ1(2 − xy) = 0(3)

λ2(1 − x) = 0(4)

λ3(1 − y) = 0(5)

(We have deliberately formulated the conditions in a slightly different way than was done for Problems2 and 4. See [18.37] in the main text.)



Suppose first that λ1 = 0. Then (1) and (2) imply

2x

x2 + 2y− x + λ2 = 0,

2

x2 + 2y− 1 + λ3 = 0

Because x and y are both ≥ 1, we get λ3 = 1 − 2/(x2 + 2y) ≥ 1 − 23 > 0 and λ2 = x − 2x/(x2 + 2y) =

xλ3 > 0. It follows from (4) and (5) that x = y = 1, which contradicts xy ≥ 2. Thus λ1 > 0 and soxy = 2.

Suppose next that λ2 > 0. Then x = 1 (because of (4)) and y = 2/x = 2, so λ3 = 0. From (2) itfollows that λ1 = 3/5. Then (1) implies that λ2 = −3/5, contradicting λ2 ≥ 0. Hence, λ2 = 0.

Suppose now that λ3 > 0. Then (5) implies y = 1, and so x = 2/y = 2. From (1) it follows thatλ1 = 4/3. Then, from (2), λ3 = −1/3 + 1 − 8/3 < 0, which is impossible. Thus λ3 = 0.

So far we have proved that λ2 = λ3 = 0, λ1 > 0, and xy = 2. Equations (1) and (2) now yield

2x

x2 + 2y= x − λ1y,

2

x2 + 2y= 1 − λ1x

It follows that x −λ1y = x(1−λ1x). But λ1 > 0 and so y = x2. Because xy = 2, we must have x3 = 2.Hence, x = 21/3 and y = 2x−1 = 22/3 is the only point which satisfies the Kuhn–Tucker conditions.(The corresponding nonnegative multipliers are λ1 = 2−1/3 − 1

3 , λ2 = λ3 = 0.)Why is this a maximum? In the feasible set, x2 +2y ≥ 3, and f ′

1(x, y) and f ′2(x, y) are both negative.

For a given value of x, f (x, y) decreases as y increases, and for a given value of y, f (x, y) decreases asx increases. It follows that the constrained maximum lies on the part of the curve xy = 2 between thetwo points (1, 2) and (2, 1). (See Fig. 18.8.4.) This is a closed and bounded set, so the extreme-valuetheorem implies that f (x, y) has a maximum on this part of the curve. This point must be a maximumpoint for f over the whole feasible set.

y

x

(1, 1)

y

1

2

3

4

x1 2 3 4

Figure 18.8.4

18.92. The Lagrangean is L = 9x + 8y − 6(x + y)2 − λ1(x − 5) − λ2(y − 3) − λ3(−x + 2y − 2) and the

Kuhn–Tucker conditions are:

L′1 = 9 − 12(x + y) − λ1 + λ3 ≤ 0 (= 0 if x > 0)(1)



L′2 = 8 − 12(x + y) − λ2 − 2λ3 ≤ 0 (= 0 if y > 0)(2)

λ1 ≥ 0 (= 0 if x < 5)(3)

λ2 ≥ 0 (= 0 if y < 3)(4)

λ3 ≥ 0 (= 0 if −x + 2y < 2)(5)

We go through the systematic procedure explained at the end of Section 18.8. (Shortcuts are possible.For instance, one could start by proving that x = 5 is impossible, then that y cannot be positive, etc.)

x < 5, y < 3, −x + 2y < 2. Then λ1 = λ2 = λ3 = 0. If x = 0, then from (1), y ≥ 3/4 > 0, so from(2), 8 − 12y = 0. Thus y = 2/3, which contradicts y ≥ 3/4. So x > 0. Then (1) implies 9 = 12(x +y).If y > 0, then (2) implies 8 = 12(x + y), a contradiction. Thus y = 0. But then 9 = 12x, so x = 3/4.So a solution candidate is (x, y) = (3/4, 0), with λ1 = λ2 = λ3 = 0.

x = 5, y < 3, −x + 2y < 2. Then λ2 = λ3 = 0 and (1) implies λ1 = 9−12(5+y) < 0, a contradiction.

x < 5, y = 3, −x + 2y < 2. Then λ1 = λ3 = 0 and (2) implies λ2 = 8−12(x +3) < 0, a contradiction.

x < 5, y < 3, −x + 2y = 2. Then λ1 = λ2 = 0 and 2y = 2 + x > 0, so y > 0 and (2) imply

8 − 12(x + y) = 2λ3. It follows that λ3 = 4 − 6[x + 12 (2 + x)] = −2 − 9x < 0, a contradiction.

x < 5, y = 3, −x + 2y = 2. Then x = 4 and λ1 = 0. Now (1) gives λ3 = 75, and then (2) givesλ2 = −226, contradicting λ2 ≥ 0.

x = 5, y < 3, −x + 2y = 2. Then the first and second equalities yield y = 3.5, which contradicts y < 3.

x = 5, y = 3, −x + 2y < 2. Then λ3 = 0. Now (1) gives λ1 = −87, contradicting λ1 ≥ 0.

x = 5, y = 3, −x + 2y = 2. Obviously impossible.

We conclude that the only admissible pair satisfying (1) to (5) is (x, y) = (3/4, 0). The extreme-valuetheorem implies that we have found the solution. (Theorem 18.4 of the next section also shows optimality.)

4. Using (a) in Theorem 18.3 with x = x implies that f (x0) −m∑

j=1λjgj (x0) ≥ f (x) −

m∑j=1

λjgj (x). But,

because x also solves [18.41], f (x) = f (x0). Thus, because λj ≥ 0 and gj (x) ≤ cj , j = 1, . . . , m,

and also because of [18.45], we have (∗)m∑

j=1λjcj ≥

m∑j=1

λjgj (x) ≥m∑

j=1λjgj (x0) =

m∑j=1

λjcj . Here the

two middle terms, being squeezed between two equal numbers, must themselves be equal. Therefore

f (x) −m∑

j=1λjgj (x) = f (x0) −

m∑j=1

λjgj (x0) ≥ f (x) −m∑

j=1λjgj (x) for all x ≥ 0. Also, if λk > 0 and

gk(x) < ck for any k, thenm∑

j=1λj (gj (x) − cj ) < 0, which contradicts (∗). Thus x satisfies (a) and (b) in

Theorem 18.3.

6. (a) As suggested by the hint, L(x∗, λ∗) ≤ L(x∗, λ) for all λ ≥ 0 implies that

(1)

m∑j=1

λ∗j (gj (x∗) − cj ) ≥

m∑j=1

λj (gj (x∗) − cj )

for all λ ≥ 0. If gk(x∗) > ck for any k, then∑m

j=1 λj (gj (x∗) − cj ) can be made arbitrarily large bychoosing λk large and λj = 0 for all j �= k. Hence, gj (x∗) ≤ cj , j = 1, . . . , m. Now, however, because



one can have λj = 0, j = 1, . . . , m, (1) implies that∑

mj=1λ∗j (gj (x∗) − cj ) ≥ 0. Yet λ∗

j ≥ 0 andgj (x∗) ≤ cj , j = 1, . . . , m, so

∑mj=1 λ∗

j (gj (x∗) − cj ) = 0.Finally, whenever gj (x) ≤ cj , j = 1, . . . , m, the inequality L(x, λ∗) ≤ L(x∗, λ∗) implies that

f (x) − f (x∗) ≤m∑

j=1

λ∗j [gj (x) − gj (x∗)] =

m∑j=1

λ∗j [gj (x) − cj ] ≤ 0

Therefore f (x) ≤ f (x∗), so x∗ solves [18.41].(b) First, part (b) of the question should be corrected to read as follows:

“Suppose that there exist x∗ ≥ 0 and λ∗ ≥ 0 satisfying both gj (x∗) ≤ cj and gj (x∗) = cj wheneverλ∗j > 0 for j = 1, . . . , m, as well as L(x, λ∗) ≤ L(x∗, λ∗) whenever x ≥ 0. Show that L(x, λ) has a

saddle point at (x∗, λ∗) in this case.”

