MAXIMA AND MINIMA

There are some functions whose maximum or minimum value can easily be determined in its domain.

For example consider the following function.

Example

Find the maximum and minimum value, if any, without using derivatives of the following function

1. f(x)=3x2+6x+8,x∈R2. f(x)=−|x−1|+5for all ,x∈R

3. f(x)=sin3x+4,for all x∈(π2

,−π /2¿

4. f(x)=sin(sinx), for all x∈R5. f(x)=|x+3|for all ,x∈R

Solution

1. f(x) = 3x2+6x+8, = 3x2+6x+3x+5 = 3(x+1)2+5Clearly, the mean value(x+1)2 =0 [being square].Thus maximum value of f(x) =5 at x= -1

2. f(x)=−|x−1|+5The|x−1|≥ 0c ≤ 0i .e max value of −|x−1|=0Thus the function f(x) has max value 5.

3. f(x)=sin3x+4The max value of sin3x is 1 and min value of sin3x is a -1Thus the max value of f(x) =1+4=5 and min value f(x) =-1+4=3

4. f(x)=sin(sinx)We know the max value of sinx = -1 and min value of sinx=-1Thus the max value of f(x) =1 and min value of f(x) = -1

5. Clearly modulus of a function cannot be –ve. Hence, the maximum value of f(x) = 0 at x= -3.

Exercise

Find the maximum and minimum value, if any, without using derivatives of the following function

1. f(x)=4x2-4x+4 on R2. f(X)= -(x-1)2+2 on R3. f(x)=sin 2x+5 on R

4. f(X)=16x2-16x+28 on R

ok

Maximum Let f(x) be the real function define on the interval I. then f(x)is said to have the maximum value in I, if there exists a point a in I such that

f(x)≤f(a) for all xϵ I

in such a case, the number f(a) is called the maximum value of f(x) in the interval I

and the point a is called a point of maximum value of f in the interval I.

Minimum Let f(x) be the real function define on the interval I. then f(x) is said to have the minimum value in I, if there exists a point a ∈ I such that

f(x)≥f(a) for all xϵ I

In such a case, the number f(a) is called

the minimum value of f(x) in the interval I and

the point a is called a point of minimum value of f in the interval I.

LOCAL MAXIMA AND LOCAL MINIMA

1. Local maximum A function f(x) is said to attain a local maximum at x=a if there exists a neighborhood (a-δ ,a+-δ ) of a such thatf(x)<f(a) for all x∈(a-δ ,a+δ ), x≠af(x)-f(a)<0 for all x∈(a-δ ,a+δ ), x≠aIn such a case, f(a) is called the local maximum value of f(x) at x=a.

2. Local minimum A function f(x) is said to attain a local minimum at x=a if there exists a Neighborhood (a-δ , a+-δ ) of a such thatf(x)>f(a) for all x∈(a-δ ,a+δ ), x≠af(x)-f(a)>0 for all x∈(a-δ ,a+δ ), x≠aThe value of the function at x= a i.e. f(a) is called the local minimum value of f(x) at x=a.The points at which attains either the local maximum or local minimum values are called the extreme values of f(x).

First Derivatives Test for finding the Local Extremum

Let y=f(x) be a function defined on the interval I and let f be derivable at an interior point c of I. Let f have an extreme value at x=c,

Case I. Let f have a local maximum value at x=c, then f is an increasing function in the left nbd of x=c i.e., for x slightly < c and a decreasing function f I the right nbd of x=c i.e. for x slightly > c. Also for x=c, the graph has a horizontal tangent.

Thus, f’(x) changes continuously from +ve to –ve as increases through c.

Case II. Let f have a local minimum value at x=c, then f is an increasing function in the left nbd of x=c i.e., for x slightly < c and a decreasing function f I the right nbd of x=c i.e. for x slightly > c. Also for x=c, the graph has a horizontal tangent.

Thus, f’(x) changes continuously from +ve to +ve as increases through c.

Working rule for first derivative test

Find the sign of f’(x) when x is slightly < c and when x is slightly >c.

If f’(x) changes sign from +ve to –ve as x increases through c, then f has a local maximum value at x=c.

If f’(x) changes sign from -ve to +ve as x increases through c, then f has a local minimum value at x=c.

