# Concept of Maxima and Minima

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#### Example – 01:

• Find the maximum and minimum values of x3 – 12x – 5
• Solution:

Let ƒ(x) = x3 – 12x – 5 ………….. (1)

Differentiating equation (1) w.r.t. x

ƒ’(x) = 3x2 – 12 ………….. (2)

Differentiating equation (2) w.r.t. x

ƒ’’(x) = 6x ………….. (3)

For maximum or minimum value ƒ’(x) = 0

3x2 – 12 = 0

∴   3(x2 – 4) = 0

∴   3(x + 2)(x – 2) = 0

∴   x + 2 = 0 or/and x – 2 = 0

∴   x = – 2 or/and x = 2

Let us consider x = 2

ƒ’’(2) = 6 x 2 = 12 > 0

Hence the function is increasing at x = 2 and has minimum value at x = 2

Substituting x = 2 in equation (1)

Minimum value = ƒ(2) = (2)3 – 12(2) – 5 = 8 – 24 – 5 = – 21

Thus point of minimum is (2, -21)

Let us consider x = – 2

ƒ’’(-2) = 6 x (- 2) = – 12 < 0

Hence the function is decreasing at x = – 2 and has maximum value at x = -2

Substituting x = – 2 in equation (1)

Maximum value = ƒ(-2) = (-2)3 – 12(-2) – 5 = – 8 + 24 – 5 = 11

Thus point of maximum is (- 2, 11)

Ans: The maximum value is 11 at x = -2 and minimum value is – 21 at x = 2

#### Example – 02:

• Find the maximum and minimum values of x3 – 9x2 + 24 x
• Solution:

Let ƒ(x) = x3 – 9x2 + 24 x ………….. (1)

Differentiating equation (1) w.r.t. x

ƒ’(x) = 3x2 –  18x + 24 ………….. (2)

Differentiating equation (2) w.r.t. x

ƒ’’(x) = 6x – 18 ………….. (3)

For maximum or minimum value ƒ’(x) = 0

3x2 –  18x + 24 = 0

∴   x2 –  6x + 8 = 0

∴   (x – 4)(x – 2) = 0

∴   x – 4 = 0 or/and x – 2 = 0

∴   x = 4 or/and x = 2

Let us consider x = 4

ƒ’’(4) = 6 x 4 – 18 = 24 – 18 = 6 > 0

Hence the function is increasing at x = 4 and has minimum value at x = 4

Substituting x = 4 in equation (1)

Minimum value = ƒ(4) = (4)3 – 9(4)2 + 24(4) = 64 – 144 + 96 = 16

Thus point of minimum is (2, -21)

Let us consider x = – 2

ƒ’’(-2) = 6 x (- 2) – 18 = -12 – 18 = –  30 < 0

Hence the function is decreasing at x = 2 and has maximum value at x = 2

Substituting x = 2 in equation (1)

Maximum value = ƒ(2) = (2)3 – 9(2)2 + 24(2) = 8 – 36 + 48 = 20

Thus point of maximum is (2, 20)

Ans: The maximum value is 20 at x = 2 and minimum value is 16 at x = 4

#### Example – 03:

• Find the maximum and minimum values of 2x3 – 3x2 – 36x + 10
• Solution:

Let ƒ(x) = 2x3 – 3x2 – 36x + 10 ………….. (1)

Differentiating equation (1) w.r.t. x

ƒ’(x) = 6x2 –  6x – 36 ………….. (2)

Differentiating equation (2) w.r.t. x

ƒ’’(x) = 12x – 6 ………….. (3)

For maximum or minimum value ƒ’(x) = 0

6x2 –  6x – 36 = 0

∴   x2 –  x – 6 = 0

∴   (x – 3)(x + 2) = 0

∴   x – 3 = 0 or/and x + 2 = 0

∴   x = 3 or/and x = – 2

Let us consider x = 3

ƒ’’(3) = 12(3) – 6 = 36 – 6 = 30 > 0

Hence the function is increasing at x = 3 and has minimum value at x = 3

Substituting x = 3 in equation (1)