Then an answer is as follows: The first inequality in [∗] is satisfied by assumption. As for the second,under the indicated hypotheses, one has

L(x∗, λ∗) − L(x∗, λ) =m∑

j=1

(λj − λ∗j )[gj (x∗) − cj ] ≤ −

m∑j=1

λ∗j [gj (x∗) − cj ] = 0

whenever λ ≥ 0. Thus, [∗] has been verified.

x

P

y

2x � y � c2

2x � y � c1

Figure 18.10.2

18.102. (a) As shown in Fig. 18.10.2, the feasible set is the intersection of the two circular disks with the non-

negative orthant. The solution is at the point P , where the circles intersect. Thus, (x + 1)2 + y2 = 4 =x2 + (y + 1)2. It follows that x = y and so 2x2 + 2x + 1 = 4. The (relevant) positive root is 1

2 (√

7 − 1),

so the solution is(

12 (

√7 − 1), 1

2 (√

7 − 1))

. (b) The Kuhn–Tucker conditions are:

2 − λ12(x + 1) − λ22x ≤ 0 (= 0 if x > 0)(1)

1 − λ12y − λ22(y + 1) ≤ 0 (= 0 if y > 0)(2)

λ1 ≥ 0 (= 0 if (x + 1)2 + y2 < 4)(3)

λ2 ≥ 0 (= 0 if x2 + (y + 1)2 < 4)(4)

The objective function is linear and thus concave. The constraint functions are both sums of convexfunctions, so they are convex. The conditions (1) to (4) are therefore sufficient for an admissible pair to



solve the problem. With x = y = 12 (

√7 − 1), equations (1) to (4) are satisfied with λ1 = (7 + 3

√7)/28

and λ2 = (3√

7 − 7)/28.(c) The change in the optimal value of 2x + y is approximately λ2 · 0.1 ≈ 0.003.

4. (a) If x∗ = a, f ′(x∗) ≤ 0; if x∗ ∈ (a, b), f ′(x∗) = 0; if x∗ = b, f ′(x∗) ≥ 0.(b) See Figs. 18.10.4.

x*�a b x

y

a x* b x

y

x*�ba x

y

(a)

Figure 18.10.4

(b) (c)

6. (a) See Fig. 18.10.6. The optimal solution is obviously at (x, y) = (1, 0). At this point the two constraintsg(x, y) = y − (1 − x)3 ≤ 0 and h(x, y) = −y ≤ 0 are both binding. The gradient vector of g is(g′

1, g′2) = (3(1 − x)2, 1) = (0, 1) at (1, 0), and the gradient of h is (h′

1, h′2) = (0, −1). Hence, the

two gradients are linearly dependent at (1, 0) (in fact, (0, −1) = (−1)(0, 1)), and so the constraintqualification is violated at this point.

3x � y � c

x1

y

1

y � (1 � x)3

Figure 18.10.6

(b) The Lagrangean is L = 3x + y − λ1(y − (1 − x)3) + λ2x + λ3y, so the Kuhn–Tucker conditions are:

L′1 = 3 − 3λ1(1 − x)2 + λ2 = 0(1)

L′2 = 1 − λ1 + λ3 = 0(2)

λ1 ≥ 0 (= 0 if y < (1 − x)3)(3)

λ2 ≥ 0 (= 0 if x > 0)(4)

λ3 ≥ 0 (= 0 if y > 0)(5)

Suppose x > 0. Then from (4), λ2 = 0, and (1) implies λ1(1 − x)2 = 1. Thus λ1 �= 0, so λ1 > 0 andthen from (3), y = (1 − x)3. From λ1(1 − x)2 = 1 it follows also that x �= 1, so y = (1 − x)3 �= 0, and


58 C H A P T E R 1 9 L I N E A R P R O G R A M M I N G

hence y > 0. So from (5), λ3 = 0. Then (2) yields λ1 = 1. Therefore (1) implies that (1 − x)2 = 1. Sox = 2, because x > 0. But for x = 2, one has y = (1 − x)3 = −1, a contradiction.

Suppose x = 0. Then (1) implies 3 − 3λ1 + λ2 = 0, so 3λ1 = 3 + λ2 > 0. Thus, from (3),y = (1 − x)3 = 1, and so by (5), λ3 = 0. Then from (2), λ1 = 1. So all the conditions (1)–(5) aresatisfied by x = 0, y = 1, with λ1 = 1 and λ2 = λ3 = 0.

So the Kuhn–Tucker conditions have a unique solution, which is not optimal. Finally, the twoconstraints g(x, y) = y − (1 − x)3 ≤ 0 and k(x, y) = −x ≤ 0 are both binding at (0, 1). But (g′

1, g′2) =

(3(1 − x)2, 1) = (3, 1) and (k′1, k

′2) = (−1, 0) are linearly independent. So the constraint qualification

is satisfied at (0, 1), where the Kuhn–Tucker conditions are met. The Kuhn–Tucker conditions can beviolated at the optimum (1, 0), because the constraint qualification is not satisfied at that point.

Chapter 19 Linear Programming

19.12. (a) No maximum exists. Consider Fig. 19.1.2. By increasing c, the dashed level curve x1 +x2 = c for the

criterion function moves to the north-east and so this function can obviously take arbitrary large values.(b) Yes, the maximum is at P = (1, 0). The level curves are the same as in (a), but the direction ofincrease is reversed.

x2

x11 652 3 4

1

4

2

3x1 � x2 � c

�x1 � x2 � �1

�x1 � 3x2 � 3700x � 1000y � c

1000 2000

1000

y

P

x


4. The LP problem is:

(∗) max 700x + 1000y subject to

⎧⎨⎩

3x + 5y ≤ 3900x + 3y ≤ 2100

2x + 2y ≤ 2200x ≥ 0 , y ≥ 0

The problem is represented geometrically in Fig. 19.1.4. The solution is obviously at P , where the two lines3x + 5y = 3900 and 2x + 2y = 2200 intersect. Solving these equations yields x = 800 and y = 300. Sothe firm should produce 800 TV sets of type A and 300 sets of type B.

19.2

2. (a) min 6u1 + 4u2 subject to

{3u1 + u2 ≥ 32u1 + 4u2 ≥ 4

u1 ≥ 0, u2 ≥ 0

(b) max 11x1 + 20x2 subject to

{x1 + 2x2 ≤ 10

3x1 + 5x2 ≤ 27x1 ≥ 0, x2 ≥ 0


C H A P T E R 1 9 L I N E A R P R O G R A M M I N G 59

19.3

2. max 300x1 + 500x2 subject to

{10x1 + 25x2 ≤ 10 00020x1 + 25x2 ≤ 8 000

x1 ≥ 0 , x2 ≥ 0

The solution can be found graphically. It is x∗1 = 0, x∗

2 = 320, and the value of the criterion function is160, 000, the same value found in Example 19.2 for the optimal value of the primal criterion function.

19.5

2. (a) Figure 19.5.2 shows a portion of the feasible set and two dashed level curves for the criterion functionZ = y1 + 2y2. We see that the minimum is attained at the point (y∗

1 , y∗2 ) = (3, 2).