Second Derivative Test for Finding Local Maxima and Local Minima

Suppose a quantity y depends on the another quantity x in a manner shown in fig. it becomes maximum at x1 and minimum at x2. At these points the tangent to the curve is parallel to the x-axis and hence its slope is tan0 = 0. Thus, at a maxima or a minima slope.

dydx

=0

Maxima and minima by using second derivative test

Just before the maximum the slope is positive, at the minimum it is zero and just after the

maximum it is negative. Thus dydx decreases at a maximum and hence the rate of change of

dydx is

negative at a maximum i.e. ddx ( dy

dx ) is the rate of change in slope. It is written as d2 ydx2 .

Condition for maxima are (a) dydx

=0 (b) d2 y

dx2 <0

Minima

Similarly, at a minimum the slope changes from negative to positive.

Hence with increases of x. the slope is increasing that means the rate of change of slope w.r.t x is positive,

Hence ddx ( dy

dx )>0

Conditions for minima are (a) dydx

=0 (b) d2 y

dx2 >0

Working Rule:

1. Find f’(x).2. Solve f’(x) = 0 within the domain to get critical point let one of the value of x = c.3. Calculate f’’(x) at x = c.4. If f’’(c) > 0 then, f(x) is minimum at x = c, and if f’’(c) < 0, then f(x) is maximum at x= c.

Ex. Find maximum or minimum values of the functions:

(A) y=25x2 + 5 – 10x (B) y= 9 – (x-3)2

Solution (A) For maximum and minimum value, we can put dydx

=0

Or dydx

=50 x−10=0∴ x=15

Further, d2 y

dx2 =50

ord2 ydx2 has positive value at x=1

5 . Therefore, y has minimum value at x=15 . Therefore y has

minimum value at x=15 . Substituting x=1

5 in the given equation, we get

ymin=25( 15 )

2

+5−10( 15 )=4

(B) y= 9 – (x-3)2 = 9 – (x)2-9+6xor, y = 6x - x2

∴ dydx

=6−2x

For minimum or maximum value of y we will substitute dydx

=0

Or 6-2x=0x=3To check whether value of y is maximum or minimum at x=3 we will have to check whether d2 ydx2 is positive or negative.

d2 ydx2 =−2

Or, d2 y

dx2 is negative at x=3. Hence, value of y is maximum value of y is,

ymax=9−(3−3 )2=9.

Maximum and minimum value in closed interval

Algorithm : Let y=f(x) be the function defined on [a, b].

Step1 find dy/dx=f’(x).

Step 2 Put f’(x) =0 and find values of x. Let c1, c2, …… cn be the values of x.

Step 3 Take the maximum and minimum values out of the values f(a), f(c1), f(c2),……… f(cn), f(b)

The maximum and minimum values obtained in step III are respectively the largest or absolute maximum and the smallest or absolute minimum values of the function.

Ex:1 The maximum and minimum values of f(x) = 2x3 – 24x +107 in the interval [1, 3].

Solution: We have f(x) = 2x3 – 24x +107

f’(x) = 6x2-24

now, f’(x) =0 => 6x2-24=0 => x=± 2

But x= -2 ∉ [1 , 3 ] . so , x=2is the only stationary point .

Now, f(1)=2-24+107=85, f(2)=2(2)3 -24(2)+107=75.

And f(3)=2(3)3 -24*3+107=89.

Hence, the maximum value of f(x) is 89 which it attains at x=3 and the minimum value is 75 which is attained at x=2.

Ex2: The maximum and minimum values of f(x) =x+sin2x in the interval [0, 2π].

Solution: We have f(x) =x+sin2x.

So, f’(x) = 1+2cos2x.

For stationary points, we have

f’(x)=0 => 1+2cos2x=0 => cos2x= -12

2x=2 π3 or, 2x =

4 π3

X=π3 or x=

2 π3

Now, f(0)=0+sin0=0

f( π3

¿=2 π3

+sin π3=2 π

3+ √3

2

f( 2π3

¿=2π3

+sin 4 π3

=2 π3

−√32

∧¿f(2π ¿=2 π+sin 4 π=2 π+0=2 π .

Of these values, the maximum value is 2 π and the minimum value is 0. Thus the maximum value of f(x) is 2 πand the minimum value is 0.

