U.A.E. University, Department of Mathematical SciencesCalculus for Business and Economics, Final Exam, Spring 2008

Name: ID No:

Instructor's Name & Section: Score: /40

Part I. Di�erentiation { Total 10 Points

1. (2 points) Find the derivative of the function. (Simpli�cation is NOT required.)

(1) (1 point) f(x) = (8x� 9)�3.

Solution. By the Chain Rule , we have

f 0(x) = �3(8x� 9)�3�1(8x� 9)0 = �3(8x� 9)�4(8) = �24(8x� 9)�4: X

(2) (1 point) f(x) = ln�7x3 � x2

�+ e�x2.

Solution. By the Chain Rule , we have

f 0(x) =(7x3 � x2)07x3 � x2 + e�x2(�x2)0 = 21x2 � 2x

7x3 � x2 + e�x2(�2x) =21x� 2x(7x� 1)

� 2xe�x2: X

2. (2 points) Consider the equation 5y2 � 8x4 + 3 = 0.

(1) (1 point) Use the implicit differentiation to �nd the derivativedydx

from the given equation.

Solution. We di�erentiate the whole equation with respect to x:

ddx

�5y2 � 8x4 + 3

�=

ddx

(0) ;ddx

�5y2

�� ddx

�8x4

�+

ddx

(3) =ddx

(0)

10ydydx� 32x3 + 0 = 0; 10y

dydx

= 32x3;dydx

=32x3

10y=

16x3

5y: X

(2) (1 point) Evaluatedydx

at the point (x; y) = (1; 1).

Solution. Putting (x; y) = (1; 1) into the result of (1) above, we get

dydx

�����(x;y)=(1;1)

=16x3

5y

�����(x;y)=(1;1)

=165: X

1

3. (3 points) Consider the function f(x) = 3x2 � 24x+ 6.

(1) (2 points) Find all the critical values of f(x).

Solution. The critical value comes from the �rst derivative of f(x):

f 0(x) = (3x2 � 24x+ 6)0 = 6x� 24 = 6(x� 4) = 0 at x = 4:

Hence, f(x) has only one critical value x = 4. X

(2) (1 point) Determine the interval where f(x) is increasing and the interval where f(x) is decreasing.

Solution. We use the following sign chart:

x Less Than 4 4 Bigger Than 4

Sign of f 0(x) � f 0(4) = 0 +

Graph of f(x) Decreases f(4) = �42 Increases

From this chart, we conclude that f(x) is increasing on (4;1) and decreasing on (�1; 4). X

4. (3 points) For the given functions, f(u) = u5 and g(x) = 2� 3x2, use the Chain Rule to compute the

derivative of (f � g) (x), i.e., the derivative of f [g(x)].

Solution. The Chain Rule implies

(f [g(x)])0 = f 0[g(x)]g0(x):

Since f 0(u) = (u5)0 = 5u4 and g0(x) = (2� 3x2)0 = �6x, so we have

(f [g(x)])0 = f 0[g(x)]g0(x) = 5[g(x)]4(�6x) = �30x[g(x)]4 = �30x(2� 3x2)4: X

2

Part II. Applications { Total 12 Points

1. (3 points) A company estimates that it will sell N(t) hair dryers after spending t thousand dollars on

advertising. Explicitly N(t) is given as follows:

N(t) = �3t3 + 450t2 � 21600t+ 1100; 40 � t � 60:

Find the interval where the rate of sales is increasing, i.e., N 0(t) is increasing?

Solution. To determine the increment of a function f(x), we look for the sign of the derivative of f(x). In

this problem, we want to determine the increment of N 0(t), so we use the sign of the derivative of N 0(t), in

other words, the sign of N 00(t).

N 0(t) = �9t2 + 900t� 21600;

N 00(t) = �18t+ 900 = �18(t� 50) = 0 at t = 50:

We use the following sign chart (Be careful! The domain of N(t) is 40 � t � 60.):

t Between 40 and 50 50 Between 50 and 60

Sign of N 00(t) + N 00(50) = 0 �Graph of N 0(t) Increases Decreases

From this chart, we conclude that N 0(t) is increasing on (40; 50) and decreasing on (50; 60). X

2. (2 points) How long will it take for 8400 Dhs to grow to 14600 Dhs at an interest rate of 9.4 percent if

the interest is compounded continuously?

Solution. The Continuously–Compounded–Interest formula implies

14600 = 8400e0:094t; i:e:; e0:094t =146008400

=7342:

When we take the natural logarithmic function on both sides of the equation, we get

ln e0:094t = ln�73

42

�; i:e:; 0:094t ln e = ln

�7342

�; i:e:; t =

10:094

ln�73

42

�� 5:88074 (years);

where ln e = 1 is used. X

3

3. (7 points) The marketing research department for a company that manufactured and sells memory chips

for microcomputers established the following price{demand, revenue and const functions, respectively:

p(x) = 75� 3x; R(x) = x(75� 3x); C(x) = 125 + 16x;

where p(x) represents the whole sale price in Dirhams at which x millions chips can be sold. All functions

have domain 1 � x � 20.

(1) (1 point) Which one of (A) and (B) below is a more plausible graph of the revenue function R(x)?

10 15 20 25

-400

-300

-200

-100

10 15 20 25

100

200

300

400

(A)

(B)

Solution. The revenue function R(x) = x(75� 3x) = �3x2 + 75x is a quadratic function with the negative

constant �3 in front of x2. Hence, the shape of the graph should be concave downward. One can use the

second derivative and the Second Derivative Test :

R0(x) = �6x+ 75; R00(x) = �6 < 0;

which implies that R(x) is concave downward by the Second Derivative Test . For these reasons, the

graph (B) is better than (A). X

(2) (3 points) Find the number of chips that will produce the maximum revenue. What is the maximum

revenue?