(b) The dual is:

max 15x1 + 5x2 − 5x3 − 20x4 subject to

x1 + x2 − x3 + x4 ≤ 1

6x1 + x2 + x3 − 2x4 ≤ 2

x1 ≥ 0, x2 ≥ 0, x3 ≥ 0, x4 ≥ 0

Because y∗1 and y∗

2 are both positive, the first two constraints are satisfied with equality. Furthermore,the two last constraints in the primal problem are satisfied with strict inequality at the optimum. Hencex∗

3 = x∗4 = 0. Thus we have x∗

1 + x∗2 = 1, 6x∗

1 + x∗2 = 2, so x∗

1 = 1/5 and x∗2 = 4/5. The maximum is

thus at (x∗1 , x∗

2 , x∗3 , x∗

4 ) = (1/5, 4/5, 0, 0).(c) If the constraint is changed to y1 + 6y2 ≥ 15.1, the solution to the primal (P) is still at the intersectionof the lines (1) and (2) in Fig. 19.5.2, but with (1) shifted up slightly. The solution to the dual is completelyunchanged. The optimal values in (P ) and (D) both increase by (15.1 − 15) · x∗

1 = 0.02.

y2

5

10

y15 10 15

(3)

y1 + 2y2 = Z0

(1)(3, 2)

(2)

(4)

Figure 19.5.2

4. The dual to problem [1] is

[2] min q1V1 + q2V2 + r1x1 + · · · + rN xN s.t.

⎧⎪⎪⎪⎨⎪⎪⎪⎩

ξ 11 q1 + ξ 1

2 q2 + r1 ≥ 1ξ 2

1 q1 + ξ 22 q2 +r2 ≥ 1

......

ξN1 q1 + ξN

2 q2 +rN ≥ 1

where the variables are all ≥ 0. The implications [i] to [vi] are all special cases of the complementary slacknessconditions [19.13] and [19.14] in Theorem 19.4.


60 C H A P T E R 2 0 D I F F E R E N C E E Q U A T I O N S

Chapter 20 Difference Equations

20.1

2. (a) Monotone convergence to x∗ from below. (b) Damped oscillations around x∗.(c) Monotonically increasing towards ∞. (d) Explosive oscillations around x∗.(e) xt = x∗ for all t . (f) Oscillations around x∗ with constant amplitude.(g) Monotonically (linearly) increasing towards ∞. (h) Monotonically (linearly) decreasing towards −∞.(i) xt = x0 for all t.

4. (a) Because the parameters are positive, yk+1 is positive provided that yk > 0. Because y0 is positive, byinduction, yt is positive for all t . (b) Substituting yt = 1/xt gives the equation xt+1 = (a/c)xt + b/c.When a = 2, b = 3, and c = 4, this equation reduces to xt+1 = (1/2)xt + 3/4. When x0 = 1/y0 = 2,the solution is xt = (1/2)t+1 + 3/2, and so yt = [(1/2)t+1 + 3/2

]−1. Clearly, yt → 2/3 as t → ∞.

6. (a) From [1],Wt+2

Wt+1= Pt+1

Pt

= γ + βWt+1

γ + βWt

, and [2] follows immediately.(b) According to [2], the

fraction Wt+1/(γ + βWt) is the same for all t . For t = 0 it is equal to W1/P0 = c, and thus [3] follows.From [20.4], the solution is

Wt =(W0 − cγ

1 − cβ

)(cβ)t + cγ

1 − cβ(1 �= cβ)

(c) The equation is stable iff |cβ| < 1. In this case, Wt → cγ

1 − cβas t → ∞.

20.2

2. According to [4] in Example 20.5, the yearly repayment is

z = 0.07 · 100, 000

1 − (1.07)−30≈ 8058.64

In the first year the interest payment is 0.07B = 7000, and so the principal repayment is ≈ 8058.64 −7000 = 1058.64. In the last year, the interest payment is 0.07b29 ≈ 8058.64[1 − (1.07)−1] ≈ 527.20,and so the principal repayment is ≈ 8058.64 − 527.20 = 7531.44.

20.4

2. xt = A + B t is a solution: xt+2 − 2xt+1 + xt = A + B (t + 2) − 2[A + B (t + 1)] + A + B t =A + Bt + 2B − 2A − 2Bt − 2B + A + B t = 0. Substituting t = 0 and t = 1 in xt = A + B t yieldsA = x0 and A+B = x1, with solution A = x0 and B = x1 − x0. So xt = A+B t is the general solutionof the given equation.

4. xt = (A + B t)2t + 1 is a solution:xt+2 − 4xt+1 + 4xt = [A + B (t + 2)]2t+2 + 1 − 4{[A + B (t + 1)]2t+1 + 1} + 4[(A + B t)2t + 1] =4A2t + 4Bt2t + 8B2t + 1 − 8A2t − 8Bt2t − 8B2t − 4 + 4A2t + 4Bt2t + 4 = 1. Substituting t = 0and t = 1 in xt = A 2t + B t2t + 1 yields A + 1 = x0 and 2A + 2B + 1 = x1, with solution A = x0 − 1and B = 1

2x1 − x0 + 12 . So xt = A 2t + B t2t + 1 is the general solution of the given equation.


C H A P T E R 2 0 D I F F E R E N C E E Q U A T I O N S 61

20.5

2. (a) The characteristic equation m2 + 2m + 1 = (m + 1)2 = 0 has the double root m = −1, so thegeneral solution of the homogeneous equation is xt = (C1 + C2t)(−1)t . We find a particular solution byinserting u∗

t = A 2t . This yields A = 1, and so the general solution of the inhomogeneous equation isxt = (C1 + C2t)(−1)t + 2t .(b) By using the method of undetermined coefficients to determine the constants A, B, and C in theparticular solution u∗

t = A5t + B cos π2 t + C sin π

2 t , we obtain A = 14 , B = 3

10 , and C = 110 . So the

general solution to the given equation is xt = C1 + C22t + 14 5t + 3

10 cos π2 t + 1

10 sin π2 t .

4. The characteristic equation is m2−4(ab+1)m+4a2b2 = 0, with solutions m1,2 = 2(ab+1±√1 + 2ab ).

The general solution is therefore Dn = C1mn1 + C2m

n2.

6. Inserting xt = ut (−a/2)t into [20.21], assuming that b = 14a2, we obtain the equation

xt+2 + axt+1 + 14a2xt = ut+2(−a/2)t+2 + aut+1(−a/2)t+1 + 1

4a2ut (−a/2)t

= 14a2(−a/2)t (ut+2 − 2ut+1 + ut ), which is 0 if ut+2 − 2ut+1 + ut = 0. The general solution of this

equation is ut = A + B t , so xt = ut (−a/2)t = (A + B t)(−a/2)t , which is the result claimed for case2 in the frame on page 751.

8. (a) The first two equations state that consumption and capital are proportional to the net national productin the previous period. The third equation states that net national product, Yt , is divided between con-sumption, Ct , and net investment, Kt − Kt−1.(b) To derive a second-order difference equation, first replace t by t + 2 in the last displayed equation inthe problem to obtain Yt+2 = Ct+2 +Kt+2 −Kt+1. But Ct+2 = cYt+1, Kt+2 = σYt+1, and Kt+1 = σYt ,so we obtain Yt+2 − (c + σ)Yt+1 + σYt = 0. Explosive oscillations occur when (c + σ)2 < 4σ andσ > 1.

10. From the quadratic formula, m2 + am + b has real zeros iff b ≤ 14a2. Combining [20.27] and [20.28],

we must prove that f (m) = am2 + bm + c = 0 has both roots in the interval (−1, 1) iff |a| < 1 + b andb < 1.

For these roots to be in the interval (−1, 1), it is necessary that f (−1) > 0, f (1) > 0, f ′(−1) < 0,and f ′(1) > 0. Thus 1 − a + b > 0, 1 + a + b > 0, −2 + a < 0, and a + 2 > 0, which is equivalent to|a| < 1 + b and |a| < 2. But then, because the roots are real, b ≤ 1

4a2 < 1.