Ex3: show that f(x) =sinx (1+cosx) is maximum at x=π3

∈theinterval [0 , π ]

Solution we have f(x) =sinx (1+cosx)

f’(x) cos x(1+cos x)-sin2x = cos x +cos2x –(1-cos2x) =2cos2x +cos x-1=2cosx-1(cosx+1)

For stationary values, we have F’(x)=0=>(2cosx-1)(cosx+1)=0=> cosx =1/2 or cos x =-1

x=π3 or x=π.

Now f(0), f(π3

¿=sin π3 (1+cos π

3 )=3√34

∧f (π )=0

Of these values, the maximum value is 3√34

. Hence f(x) attains the maximum value 3√34

at x=π3

.

Ex 4: Find both the maximum and minimum values of f(x) =3x4 -8x3 + 12 x2-48x+1 on interval [1, 4]

Solution: f(x) =3x4 -8x3 + 12 x2-48x+1then

f’(x) =12x3-24x2+24x-48 and f’’(x)=36x2-48x+24

Now, f’(x) =0 => 12x3-24x2+24x-48=0

x3-2x2+2x-4=0 x2(x-2) +2(x-2)=0 (x-2)(x2+2)=0 X=2

For x=2, f”(x)=36x2 -48(2)+24=72>0So, x=2 is sa point of local minimum.F(2) =-59 f(1) =- 40 and f(4) =257So , the minimum and maximum values of f(x) on [1, 4] are -59 and 257 respectively.

Application of Maxima and Minima to problems

Working rule

(i) In order to illustrate the problem, draw a diagram, if possible. Distinguish clearly between the variable and constants.

(ii) If y is the quantity to be maximized or minimized, express y in terms of a single independent variable with the help of given data.

(iii) Get dy/dx and d2y/dx2, then equate dy/dx=0 get the value of x.(iv) Determine the sine of d2x/dx2 at x and proceed.

Example1: Show that the area of a rectangle of given perimeter is maximum when the rectangle of a square.

Solution: Perimeter is given here. Area is to be maximized.

Let the given perimeter be p. Let the dimension of rectangle be x and y, then

2x+ 2y = p (1)

Let A denote the area of the triangle, then

A = xy [as stated in working rule]

To express A in term of one independent variable, say x, we have

2y=p-2x [from (1)]

y=1/2(p-2x)

since x>0, y>0, ∴0<x<p/2

A=x.1/2(p-2x)=1/2(px-2x2)

dAdx

=12

( px−2 x2 )

d2 Ad x2 =

12

(−4 )=−2

For max, or min, dAdx

=0

½(p-4x)=0 X=p/4

Also d2 Ad x2 ]

x=p /4

=−2<0

A is maximum when x=p/4 and ∴ y=12 ( p−2. p

4 )= p4

.

Area is maximum when x=y=p/4 Area is maximum when the rectangle is square.

Example 2: Show that the rectangle of maximum area that can be inscribed in a circle is a square.

Solution Here circle is given

Hence its radius ‘a’ is to be consider as constant.

Let x be the length and y be the breadth of rectangle.

Area = xy

To convert Area in one independent variable we use the formula.

X2+y2=(2a)2

x2+y2=4a2

y2=4a2-x2

Now A=xy=x √4a2−x2

For the sake of simplicity we consider Z = A2 = x2(4a2 – x2)=4a2x2- x4

Naturally when Z is max A is also max. [To avoid differentiation of square root we consider Z = A2]

(As stated in the working rule) differentiate with respect to x

dZdx

=8 a2 x−4 x3

again diff . , we get

d2 Zd x2 =8 a2−12 x2

for max ¿min

dZdx

=0

8a2−12x2=0

4 x (2a2−x2 )=0 , x≠ 0

2a2 – x2 =0 X2 =2a2 X=√2a

d2 Zd x2 =8a2−12 12× 2 a2=−16 a2<0

∴Z is max i . e A is max

when x2=2a2 2 x2=4 a2

= x2+y2 [from eqn (1)] x2 = y2 x = y.

Hence rectangle is a square.(proved)

Example 3 Show that a closed right circular cylinder of given total surface area and maximum volume is such that its height is equal to the diameter of its base.

Solution: Let h be the height and r, the radius of the circular base.

