Chap 5 Max Min

8/4/2019 Chap 5 Max Min

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Differential Calculus Lecture Notes by Dr. M. A. Alim

Chapter 5: Maximum and Minimum of Functions

Maxima y

Function y = f(x)

Minima O x Figure 5.1: Maxima and Minima of the function y = f(x)

5.1 Definition

A function f(x) is said to have a maximum for a value a of x if f(a) is greater than any other value

that the function can have in the small neighbourhood of x = a.

Similarly a function f(x) is said to have a minimum for a value a of x if f(a) is less than any other

value that the function can have in the small neighbourhood of x = a.

Note: Maximum value of f(x) is not necessarily the greatest value. A function may have severalmaximum values i.e. a curve may have several maxima. Any of the maxima or minima is

known as local maximum or minimum. Maximum of all the maxima is the absolute maximum,

Minimum of all the minima is the absolute minimum.

Minimum value of f(x) is not necessarily the smallest value. A function may have several

minimum values i.e. a curve may have several minima.

(i) The function f(x) is maximum at x = a if f′(a) = 0 and f′′ (a) is negative.

(ii) The function f(x) is minimum at x = a if f′(a) = 0 and f′′ (a) is positive.

5.2 General conception of maximum and minimumFor the first derivative If y = f(x) is a function and

1. If dy dx

> 0

then the function f(x) is increasing

2. If dy dx

< 0

then the function f(x) is decreasing

3. If dy dx

= 0

then tangent to the function f(x) is parallel to the x-axis.



Chapter 5: Maximum and Minimum

For the second derivative

1. If

d 2

y dx

2

> 0

then the function f(x) is concave upward

2. If

d 2

y dx2

< 0

then the function f(x) is concave downward

3. If dy dx

= 0

then tangent to the function f(x) has a point of inflection.

Point of inflection is the location where the curve changes its concavity from upward to

downward or from downward to upward.

At the minimum point the function is concave upward and at the maximum point the function is

concave downward.

5.2 Working Rule

1. If the function f(x) be given, find f′(x) and equate it to zero. Solve this equation for x. Let

its roots be a

1,

a2,

a

3

……

2. Find f′′ (x) and hence find and f′′ (a



1

……

3. If f′′ (a

1

), f′′ (a

2

) f′′ (a

3

) a

3 ) is negative we have a maximum of f(x) at the

point x = a

1

. If f′′ (a1

) is positive we have a minimum of f(x) at the point x = a

1

.

Similarly for all other critical points a

2,

……

5.3 Working Rule for the exceptional cases

1. If f′′ (a

1

a

3

) = 0, find f′′′ (x) and then nor minimum at x = a

1

. If f′′′ (a

1

) f′′′ = 0; (a

find 1

). If f iv(x) f′′′ (a

and 1



) ≠ then 0, then f iv (a

there 1

). If is f neither iv (a

1

maximum

then f(x) is maximum at x = a

1

and ) is negative,

if f iv (a

1

) is positive, then f(x) is minimum at x = a

1

. 2.

If f iv (a

1

) = 0; find f v(x) and then f v (a

1

) and so on.

Example 1 Investigate the maximum and minimum values of f(x) = 2x

3

-15x

2

+ 36x + 10. Solution: Given,

f(x) = 2x3 -15x2 + 36x + 10

Then f ′ ( x ) = 6 x 2

− 30 x + 36 and f ′′ ( x ) = 12 x − 30 Now, f′(x) = 0

⇒ 6 x 2

− 30 x+ 36

=

or, x 2

− 5 x



+ 6 =

0 or, ( x -2)( x

-3)

= 0 x

=

2, 3

0

∴

At the point x = 2

f ′′ ( x ) = f ′′ (2) = 12(2) − 30 = − 6 , which is negative.

2




∴ f(x) is maximum at x = 2 and its maximum value is f(2) = 2(2)3 -15(2)2 + 36(2) + 10=38

Maximum point is (2,38)

At x = 3

f ′′ ( x ) = f ′′ (3) = 12(3) − 30 = 6 , which is positive.

∴ f(x) is minimum at x = 2 and its minimum value is f(3) = 2(3)3 -15(3)2 + 36(3) + 10=37

Minimum point is (3,37)

Example 2

If y = (x-1)4 (x+1)5 , find the values of x for which y is maximum or minimum Solution: Given,

y = (x-1)4 (x+1)5 , then

dy dx

= 4( x − 1) 3 ( x + 1) 5 + ( x − 1) 4

⋅ 5( x

+

1)

4

= ( x − 1) 3 ( x + 1) 4

(9 x

−

1)

Differentiating, we get

d 2

2 3 2 dx

2

3

2 3 2 3

4

2 3 2 3 2 4

2 2 y

= 8( x − 1) ( x + 1) (9 x − 2 x

−



1)

d dx

y

= 24( x − 1)( x + 1) (21 x − 7 x − 7 x+

1)

d dx

y

= 24( x + 1) (21 x − 7 x − 7 x + 1) + 48( x − 1)( x + 1)(21 x − 7 x − 7 x

+

1)+ 168( x − 1)( x + 1) (9 x − 2 x

−

1)

Now equating

dy dx

to zero, we get x = 1, -1 and 1/9

When x = 1,d 2 y dx 2 = 0, d 3

y dx

3

=

0

and

d 4

y dx

4

= 24(2) 2

(21 − 7 − 7 + 1)

which is positive.



So, there is a minimum at x = 1.

When x = -1,

d 2 y dx 2 = 0, d 3 y dx 3 = 0, d 4

y dx

4

=

0

and

d 5

y dx

5

≠

0

3




so, there is neither maximum nor minimum at x = -1.