Solution. The revenue function is a quadratic function, which has the maximum/minimum value at the

vertex. So we �nd the vertex either by completing the square or by using the derivative. Let us use the

latter method:

R(x) = x(75� 3x) = �3x2 + 75x; R0(x) = �6x+ 75 = �3(2x� 25) = 0 at x =252

= 12:5:

4

So R(x) has the critical value x = 25=2 at which R(x) has the value

R�25

2

�=

18754� 468:75:

Thus, the quadratic function R(x) has the vertex�25

2;1875

4

�� (12:5; 468:75) and the vertex form

R(x) = �3x2 + 75x = �3�x� 25

2

�2+

18754

:

Therefore, the number of chips producing the maximum revenue is x = 25=2 and the maximum revenue is

R (25=2) = 468:75. X

(3) (1 point) What is the whole sale price per chip that produces the maximum revenue?

Solution. From the result in (2) above, the maximum revenue occurs at x = 25=2. The whole sale price

per chip is obtained by the price–demand function p(x) = 75� 3x and thus,

p�25

2

�= 75� 3

�252

�=

752

= 37:5

gives the whole sale price per chip producing the maximum revenue. X

(4) (2 points) Find the break even points.

Solution. Break{even points occur when R(x) = C(x). So we solve the equation R(x)� C(x) = 0 for x.

0 = R(x)� C(x) = �3x2 + 75x� (16x+ 125) = �3x2 + 59x� 125; i:e:; 3x2 � 59x+ 125 = 0:

The Quadratic Formula implies that the equation has the roots

x =59�p1981

6� 2:41526 or 17:2514:

Therefore, we have break{even points at x � 2:41526 and x � 17:2514. X

5

Part III. Limits { Total 4 Points

(4 points) Use L’Hopital’s Rule to �nd the limit: limx!1

x3 � 1lnx

.

Solution. As x goes to 1, the denominator lnx goes to ln 1 = 0 and the numerator x3�1 goes to 13�1 = 0.

That is, we have the form 0=0. So by L’Hopital’s Rule , we deduce

limx!1

x3 � 1lnx

= limx!1

(x3 � 1)0(lnx)0 = lim

x!1

3x2

1=x= lim

x!1

�3x3

�= 3: X

Part IV. Integration { Total 8 Points

1. (4 points) Compute the integral.

(1) (2 points)Z

5�t2 � 5t� 2

�dt.

Solution. A simple computation showsZ5�t2 � 5t� 2

�dt = 5

Z �t2 � 5t� 2

�dt = 5

�Zt2 dt� 5

Zt dt�

Z2 dt

�= 5

"t3

3� 5t2

2� 2t+ C

#= 5

"t3

3� 5t2

2� 2t

#+D;

where C and D are constants of integration. X(2) (2 points)

Z �2

x� 7ex

!dx.

Solution. A simple computation showsZ �2

x� 7ex

!dx = �2

Z 1xdx� 7

Zex dx = �2 ln jxj � 7ex + C;

where C is the constant of integration. X2. (4 points) Use the Integration–by–Substitution formula to �nd the integral:

Z8x7

�x8 + 4

�dx.

Solution. Let u = x8 + 4. Then we get

dudx

=ddx

�x8 + 4

�= 8x7; i:e:; du = 8x7dx:

Thus, the given integral becomesZ8x7

�x8 + 4

�dx =

Zudu =

u2

2+ C =

(x8 + 4)2

2+ C;

where C is the constant of integration. X

6

Part V. Matrices and Linear Systems { Total 6 Points

1. (2 points) For matrices A and B given by

A =

0@ 1 3

5 7

1A ; B =

0@ 2 4

6 8

1A ;compute the product AB.

Solution. By the de�nition of the matrix multiplication, we get

AB =

0@ 1 3

5 7

1A0@ 2 4

6 8

1A =

0@ 1(2) + 3(6) 1(4) + 3(8)

5(2) + 7(6) 5(4) + 7(8)

1A =

0@ 20 28

52 76

1A : X

2. (4 points) Consider the system: 8<: 2x+ 3y = 6

4x+ 7y = 8(F)

When we introduce matrices A, X and C de�ned by

A =

0@ 2 3

4 7

1A ; X =

0@ x

y

1A ; C =

0@ 6

8

1A ;the given system (F) can be expressed as0@ 2 3

4 7

1A0@ x

y

1A =

0@ 6

8

1A ; i:e:; AX = C:

Multiplying by the inverse of A on both sides of the equation, we deduce

A�1AX = A�1C; i:e:; IX = A�1C; i:e:; X = A�1C;

where I is the 2 � 2 identity matrix and A�1 is the inverse matrix of A. Find the solution of the system

(F), x and y, by computing A�1C.

Solution. By the de�nition of the inverse matrix, we have

A�1 =1

2(7)� 3(4)

0@ 7 �3

�4 2

1A =12

0@ 7 �3

�4 2

1A =

0@ 72 �3

2

�2 1

1A :Putting this into X = A�1C, we get0@ x

y

1A = X = A�1C =

0@ 72 �3

2

�2 1

1A0@ 6

8

1A =

0@ 72(6)� 3

2(8)

�2(6) + 1(8)

1A =

0@ 9

�4

1A :Therefore, we deduce the solution of the system (F), x = 9 and y = �4. X

7