Let S be the total surface area, then

S = 2 πrh+2 π r2=¿h=S−2 π r2

2 πr(1)

[To express V in term of one independent variable]

Volume V=π r2h=π r2[ S−2 π r2

2 πr ]=r [ S−2 π r 2

2 πr ]=12(rS−2 π r 3)

Now as stated in working rule get dVdr

=12

. (S−6 π r2 ) ,

d2Vd r2 =1

2(−12 πr )=−6 πr

For max or min, dVdr =0.

S=6 π r2

r=√ S6 π

Also d2V

d r2 =−6 πr<0

∴V is maximumwhenr=√ S6 π

∧¿

∴h=6 π r2−2π r2

2πr=2 r (∵S=6 π r2)

Hence, the volume of the cylinder is maximum when its height is equal to the diameter of its base.

Ex 4: Find the maximum and minimum values of f(x) =x1x and hence deduce that eπ >πe .

Solution: y=x1x

log y=1x

log x

Differentiate both side w.r.t to x

1y

dydx

=x . 1

x−logx

x2

dydx

=y [1−logx ]

x2

d2 ydx2 =dy

dx[ 1−logx ]

x2 + yx2(0−1

x )−(1−logx )2 x

x4

¿ dydx

[1−logx ]x2 + y [−1−2 (1−logx ) ]

x3

For max or min dydx

=0=>y [ 1−logx ]

x2 =0

y (1-log x) =0

1-log x=0

log x = 1

x=e

d2 ydx2 =0+

e1e [−1−0 ]

e3 [as at x=e , dydx

=0∧loge=1] ¿−ve

∴ y is max when x=e∧max value of y=e1e

Now, it is known that x1x is greater than any value of x except x=e.

Hence, the value of x1x is greater at x=e than at x=π

i.e. e1e>π

1π .

=> eπ >πe .

Ex5. Show that the maximum value of (1/x)x is e1/e.

Solution: let y=( 1x )

x

=x−x=¿ logy=−xlogx

1y

dydx

=−(1+ logx )=¿ dydx

=− y (1+logx )

¿ , d2 ydx2 =−dy

dx(1+logx )− y

x= y (1+logx )2− y

x

d2 ydx2

=x− x (1+logx )2− x−x

x=x− x (1+logx )2−x− x−1

For the maximum and minimum, we must have

dydx

=0=¿− y (1+logx )=0=¿1+logx=0=¿logx=−1

¿>x=e−1=1e [∵log e A=B=¿ A=eB ]

Also ,( d2 ydx2 )x=1

e

=−( 1e )

−1e (1+log 1

e )2

−( 1e )

−1e

−1

¿ (e−1 )−1e (1−loge )2−( e−1)

−1e −1

¿ (e )1e (1−1 )2−(e )

−1e +1

<0

So, x=1/e is a point of local maximum. The local maximum value of y is given by y=(e)1/e.

Ex 6. Show that the height of the right circular cylinder of the max volume that can be inscribed in a

given right circular cone of height h is 13 h.

Solution: Let the radius of the cylinder be x and height be y.

The volume,V=πx2 y

To express V in terms of single variable we get a relation that ∆ ABC∼∆ ADE

ABOA

=BCDE

h− yh

= xR

hR− yR=hx

hR−hx= yR=¿ y=hR−hxR -------------------------------------- (1)

V=πx2 y=πx2 [ hR−hx ]R

=πR

[hR x2−h x3 ]

dVdx

= πR

[hR 2 x−3 hx2 ]

d2Vdx2 = π

R[ 2 hR−6hx ]

For max or min dVdx

=0=¿ πR

[hR 2x−3h x2 ]=0

[hR 2 x=3h x2 ]

[ 2R=3 x ]

x=2 R3

.

d2 ydx2 at x=2 R

3= π

R [2hR−6 h 2 R3 ]=2 hR π

R=−2 hπ<0

∴V is max when x=2 R3

¿eqn (1 ) y=hR−h 2 R

3R

= hR3 R

=h3

.( proved)

Ex 7: Show that the semi vertical angle of a cone of maximum volume and given slant height is tan−1 √2

Sol: let radius of cone be x and height be y. then volume V=13

πx2 y

To express V in term of single variable we use the formula

x2+ y2=l2 ------------------------------------------- (1)

∴ x2=l2− y2

∴V =13

π (l2− y2) y

¿ 13

π (l2 y− y3)

dVdy

=13

π (l2 y− y3)

d2Vdx2 =1

3π (−6 y )=−2 πy

3<0

∴ v is max

Provided dVdy

=0

l2=3 y2

From eqn (1) x2+ y2=3 y2=> x2=2 y2

x2

y2=2=¿ xy=√2

tan∝=√2=¿∝=tan−1√2 .