Conversely, if |a| < 1 + b and b < 1, then f (−1) > 0, f (1) > 0, and |a| < 2. So f ′(−1) =−2 + a < 0 and f ′(1) = 2 + a > 0.

12. (a) Because u(1)t and u

(2)t are two linearly independent solutions of the homogeneous (!) equation, Au

(1)t +

Bu(2)t is the general solution of the homogeneous equation. So, to prove that [∗∗] is the general solution of

[∗], it suffices to prove that U∗t = −u

(1)t

t−1∑i=0

ciu(2)i+1

Di+1+u

(2)t

t−1∑i=0

ciu(1)i+1

Di+1is a particular solution of [∗]. Insertion

in [∗], followed by tedious algebraic computation, eventually yields the result.

(b) pt = Art1 + Brt

2 +t−1∑i=0

rt−i−12 − rt−i−1

1

r2 − r1ui+2, t = 0, 1, . . .

(First write the equation in the form pt+2 + λ1pt+1 + λ2pt = ut+2, t = 0, 1, 2, . . .. Assuming that r1, r2

are real(!) and different roots of the characteristic polynomial, two linearly independent solutions of thehomogeneous equation are u

(1)t = rt

1 and u(2)t = rt

2. So Di+1 = ri+11 ri+2

2 −ri+21 ri+1

2 = ri+11 ri+1

2 (r2 −r1).


62 C H A P T E R 2 1 D I F F E R E N T I A L E Q U A T I O N S

Then

−rt1

t−1∑i=0

ui+2ri+12

ri+11 ri+1

2 (r2 − r1)+ rt

2

t−1∑i=0

ui+2ri+11

ri+11 ri+1

2 (r2 − r1)=

t−1∑i=0

rt−i−12 − rt−i−1

1

r2 − r1ui+2

is the particular solution in [∗∗] to the nonhomogeneous equation.)

Chapter 21 Differential Equations

21.1

2. (a) If x = Ct2, then x = 2Ct , and so t x = 2Ct2 = 2x. (b) x = Ct2 passes through (1, 2) if 2 = C ·12,so C = 2. Hence, x = 2t2 is the desired solution.

4. (a) Replace the problem by: If x = x(t) satisfies x2 = 2√

1 − t2, then x = −t

x√

1 − t2.

(The original problem makes no sense. The differential equation only makes sense if t ∈ (−1, 1),

and then x2 + 2√

1 − t2 = 0 is only satisfied for x = 0.) Differentiating each side of x2 = 2√

1 − t2

w.r.t. t yields 2xx = −2t/√

1 − t2, and so x = −t/x√

1 − t2.(b) Differentiating each side yields 1

2et22t − e−x x(x + 1) + e−x x = 0, and the result follows.

(c) Differentiating each side yields −x2 + 2(1 − t)xx = 3t2. Simplifying and using(1 − t)x2 = t3 yields the given equation.

6. The point of this problem is that you do not have to solve the differential equation (which would involvethe inverse tangent function, anyway). From x = (1+x2)t , we see at once that x < 0 for t < 0 and x > 0for t > 0. Thus t = 0 is a global minimum point for x(t), and because x(0) = 0, one has x(t) ≥ 0 for allt . Differentiating the given equation w.r.t. t yields x = 2xxt + (1 + x2) = 2xt (1 + x2)t + (1 + x2) =(1 + x2)(2xt2 + 1). Clearly x > 0 for all t , so x = x(t) is convex. (In fact, the equation does not have asolution on all of �, so x(t) only has a minimum on an open interval around 0.)

21.2

2. See Fig. 21.2.2. (The solutions are the circles t2 + x2 = C, t �= 0, with C an arbitrary nonnegativeconstant. This can be shown by direct differentiation, or by solving the separable equation x = −t/x

using the method set out in the next section.)

x

1t

1

t2 + x2 = 4

Figure 21.2.2


C H A P T E R 2 1 D I F F E R E N T I A L E Q U A T I O N S 63

21.32. (a) Direct integration yields x = 1

4 t4 − 12 t2 + C. (b) x = tet − et − 1

2 t2 + C (c) ex(dx/dt) = t + 1, orex dx = (t+1) dt , so that

∫ex dx = ∫ (t+1) dt and ex = 1

2 t2+t+C. The solution is x = ln( 12 t2+t+C).

4. In both cases N depends on both N and t . (In the first figure, for instance, N(t1) = N(t2), but N(t1) �=N(t2).)

6. (a) x = Cta (b) x = Ctbeat (c) x = Cbtb/(1 − aCtb)

8. If z = x/t , then x = zt , so x = zt + z, and the equation x = g(x/t) becomes the separable equation

t z = g(z)−z. The special equation is x = x3 + t3

3tx2= 1

3

x

t+ 1

3

(x

t

)−2. The suggested substitution leads

to t z = 13z−2 − 2

3z. Separating the variables and integrating yields∫

dz

z−2 − 2z= 1

3

∫dt

t. The integral

on the LHS is∫

z2 dz

1 − 2z3. It can be found by introducing the new variable u = 1 − 2z3, implying that

du = −6z2dz. The final answer is x = 3√

12 t3 + Ct .

21.4

2. (a) Because g(ξ) = 1/ξ , f (s) = s, x0 = 1, and t0 = √2, formula [21.3] yields

x∫1

ξdξ =t∫

√2

s ds,

or∣∣x1

12ξ 2 = ∣∣t√

212 s2, or 1

2x2 − 12 = 1

2 t2 − 1. Thus, the required solution is x = √t2 − 1. (Actually,

it is probably easier to find the general solution x2 − t2 = C first, and then determine the appropriateconstant.)(b) Formula [21.3] gives

∫ x

0 (ξ + 1)−2 dξ = ∫ t

0 e−2s ds, so∣∣x0 − 1/(ξ + 1) = − 1

2

∣∣t0e

−2s , implying that

1 − 1/(x + 1) = 12 (1 − e−2t ). Hence, x = 1 − e−2t

1 + e−2t.

4. Using the given identity, the separable differential equation [1] becomes

∫ (1

y+ αy�−1

1 − αy�

)dy =

∫dx

x

Integration yields ln y − (1/�) ln |1 − αy�| = ln x + C1. Multiplying both sides by � yieldsln y� − ln |1 − αy�| = ln x� + C1�, or

lny�

|1 − αy�| = ln eC1�x�, hencey�

1 − αy�= Cx� with C = ±eC1�

Putting β = 1/C and solving for y yields [2].

21.52. x = Ce−t/2 + 1

2 . The equilibrium state x∗ = 1/2 is stable. See Fig. 21.5.2.

4. dx/dt = b − ax, so∫

dx/(b − ax) = ∫ dt + A for some constant A. This implies that (−1/a) ln |b −ax| = t + A, so ln |b − ax| = −at − aA, and |b − ax| = e−at−aA = e−at e−aA. It follows that (∗)

b − ax = ±e−at−aA = ±e−aAe−at = Ce−at for C = ±e−aA. The conclusion follows by solving (∗)

for x.


64 C H A P T E R 2 1 D I F F E R E N T I A L E Q U A T I O N S

x

−1

1

t

x = 12 e−t/2 + 1

2 (C = 12 )

x = −e−t/2 + 12 (C = −1)

Figure 21.5.2

6. From x = X/N , by logarithmic differentiation, x/x = X/X − N/N . Moreover, [2] implies thatX/X = aN/N , so x/x = (a − 1)N/N = (a − 1)[α − β(1/x)]. It follows that the differential equationfor x is x = (a − 1)αx − (a − 1)β. The solution of this equation is x(t) = [x(0) − β/α]eα(a−1)t + β/α.Then [2] and x = X/N together imply that N(t) = [x(t)/A]1/(a−1), X(t) = A[N(t)]a . For 0 < a < 1,x(t) → β/α, N(t) → (β/αA)1/(a−1), and X(t) → A(β/αA)a/(a−1) as t → ∞.