When x = 1/9,

d 2

y dx

2

= 8(8/9) 2 (10/9) 3

− (10/9)

which is negative.

So, there is a maximum at x = 1/9.

Example 3

Find the maximum and minimum values of f(x) = x4 -8x3 +22x2 -24x + 5. Sketch the graph of

f(x).

Solution: Given, f(x) = x

4

-8x

3

+22x

2

-24x + 5

Then f ′ ( x ) = 4 x 3 − 24 x 2 + 44 x − 24 and f ′′ ( x ) = 12 x 2

− 48 x + 44 Now, f′(x) = 0

⇒ 4 x 3 − 24 x 2

+ 44 x

− 24

=or, x 3 − 6 x 2

+ 11 x

− 6 =

0 or, ( x -1)( x -2)( x



-3)

= 0 x

=

1, 2, 3

0

∴

At the point x = 1

f ′′ ( x ) = f ′′ (1) = 12(1) 2 − 48(1) + 44 = 8 , which is positive.

∴ f(x) is minimum at x = 1 and its minimum value is f(1) = (1)4 -8(1)3 +22(1)2 -24(1) + 5 = 1 -

8+22-24+ 5 = -4

At x = 2

f ′′ ( x ) = f ′′ (2) = 12(2) 2 − 48(2) + 44 = 48 − 96 + 44 = − 4 , which is negative.

∴ f(x) is maximum at x = 2 and its maximum value is f(2) = (2)4 -8(2)3 +22(2)2 -24(2) + 5 = 16

-64+88-48+ 5 = -3

At x = 3

f ′′ ( x ) = f ′′ (3) = 12(3) 2 − 48(3) + 44 = 108 − 144 + 44 = 8 ′

, which is positive.

∴ f(x) is minimum at x = 3 and its minimum value is

f(3) = (3)4

-8(3)

3

+22(3)

2

-24(3) + 5 = 81 -216+ 198-72+ 5 = -4

Example 4 Sum of two numbers is 12. Find when the product is maximum. Solution: Let the

numbers be x and 12 – x . so the product is x (12 – x) = 12x- x2 Now let f(x) = 12x- x

2

Then f ′ ( x ) = 12 − 2 x and f ′ ( x ) = −

2 4




Now, f′(x) = 0 ⇒ 12 − 2 x

=

0 ∴ x

=

6

At the point x = 6

f ′′ ( x ) = f ′′ (6) = − 2 , which is negative.

∴ f(x) is maximum at x = 6. So, the numbers are 6 and 12-6 = 6

The maximum value is f(6) = 6(12-6)=36

∴ The product is maximum when two numbers are equal.

Example 5 Show that the radius of the right circular cylinder of greatest curved surface, which

can be inscribed in a given cone is half that of the base of the cone.

Solution: Let h be the height and α the semi-vertical angle of the given cone. Let x be the radius

of the base of the right circular cylinder.

Then OC = OC′ = x, where O and O′ are the centers of the ends of the cylinder (see figure) In the

ΔVOC′ , VO′ = O′C′cot α = x cot α. Now let f(x) = 12x- x2

∴

The height of the cylinder, h′ = OO′ = VO -VO′ = h - x cot α.Let S be the curved surface of the cylinder, then we get

V S = 2πrh′ =

2πx(h - x cot α)

= 2πhx - 2πx2 cot α

α

∴

dS dx = 2 π h − 4 π x

cot α ; d 2

S dx

2

= − 4 π cot

α = −ve



D′

O′ C′ Equating

dS dx

x =

h tan α B O

A

As

= 0, we have 1 2

D

C

d 2

S dx

2

is negative

∴ S is maximum when x =

1 2 h tan α i.e. when x = 1 2 ( OA ) = 1

2

(radius of the base of the cone ) . (Proved)

Example 6

Find the dimension of largest rectangle, which can be inscribed in the ellipse

x 2 a 2 + b y

2 2

= 1

.

Solution: The given ellipse is

x 2 a 2 + b y

2 2

= 1

.

5




y 2 b 2 = 1 − a x 2 2 or y 2 = b

2

2 22 2

1

−

a x

or

y = ± b a

a −

x

Let 2x be the length of the rectangle

So, area of the rectangle is, A = | 2x2y| = 4xy , when x > 0, y > 0

or, A = 4b a

x a 2

− x

2 , or

dA dx =

4 a b a 2 −

2 x

2

a 2 −

x

2

,

y

P(x, y)

Now, 0

x =



a

O

y

x

and

dA dx

= ⇒

2

d 2 A dx 2

= − 16 a

b

< 0

at

x =

a

2

∴A is maximum at

x = a

2

then

y =

b

2

∴Sides of the largest rectangle are 2 x = 2 a 2 = 2a and 2 y = 2 b 2

=

2 b 4.5 Problemson Maxima and Minima

1. Discuss the maximum and minimum of f(x) = x

5

- 5x

4



+5x

3

-1. 2. Sum of two numbers is 6.

Find the numbers if the sum of their squares is minimum.

3. Find the volume of greatest right circular cone that can be inscribed in a sphere of radius a.

(Ans. h = 4a/3, r2 = 2ah- h2 and V = 1/3πr2h ) 4. Show that the volume of greatest cylinder

which can be inscribed in a cone of height h and

the semi-vertical angle α is 4/27πh3 tan2 α.

5. Show that the function

f ( x ) = x

+ 1 x

has a local maximum and a local minimum values and

that the minimum value is greater than the maximum value. Then sketch the graph of f(x). 6. A

square piece of tin 18 meter on each side is to be made into a box without top by cutting a square

from each corner and folding up the flaps to form the sides. What size corners should be cut in

order that the volume of the box be as large as possible? ( Ans. 3×3 = 9 sq-meter)

6

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