Ex 8: Prove that the volume of the largest cone that can be inscribed in a sphere of radius R is 8/27 of the volume of the sphere.

Solution: Let radius of cone be x and height be y. then volume V=13

πx2 y

To express V in term of single variable we use the formula

∵ x2+( y−r )2=R2

∵ x2=R2−( y−r )2 ------------------------- (1)

V=13

π ¿¿

=π3 [(R¿¿2− y2+2 yR−R2)¿ y ]

=π3

[− y3+2 y2 R ] ------------------------- (2)

dVdy

=π3

[−3 y2+4 yR ]

d2Vdx2 = π

3[−6 y+4 R ]

Provided dVdy

=0

π3

[−3 y2+4 yR ]=0

4 yR=3 y2

R¿ 3 y2

4 y=

3 y4

d2Vdx2 = π

3 [−6 43

R+4 R] ∴ v is max when y=4

3R

Max volume from eqn (2)

=π3 [−64 R3

27+

16 R3

9 ]=π

3 [−64 R3+96 R3

27 ]¿ 4

3π R3( 8

27 ) . =

827 volume of the given sphere.

Ex: 9 A wire of length 28 m is to be cut into two pieces. One of the pieces is to be made into a square and the other into a circle. What should be the length of the two pieces so that the combined area of the square and the circle is minimum?

Solution: let x and 28-x be the length the two pieces by which square and circle are

Area of square =( x4 )

2

= x2

16.

Area of circle formed from the wire of length 28-x=π ( 28−x2 π )

2

[2 πr=28−x ]

¿ (28−x )2

4 π

Combined Area= x2

16+

(28−x )2

4 π

dAdx

= x8−2(28−x )

4 π

d2 Adx2 =1

8− 1

2π

For the point of max or min

dAdx

=0=¿ x8− (28−x )

2 π=0

=x= 1124+π

Hence the volume is minimum at x=1124+π

Exercise 11. Determine two positive numbers whose sum is 15 and the sum of whose square is minimum.2. Find the number whose sum is 15 and the square of one multiplied by the cube the other is

maximum.3. Prove that the right circular cone of maximum volume which can be inscribed in a sphere of

radius a has its altitude equal to 4/3 a.4. Determine the points on the curve y=1/4 x2 which are nearest to the point (0, 5).5. Of all the rectangles, each of which has perimeter 40 cm, find the one having maximum area.

Also find that area.

6. How should a wire 20 cm long be divided into two parts, if one part is to be bent into a circle, the other is to be bent into a square and the two plane figures are to have areas the sum of which is minimum?

7. Show that a right-circular cylinder of a given volume, open at the top, has minimum total surface area, provided its height is equal to the radius of its base.

Exercise 2

1. Find the two number whose sum is 15 and the square of one multiplied by the cube of the other is maximum.

2. Find the two positive numbers whose sum is 15 and the sum of whose square is minimum.3. If the sum of the lengths of he hypotenuse and another side of a right-angled triangle is given,

show that the area of the triangle is maximum when the angle between them is π3

.

4. Find the area of the largest isosceles triangle having perimeter is 18 metres. 5. A window is in the form of a rectangle, surmounted by asemi-circle. If the perimeter be 30

metres, find the dimension so that the greatest possible amount of light may be admitted.6. Find the altitude of the right circular cylinder of maximum volume that can be inscribed in a

sphere of radius r.7. If 48 square meters of sheet metal are to be used in the construction of an open tank with a square

base, find the dimension so that capacity is the greatest possible.8. A rectangular sheet of tin 45 cm by 24 cm is to be made into a box without top, by cutting off

equal from the corner and folding up the flaps. What should be the side of the square to be cut off so that the volume of the box is maximum possible?

9. Determine the points on the curve y14

x2 which are nearest to the point (0, 5).

10. Of all the rectangle, each of which has perimeter 40 cm, find the one having maximum area. Also find that area.