21.6

2. Formula [21.12] yields

x =∫ t

0s(1 + s2)e

∫ t

s2ξ dξ ds =

∫ t

0s(1 + s2)et2−s2

ds

By substituting u = −t2, then integrating by parts, we eventually derive the solution x(t) = et2 −1− 12 t2.

It follows that limt→∞ x(t) does not exist because x(t) → ∞.

4. Multiply the given equation by x−n to obtain x−nx = Q(t)x1−n + R(t). From z = x1−n we findz = (1 − n)x−nx, so the new equation is 1

1−nz = Q(t)z + R(t), which is a linear differential equation

in the unknown function z.

6. Using the technique suggested by Problem 4 yields the solutionK = {Ce−αδ(1−b)t + αAna

0(1 − b)e(av+ε)t /(av + ε + αδ(1 − b))}1/(1−b)

21.7

2. (a) x = (x − 1)(x + 1)2. Here x = 1 is unstable, whereas x = −1 is neither stable nor unstable. (Itis stable on the right, but unstable on the left.) See Fig. 21.7.2(a). (b) No equilibrium states. See Fig.21.7.2(b). (c) The only equilibrium state x = 0 is unstable. See Fig. 21.7.2(c).

2

1

�1

�2

x

2�2 x

2

1

�1

�2

x

1�1 x

2

1

�1

�2

x

1�3 �2 x



C H A P T E R 2 1 D I F F E R E N T I A L E Q U A T I O N S 65

4. (a) x(t) = (1 + Aet )/(1 − Aet ), where A = (x0 − 1)/(x0 + 1). Another solution is x(t) = −1 for all t .For x0 < 1, x(t) → −1 as t → ∞. For x0 > 1, which occurs when 0 < A < 1, one has x(t) → ∞ ast → (− ln A)−, and x(t) → −∞ as t → (− ln A)+. For x0 = 1, x(t) ≡ 1. See Fig. 21.7.4(a) for someintegral curves. (b) x = −1 is stable; x = 1 is unstable. See Fig. 21.7.4(b).

x

t

x ≡ 1

x ≡ −1

x

1

2

x−2 −1 1 2

Figure 21.7.4(a) Figure 21.7.4(b)

21.8

2. (a) It is easy to verify by direct differentiation that u1 = et and u2 = tet both satisfy x − 2x + x = 0. Iftet = ket for all t , then t = k for all t , which is absurd. The general solution is therefore x(t) = Aet+Btet ,where A and B are arbitrary constants. (b) A particular solution of the equation is obviously u∗(t) = 3,so the general solution of the nonhomogeneous equation is x(t) = Aet + Btet + 3.

4. By direct differentiation, u1 = eat and u2 = eat/(1−α) are easily seen to be solutions. They are notproportional, so the general solution is x(t) = Aeat +Beat/(1−α), where A and B are arbitrary constants.

21.9

2. (a) x = C1et + C2e

−t − 12 sin t . Not stable. (b) x = C1e

t + C2e−t − 1

2 te−t . Not stable.(c) x = Ae5t + Bte5t + 2

75 t + 3125 . Not stable.

4. x = Ae2t + Be7t/4 + 1

14t + 15

142+ 1

65(2 sin t + 3 cos t)

6. (a) If x = tf (1/t), then x = f (1/t) − (1/t)f ′(1/t) and x = (1/t3)f ′′(1/t). But then x + t−n−2x =(1/t3)f ′′(1/t)+ t−n−2tf (1/t) = (1/t3)[f ′′(1/t)+ t−n+2f (1/t)] = 0, because when x = f (t) satisfiesx + tn−2x = 0, then f ′′(u)+un−2f (u) = 0 at u = 1/t in particular, and so f ′′(1/t)+ t−n+2f (1/t) = 0.(b) Let n = 2. Then x + tn−2x = x + x = 0, with general solution x = A sin t + B cos t . From (a), itfollows that x+t−n−2x = x+t−4x = 0, or t4x+x = 0, has the solution x = t[A sin(1/t)+B cos(1/t)].

8. For λ = γ (a − α) > 0 the solution is p(t) = C1ert + C2e

−rt − k/λ, where r = √λ; for λ = 0 the

solution is p(t) = C1t + C2 + 12kt2; for λ < 0 the solution is p(t) = C1 cos st + C2 sin st − k/λ, where

s = √−λ . In no case is the solution ever stable.

10. If x = uert , then x = ert (u + ru) and x = ert [u + 2ru + r2u]. Hence, x + ax + bx = ert [u + (2r +a)u + (r2 + ar + b)] = ert u because r = −a/2 and r satisfies the characteristic equation. We concludethat x = uert satisfies the given equation provided u = 0. Then u = At +B, and the conclusion follows.


66 A P P E N D I X B S U M S , P R O D U C T S , A N D I N D U C T I O N

Appendix B Sums, Products, and Induction

B.12. (a) 2

√0 + 2

√1 + 2

√2 + 2

√3 + 2

√4 = 2(1 + √

2 + √3 + 2) = 2(3 + √

2 + √3)

(b) (x + 0)2 + (x + 2)2 + (x + 4)2 + (x + 6)2 = 4(x2 + 6x + 14)

(c) a1ib2 + a2ib

3 + a3ib4 + · · · + anib

n+1 (d) f (x0)�x0 + f (x1)�x1 + f (x2)�x2 + · · · + f (xm)�xm

4.2 · 3 + 3 · 5 + 4 · 7

1 · 3 + 2 · 5 + 3 · 7· 100 = 6 + 15 + 28

3 + 10 + 21· 100 = 49

34· 100 ≈ 144.12

6. (a) The total number of people moving from region i. (b) The total number of people moving to region j .

B.22. (a + b)6 = a6 + 6a5b + 15a4b2 + 20a3b3 + 15a2b4 + 6ab5 + b5. (The coefficients are those in the sixth

row of Pascal’s triangle in the text.)

4. (a)

(8

3

)= 8 · 7 · 6

1 · 2 · 3= 56. Also,

(8

8 − 3

)=(

8

5

)= 56;(

8

3

)+(

8

3 + 1

)= 56 + 8 · 7 · 6 · 5

1 · 2 · 3 · 4= 126 and

(8 + 1

3 + 1

)=(

9

4

)= 9 · 8 · 7 · 6

1 · 2 · 3 · 4= 126.

(b)

(m

k

)= m(m − 1) · · · (m − k + 1)

k!= m!

(m − k)!k!=(

m

m − k

)and(

m

k

)+(

m

k + 1

)= m!

(m − k)!k!+ m!

(m − k − 1)!(k + 1)!= m!(k + 1 + m − k)

(m − k)!(k + 1)!= (m + 1)!

(m − k)!(k + 1)!=(

m + 1

k + 1

)

6.n−1∑i=0

(a + id) = a + (a + d) + · · · + [a + (n − 2)d] + [a + (n − 1)d]

= [a + (n − 1)d] + [a + (n − 2)d] + · · · + (a + d) + a = 12 [2a + (n − 1)d]n = na + n(n−1)d

2 .An alternative proof is this:n−1∑i=0

(a + id) =n−1∑i=0

a + d

n−1∑i=0

i = na + 1

2[1 + (n − 1)](n − 1) = na + n(n − 1)d

2.

B.32.∑i

j=1 aij is the sum of all the i numbers in the ith row, so in the first double sum we sum all the sums inthe m rows.

∑mi=j aij is the sum of all the m − j + 1 numbers in the j th column, so in the second double

sum we sum all the sums in the m columns.