Minimum value = ƒ(3) = 2(3)3 – 3(3)2 – 36(3) + 10 = 54 – 27 – 108 = -71

Thus point of minimum is (3, -71)

Let us consider x = – 2

ƒ’’(2) = 12(-2) – 6 = – 24 – 6 = – 30 < 0

Hence the function is decreasing at x = – 2 and has maximum value at x = – 2

Substituting x = – 2 in equation (1)

Maximum value = ƒ(-2) = 2(-2)3 – 3(-2)2 – 36(-2) + 10 = – 16 – 12 + 72 + 10 = 54

Thus point of maximum is (-2, 54)

Ans: The maximum value is 54 at x = – 2 and minimum value is – 71 at x = 3

#### Example – 04

• Find the maximum and minimum values of x3 + 3x2 – 2
• Solution:

Let ƒ(x) = x3 + 3x2 – 2  ………….. (1)

Differentiating equation (1) w.r.t. x

ƒ’(x) = 3x2 +  6x  ………….. (2)

Differentiating equation (2) w.r.t. x

ƒ’’(x) = 6x + 6 ………….. (3)

For maximum or minimum value ƒ’(x) = 0

3x2 +  6x = 0

∴   3x(x  +  2) = 0

∴   x  = 0 or/and x + 2 = 0

∴   x = 0 or/and x = – 2

Let us consider x = 0

ƒ’’(-2) = 6(0) + 6 = 6  > 0

Hence the function is increasing at x = 0 and has minimum value at x = 0

Substituting x = 0 in equation (1)

Minimum value = ƒ(0) = (0)3 + 3(0)2 – 2 = 0 + 0 – 2 = -2

Thus point of minimum is (0, -2)

Let us consider x = – 2

ƒ’’(-2) = 6(-2) + 6 = -12 + 6 = – 6 < 0

Hence the function is decreasing at x = – 2 and has maximum value at x = – 2

Substituting x = – 2 in equation (1)

Maximum value = ƒ(-2) = (-2)3 + 3(-2)2 – 2 = – 8 + 12 – 2 = 2

Thus point of maximum is (-2, 2)

Ans: The maximum value is 2 at x = – 2 and minimum value is – 2 at x = 0

#### Example – 05:

• Find the maximum and minimum values of 3x3 – 9x2 – 27x + 15
• Solution:

Let ƒ(x) = 3x3 – 9x2 – 27x + 15  ………….. (1)

Differentiating equation (1) w.r.t. x

ƒ’(x) = 9x2 –  18x – 27 ………….. (2)

Differentiating equation (2) w.r.t. x

ƒ’’(x) = 18x – 18 ………….. (3)

For maximum or minimum value ƒ’(x) = 0

9x2 –  18x – 27 = 0

∴   x2 –  2x – 3 = 0

∴   (x – 3)(x + 1) = 0

∴   x – 3 = 0 or/and x + 1 = 0

∴   x = 3 or/and x = – 1

Let us consider x = 3

ƒ’’(3) = 18(3) – 18 = 54 – 18 = 36 > 0

Hence the function is increasing at x = 3 and has minimum value at x = 3

Substituting x = 3 in equation (1)

Minimum value = ƒ(3) = 3(3)3 – 9(3)2 – 27(3) + 15 = 81 – 81 -81 +15 = – 66

Thus point of minimum is (3, – 66)

Let us consider x = – 1

ƒ’’(2) = 18(-1) – 18 = – 18 – 18 = – 36 < 0

Hence the function is decreasing at x = – 1 and has maximum value at x = – 1

Substituting x = – 1 in equation (1)

Maximum value = ƒ(-1) = 3(-1)3 – 9(-1)2 – 27(-1) + 15 = – 3 – 9 + 27 + 15 = 30

Thus point of maximum is (-1, 30)

Ans: The maximum value is 30 at x = – 1 and minimum value is – 66 at x = 3

#### Example – 06:

• Find the maximum and minimum values of 2x3 – 21x2 + 36x – 20
• Solution:

Let ƒ(x) = 2x3 – 21x2 + 36x – 20  ………….. (1)

Differentiating equation (1) w.r.t. x

ƒ’(x) = 6x2 –  42x + 36 ………….. (2)

Differentiating equation (2) w.r.t. x

ƒ’’(x) = 12x – 42 ………….. (3)

For maximum or minimum value ƒ’(x) = 0

6x2 –  42x + 36 = 0

∴   x2 –  7x + 6 = 0

∴   (x – 6)(x – 1) = 0

∴   x – 6 = 0 or/and x – 1 = 0

∴   x = 6 or/and x = 1

Let us consider x = 6

ƒ’’(6) = 12(6) – 42 =72 – 42 = 30 > 0

Hence the function is increasing at x = 6 and has minimum value at x = 6

Substituting x = 6 in equation (1)

Minimum value = ƒ(3) = 2(6)3 – 21(6)2 + 36(6) – 20 = 432 – 756 = 216 – 20 = – 128

Thus point of minimum is (6, – 128)

Let us consider x = 1

ƒ’’(2) = 12(1) – 42 = 12 – 42 = – 30 < 0

Hence the function is decreasing at x = 1 and has maximum value at x = 1

Substituting x = 1 in equation (1)

Maximum value = ƒ(1) = 2(1)3 – 21(1)2 + 36(1) – 20 = 2 – 21 +36 – 20 = – 3

Thus point of maximum is (1, -3)

Ans: The maximum value is -3 at x =  1 and minimum value is – 128 at x = 6

#### Example – 07:

• Find the maximum and minimum values of x2 + 16/x2
• Solution:

Let ƒ(x) = x2 + 16/x2

ƒ(x) = x2 + 16 x -2  ………….. (1)

Differentiating equation (1) w.r.t. x

ƒ’(x) = 2x – 32x -3 ………….. (2)

Differentiating equation (2) w.r.t. x

ƒ’’(x) = 2 + 96 x -4 ………….. (3)

For maximum or minimum value ƒ’(x) = 0

2x – 32x -3 = 0

∴  2x = 32x -3

∴  2x = 32/x3

∴  x4 = 16

∴   x = ± 2

Let us consider x = 2

ƒ’’(2) = 2 + 96 x (2) -4 = 2 + 96/16 = 2 + 6 = 8 > 0

Hence the function is increasing at x = 2 and has minimum value at x = 2

Substituting x = 2 in equation (1)

Minimum value = ƒ(2) = 22 + 16/22 = 4 + 4 = 8

Thus point of minimum is (2, 8)

Let us consider x = – 2

ƒ’’(2) = – 2 + 96 x (- 2) -4 = – 2 + 96/16 = – 2 + 6 = 4 > 0

Hence the function is increasing at x = – 2 and has minimum value at x = – 2

Substituting x = – 2 in equation (1)

Minimumm value = ƒ(-2) =(-2)2 + 16/(-2)2 = 4 + 4 = 8

Thus point of minimum is (- 2, 8)

Ans: The minimum value is 8 at x = ±2

#### Example – 08:

• Find the maximum and minimum values of x.logx
• Solution:

Let ƒ(x) = x.logx  ………….. (1)

Differentiating equation (1) w.r.t. x

ƒ’(x) = x(1/x) + logx. (1)

ƒ’(x) = 1 + logx ………….. (2)

Differentiating equation (2) w.r.t. x

ƒ’’(x) = 1/x  ………….. (3)

For maximum or minimum value ƒ’(x) = 0

1 + logx = 0

∴  log x = -1

∴   x = e-1 = 1/e

ƒ’’(1/e) = 1/(1/e) = e > 0

Hence the function is increasing at x = 1/e and has minimum value at x = 1/e

Substituting x = 1/e in equation (1)

Minimum value = ƒ(1/e) = (1/e). log(1/e) = – (1/e). log(e) =  – (1/e)

Thus point of minimum is (1/e, – 1/e)

Ans: The minimum value is – 1/e at x = 1/e

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