B.4

2. (a)n∏

k=1

2k

2k − 1

2k

2k + 1=

n∏k=1

4k2

4k2 − 1= 4 · 12

4 · 12 − 1· 4 · 22

4 · 22 − 1· · · 4n2

4n2 − 1

(b)n∏

i=1

ai

bi

= a1

b1

a2

b2· · · an

bn

(c)n−1∑i=1

( n∏s=i+1

as

)bi =

n−1∑i=1

(ai+1ai+2 · · · an

)bi =

(a2a3 · · · an)b1 + (a3a4 · · · an)b2 + · · · + anbn−1


A P P E N D I X C T R I G O N O M E T R I C F U N C T I O N S 67

B.52. We prove only [B.7]; the proof of [B.6] is very similar, but slightly easier. For n = 1 the LHS and the

RHS of [B.7] are both equal to 1. Suppose [B.7] is true for n = k, so thatk∑

i=1

i3 = 13 + 23 + 33 + · · · + k3 =[k(k + 1)

2

]2

Then

k+1∑i=1

i3 = 13 + 23 + 33 + · · · + k3 + (k + 1)3 =[k(k + 1)

2

]2

+ (k + 1)3 = (k + 1)2( 14k2 + k + 1)

But this last expression is equal to(k + 1)2(k2 + 4k + 4)

4=[ (k + 1)(k + 2)

2

]2, which proves that [B.7]

is true for n = k + 1. By induction, we have proved [B.7].

4. The claim is true for n = 1. Suppose k3 + (k + 1)3 + (k + 2)3 is divisible by 9. Then (k + 1)3 + (k +2)3 + (k + 3)3 = (k + 1)3 + (k + 2)3 + k3 + 9k2 + 27k + 27 = k3 + (k + 1)3 + (k + 2)3 + 9(k2 + 3k + 3)

is divisible by 9, because the first three terms are divisible by 9 by the induction hypothesis, whereas thelast term is also obviously divisible by 9.

6. s1 = 41, s2 = 43, s3 = 47, s4 = 53, s5 = 61 are all prime. For n = 41, we have n2 − n + 41 = (41)2,which is not prime!

Appendix C Trigonometric Functions

C.12. Look at Fig. C.1 on page 866 of the text. If x is changed to −x, the point P−x will have coordinates (u, −v),

so sin(−x) = − sin x, and cos(−x) = cos x. Then by definition [C.2], tan(−x) = sin(−x)/ cos(−x) =− sin x/ cos x = − tan x.

4. cos(y − π/2) = sin y follows directly from [C.8]. Then, from the hints in the question, as well as [C.8]and the result of Problem 2, it follows that

sin(x + y) = cos[x + (y − π/2)] = cos x cos[−(y − π/2)] + sin x sin[−(y − π/2)]

= sin x cos y + cos x sin y

Also, using the result of Problem 2 again, this last equation implies that

sin(x − y) = sin x cos(−y) + cos x sin(−y) = sin x cos y − cos x sin y

6. (a) 1/2. (Either draw a figure similar to Figure C.3 on page 868, or use the formula for sin(x − y) inProblem 4.) (b) −√

3/2 (c) −√2/2 (d) −√

2/2 (e)√

3/3 (f) sin π/12 = sin(π/3 − π/4) =sin π/3 cos π/4 − cos π/3 sin π/4 = 1

4 (√

6 − √2).

8. Note that x + y = A and x − y = B imply x = 12 (A + B) and y = 1

2 (A − B). The claimed formulathen follows easily from the hint.


68 A P P E N D I X C T R I G O N O M E T R I C F U N C T I O N S

10. (a) See Fig. C.1.10(a). Period π , amplitude 1. (b) See Fig. C.1.10(b). Period 4π , amplitude 3.(c) See Fig. C.1.10(c). Period 2π/3, amplitude 2.

y

−1

1

x

π

−π

y

−3

−2

−1

1

2

3

x

−2π 4π

y

2

4

x−π π 2π

Figure C.1.10(a) Figure C.1.10(b) Figure C.1.10(c)

12. (a) y = 2 sin 14x (b) y = 2 + cos x (b) y = 2e−x/π cos x

14. The given equality is (cos x − cos y)2 + (sin x − sin y)2 = 1+1−2 cos(x −y). By expanding and using[C.7], we easily find formula [C.8].

C.2

2. y = tan x = sin x

cos xgives y ′ = cos x cos x − sin x(− sin x)

cos2 x= cos2 x + sin2 x

cos2 x= 1 + tan2 x. Since

cos2 x + sin2 x = 1, we get y ′ = 1/ cos2 x as an alternative answer.

4. (a) a sin ax (b) a sin bt + abt cos bt (c) −a cos(at + b) sin[sin(at + b)] cos{cos[sin(at + b)]}6. f ′(x) = 3(sin x − x − 1)2(cos x − 1). In the open interval I = (0, 3π/2), f ′(x) = 0 only at cos x = 1,

when x = π/2. (It is easy to see that sin x < x + 1 for all x ≥ 0 because the function defined byg(x) = x + 1 − sin x for all x ≥ 0 satisfies g(0) = 1 and g′(x) = 1 − cos x ≥ 0 for all x ≥ 0.) Thus,the only possible extreme points are 0, π/2, and 3π/2. Comparing the function values at these points,we find that f (x) has its maximum −1 at x = 0, and its minimum −(2 + 3π/2)3 at x = 3π/2. (Theextreme-value theorem ensures that extreme points do exist.)

8. Implicit differentiation yields 1 · cos y + x(− sin y)y ′ − y ′ sin x − y cos x = 0, so y ′ = cos y − y cos x

sin x + x sin y.

At (π, π/2), y ′ = 1/2, so the equation for the tangent is y = x/2.

10. (a) − cos x + C (b)∫ π/2

0 cos x dx = ∣∣π/20 sin x = sin(π/2) − 0 = 1 (c) Integrating by parts yields

I = ∫ sin2 x dx = sin x(− cos x)−∫ cos x(− cos x) dx = − sin x cos x+∫ cos2 x dx = − sin x cos x+∫(1−sin2 x) dx. Hence, I = − sin x cos x+x−I+C. Solving for I gives I = − 1

2 (sin x cos x−x)+C1.(d)∫ π

0 x cos x dx = ∣∣π0 x sin x − ∫ π

0 sin x dx = 0 + ∣∣π0 cos x = cos π − cos 0 = −2.

12. (a) π/4 (b) π/2 (c) π/6 (d) π/3. (You can read all these values off from Table C.1.)

14. Make use of [C.20] and the chain rule to derive y ′ = 1

1 +(

ex − e−x

2

)2 · ex + e−x

2. Then simplify.

C.3

2. See Fig. C.3.2.


69

4. (a) 2√

3(cos π/3 + i sin π/3) (b) cos π + i sin π (c) 4(cos 4π/3 + i sin 4π/3)(d)

√2(cos 7π/4 + i sin 7π/4)

Imaginary axis

−2i

−i

i

2i

3i

4i

Real axis−1 1 2 3 4

w = 1 + 3i

z + w = 3 + i

z = 2 − 2i

Figure C.3.2


70

TEST I (Elementary Algebra)

A certain familiarity with elementary algebra is an essential prerequisite for reading the textbook (and forunderstanding most modern economics texts). This test is designed for students and instructors to discoverwhether the students have the proper background. (In a number of countries, many beginning economicsstudents’ background in elementary algebra appears to have become much weaker during the last few years.In fact, lecturers using this test (or similar ones) have been shocked by the results, and have had to readjusttheir courses.)

At the head of each problem, immediately after the number, the relevant sections of the introductorychapters in the book are given in parentheses, followed in square brackets by the number of points for acorrect answer to each separate part of the problem. In a 20–30 minute test, any student who scores less than50 (out of 100) has serious problems with elementary algebra. Such students definitely need to review therelevant section of Chapters 1 and 2, or consult other elementary material.

The correct answers are given on a separate page following the test.

1 (1.2) [Points: (a) 2; (b) 2; (c) 3; (d) 3] Calculate/simplify:

(a)73 · 72

74(b) (5.5 − 3.5)3 (c)

(−2

5

)(−2

5

)(−2

5

)(d)

219 − 217

219 + 217

2 (1.2–1.4) [Points: (a) 2; (b) 2; (c) 2; (d) 4]

(a) If 2x2y = 5, then 4x4y2 = ? (b) 11 % of 3500 is? (c)√

132 − 122 = ?

(d) Rationalize the denominator of

√3 + √

2√3 − √

2(i.e. find a new fraction that is equal but has no square

root in the denominator).

3 (1.3) [Points: (a) 2; (b) 2; (c) 2] Expand:

(a) (x + 2y)2 (b) (2x − 3y)2 (c) (a + b)(a − b)

4 (1.3) [Points: (a) 2; (b) 3; (c) 3; (d) 4] Expand and simplify:

(a) 5a − (3a + 2b) − 2(a − 3b) (b) (x + 2)2 + (x − 2)2 − 2(x + 2)(x − 2)

(c) (1 − x)2(1 + x)2 (d) (2 − a)3

5 (1.2) [Points: 4] If the GNP of a certain country in 2000 was 8 billion dollars, write down an expressionfor the GNP 6 years later if it increases by 5% each year.

6 (1.3) [Points: (a) 3; (b) 3; (c) 4] Factorize:

(a) 5a2b + 15ab2 (b) 9 − z2 (c) p3q − 4p2q2 + 4pq3


71

7 (1.4) [Points: (a) 2; (b) 2; (c) 2] Expand and simplify to a single fraction:

(a)1

2− 1

3(b)

6a

5− a

10+ 3a

20(c)

1

2− 1

3

1

4− 1

6

8 (1.5) [Points: (a) 2; (b) 2; (c) 2; (d) 2] Calculate/simplify:

(a) 251/2 (b) (x1/2y−1/4)4 (c) 3√

27a6 (d) p1/5(p4/5 − p−1/5)

9 (2.1) [Points: (a) 2; (b) 2; (c) 2] Solve the following equations for the unknown x:

(a)3

5x = −6 (b)

1

x − 1= 3

2x + 3(c)

√3 − x = 2

10 (1.6) [Points: (a) 2; (b) 3; (c) 3] Solve the following inequalities:

(a) −3x + 2 < 5 (b)x − 1

x + 3≤ 0 (c) x3 < x

11 (2.3) [Points: (a) 3; (b) 3; (c) 3] Solve the following equations:

(a) 3x − 9x2 = 0 (b) x2 − 2x − 15 = 0 (c) 2P 2 = 2 − 3P

12 (2.4) [Points: (a) 3; (b) 4; (c) 4] Solve the following systems of equations:

(a)2x − y = 5

x + 2y = 5(b)

1.5p − 0.5q = 14

2.5p + 1.5q = 28(c)

3

p+ 3

q= 3

3

p− 1

q= 7


72

TEST II (Elementary Mathematics)

Students who now enter university or college courses in economics tend to have a wide range of math-ematical backgrounds and aptitudes. At the low end, they may have no more than a shaky command ofelementary algebra. Or, at the high end, they may already have a ready facility with calculus, thoughoften it is some years since economics students took their last formal mathematics course. Experiencesuggests therefore that right from the start of the course, it is very important that the instructor, as wellas each individual student, should get some impression of what the student knows well, what is vaguelyfamiliar, and what seems to be more or less forgotten or perhaps never learned at all.

The present test is meant to test the students’ actual knowledge of some elementary mathematicaltopics of interest to economists. The level is nevertheless more advanced than for Test I and the topicscovered are discussed in the earlier chapters of the main text. The maximum total score is 100.

1 [Points: (a) 2; (b) 2; (c) 2; (d) 2]

(a) 25 + 25 = 2x �⇒ x = (b) 3−15 + 3−15 + 3−15 = 3y �⇒ y =

(c)√

132 − 52 = (d)226 − 223

226 + 223= z

9�⇒ z =

2 [Points: (a) 1+1; (b) 2+2]

(a) Find the slopes of the following straight lines:

(i) y = − 32x + 4 (ii) 6x − 3y = 5

(b) Find the equation of the straight line that:

(i) passes through (−2, 3) and has slope −2.(ii) passes through both (a, 0) and (0, b).

3 [Points: 5] Fill in the following table and sketch the graph of y = −x2 + 2x + 4.

x −2 −1 0 1 2 3 4

y = −x2 + 2x + 4

4 [Points: 4+4] Determine the maximum/minimum points for:

(a) y = x2 − 4x + 8 (b) y = −2x2 + 16x − 14

5 [Points: 4+4] Perform the following polynomial divisions:

(a) (2x3 − 3x + 10) ÷ (x + 2) (b) (x4 + x) ÷ (x2 − 1)

6 [Points: 5] f (x) = 43x3 − 1

5x5. For what values of x is f ′(x) = 0?

7 [Points: 3+3+3+3] Sketch the graph of a function f in each of the following cases:

(a) f ′(x) > 0 and f ′′(x) > 0, (b) f ′(x) > 0 and f ′′(x) < 0

(c) f ′(x) < 0 and f ′′(x) > 0, (d) f ′(x) < 0 and f ′′(x) < 0


73

8 [Points: 3+3]

(a) The cost in dollars of extracting T tons of a mineral ore is given by C = f (T ). Give an economicinterpretation of the statement that f ′(1000) = 50.

(b) A consumer wants to buy a certain item at the lowest possible price. Let P(t) denote the lowest pricefound after searching the market for t hours. What are the likely signs of P ′(t) and P ′′(t)?

9 [Points: 2+2+2+2] Write down the general rules for differentiating the following:

(a) y = f (x) + g(x) �⇒ y ′ = (b) y = f (x)g(x) �⇒ y ′ =(c) y = f (x)/g(x) �⇒ y ′ = (d) y = f (g(x)) �⇒ y ′ =

10 [Points: 2+2+2+2+2+2+2+2] Differentiate the following functions:

(a) y = x2 (b) y = x5/5 (c) y = x

x + 1(d) y = (x2 + 5)6

(e) y = ex (f) y = ln x (g) y = 2x (h) y = xx

11 [Points: 2+2+2+2+2] Which of the following statements are correct?

(a) The rule which converts a temperature measured in degrees Fahrenheit into the same temperaturemeasured in degrees Celsius is an invertible function.

(b) A concave function always has a maximum.

(c) A differentiable function can only have an interior maximum at a stationary point for the function.

(d) If f ′(a) = 0, then a is either a local maximum point or a local minimum point.

(e) The conditions f ′(a) = 0 and f ′′(a) < 0 are necessary and sufficient for a to be a local maximumpoint for f .

12 [Points: 2+2+2+2] In each of the following cases, decide whether the given formula is correct or not:

(a)∫

x2 dx = 1

3x3 + C

(b)∫

[f (x) + g(x)] dx =∫

f (x) dx +∫

g(x) dx

(c)∫

f (x)g(x) dx =∫

f (x) dx

∫g(x) dx

(d)∫ b

a

x dx = b − a


74

Answers to TEST I

1. (a)73 · 72

74= 73+2

74= 75

74= 75−4 = 71 = 7 (b) (5.5 − 3.5)3 = 23 = 8

(c)

(−2

5

)(−2

5

)(−2

5

)= −8

125(= −0.064) (d)

219 − 217

219 + 217= 217(22 − 1)

217(22 + 1)= 3

5

2. (a) 4x4y2 = (2x2y)2 = 25 (b) 11 % of 3500 is 3500 · 11/100 = 3500 · 0.11 = 385

(c)√

132 − 122 = √(13 + 12)(13 − 12) = √

25 = 5 (or√

132 − 122 = √169 − 144, etc.)

(“√

132 − 122 = √132 − √

122 = 13 − 12 = 1” is a SERIOUS mistake.)

(d)

√3 + √

2√3 − √

2= (

√3 + √

2)(√

3 + √2)

(√

3 − √2)(

√3 + √

2)= 3 + 2

√3√

2 + 2

3 − 2= 5 + 2

√6

3. (a) (x + 2y)2 = x2 + 4xy + 4y2 (b) (2x − 3y)2 = 4x2 − 12xy + 9y2

(c) (a + b)(a − b) = a2 − b2

4. (a) 5a − (3a + 2b) − 2(a − 3b) = 5a − 3a − 2b − 2a + 6b = 4b

(b) (x + 2)2 + (x − 2)2 − 2(x + 2)(x − 2) = [(x + 2) − (x − 2)]2 = 42 = 16

(c) (1 − x)2(1 + x)2 = [(1 − x)(1 + x)]2 = (1 − x2)2 = 1 − 2x2 + x4

(d) (2 − a)3 = (2 − a)2(2 − a) = (4 − 4a + a2)(2 − a) = 8 − 4a − 8a + 4a2 + 2a2 − a3 =−a3 + 6a2 − 12a + 8

5. 8(1.05)6 billion dollars.

6. (a) 5a2b + 15ab2 = 5ab(a + 3b) (b) 9 − z2 = (3 − z)(3 + z)

(c) p3q −4p2q2 +4pq3 = pq(p2 −4pq +4q2) = pq(p−2q)2. (One point for the first equality.)

7. (a)1

2− 1

3= 3

2 · 3− 2

3 · 2= 3

6− 2

6= 1

6(b) 20 is the common denominator, so

6a

5− a

10+ 3a

20= 4 · 6a

20− 2a

20+ 3a

20= 24a − 2a + 3a

20= 25a

20= 5a

4

(c)12 − 1

314 − 1

6

=6

12 − 412

312 − 2

12

=2

121

12

= 2 (or 12 − 1

3 = 2( 14 − 1

6 ), so the ratio is 2).

8. (a) 251/2 = 5 (b) (x1/2y−1/4)4 = x(1/2)·4y(−1/4)·4 = x2y−1 (c) 3√

27a6 = 3√

27 3√

a6 = 3a2

(d) p1/5(p4/5 − p−1/5) = p1/5p4/5 − p1/5p−1/5 = p1/5+4/5 − p1/5−1/5 = p − 1

9. (a) 3x = −30, so x = −10

(b) 2x + 3 = 3(x − 1), so x = 6. (Neither denominator is 0 when x = 6.)

(c) If√

3 − x = 2, then 3 − x = 4, so x = −1. This is indeed a solution, as is easily checked.

10. (a) −3x + 2 < 5 or −3x < 3, so that x > −1. (Recall that an inequality is reversed if multipliedby a negative number.)

(b) −3 < x ≤ 1. (Use a sign diagram. Note that the fraction is undefined if x = −3.)

(c) x < −1 or 0 < x < 1. (x3 < x, or x3 − x < 0, and so x(x2 − 1) < 0, or x(x − 1)(x + 1) < 0.Then use a sign diagram.)

11. (a) 3x(1 − 3x) = 0, so x = 0 or x = 1/3 (b) x = −3, x = 5 (c) P = −2 or P = 1/2

12. (a) x = 3, y = 1 (b) p = 10, q = 2 (c) Put x = 1/p, y = 1/q. Then 3x + 3y = 3 and3x − y = 7. The solution to this system is x = 2 and y = −1. But then p = 1/2 and q = −1.


75

Answers to TEST II

1. (a) 25 + 25 = 2 · 25 = 26, so x = 6.

(b) 3−15 + 3−15 + 3−15 = 3 · 3−15 = 3−14, so y = −14.

(c)√

132 − 52 = √169 − 25 = √

144 = 12.

(d)226 − 223

226 + 223= 223(23 − 1)

223(23 + 1)= 23 − 1

23 + 1= 7

9, so z = 7.

2. (a) (i) −3/2 (ii) 2 (b) (i) y = −2x − 1 (ii) y = (−b/a)x + b

3.x −2 −1 0 1 2 3 4

y = −x2 + 2x + 4 −4 1 4 5 4 1 −4

The graph of the function is shown in the figure below.

y

x

y

x

y

−4

−3−2

−1

12

3

45

x−4 −3 −2 −1 1 2 3 4

y = −x2 + 2x + 4

4. (a) Minimum 4 for x = 2. (Follows from x2 − 4x + 8 = x2 − 4x + 22 + 8 − 22 = (x − 2)2 + 4,or by using calculus.)

(b) Maximum 18 for x = 4. (Follows from −2x2 + 16x − 14 = −2(x2 − 8x + 7) =−2(x2 − 8x + 42 + 7 − 42) = −2[(x − 4)2 − 9] = −2(x − 4)2 + 18, or by using calculus.)

5. (a)2x3 − 3x + 10

x + 2= 2x2 − 4x + 5

(b)x4 + x

x2 − 1= x2 + 1 + x + 1

x2 − 1= x2 + 1 + 1

x − 1

6. f ′(x) = 4x2 − x4 = x2(4 − x2) = x2(2 − x)(2 + x) = 0 for x = 0 and for x = ±2.

7. y

x

y = f (x)

(a)

y

x

y = f (x)

(b)

y

x

y = f (x)

(c)

y

x

y = f (x)

(d)


76

8. (a) The marginal cost is 50 when output is 1000 tons. (Or: The cost of extracting one ton more than1000 tons is approximately 50 dollars.)

(b) P ′(t) ≤ 0, because more search never leads to a higher best price, and generally leads to a lowerone. However, P ′′(t) is likely to be positive, because the longer you search the smaller gain in priceyou will obtain. (P(t) is decreasing and convex.)

9. (a) y = f (x) + g(x) ⇒ y ′ = f ′(x) + g′(x)

(b) y = f (x)g(x) ⇒ y ′ = f ′(x)g(x) + f (x)g′(x)

(c) y = f (x)/g(x) ⇒ y ′ = f ′(x)g(x) − f (x)g′(x)

(g(x))2

(d) y = f (g(x)) ⇒ y ′ = f ′(g(x))g′(x)

10. (a) y = x2 ⇒ y ′ = 2x (b) y = 15x5 ⇒ y ′ = x4 (c) y = x

x + 1⇒ y ′ = 1

(x + 1)2

(d) y = (x2 + 5)6 ⇒ y ′ = 6(x2 + 5)52x = 12x(x2 + 5)5 (e) y = ex ⇒ y ′ = ex

(f) y = ln x ⇒ y ′ = 1/x (g) y = 2x ⇒ y ′ = 2x ln 2 (h) y = xx ⇒ y ′ = xx(ln x + 1)

11. (a) Correct. (b) Wrong. (Consider y = −ex .) (c) Correct. (d) Wrong. (a could be an inflec-tion point.) (e) Wrong. (The conditions are sufficient, but not necessary. For example, f (x) =−x4 has a maximum at x = 0 and yet f ′(0) = f ′′(0) = 0.)

12. (a) Correct. (b) Correct. (c) Wrong, because ddx

(∫f (x) dx

∫g(x) dx

)= f (x)

∫g(x) dx + g(x)

∫f (x) dx �= f (x)g(x), except for very special functions f and g.

(d) Wrong, because∫ b

ax dx = ∣∣b

a12x2 = 1

2 (b2 − a2), which equals b − a only when a + b = 2 ora = b.


even number solutions

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