# Maximum and Minimum Value of Function

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#### Example – 01:

• If x + y = 3 show that the maximum value of x2 y is 4
• Solution:

Given x + y = 3

hence y = 3 – x

∴  x2 y = x2(3 – x) = 3x2 – x3

Let ƒ(x) = 3x2 – x3 ………….. (1)

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

ƒ’(x) = 6x – 3x2 ………….. (2)

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

ƒ’’(x) = 6 – 6x  ………….. (3)

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

6x – 3x2 = 0

∴   3x(2 – x) = 0

∴   x = 0 or/and 2- x = 0

∴   x = 0 or/and x = 2

Let us consider x = 0

ƒ’’(0) =6 – 6(0) = 6 – 0 = 6 > 0

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

Let us consider x = 3

ƒ’’(2) = 6 –  6(2) = 6 – 12 = -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) = 3(2)2 – (2)3 = 12 – 8 = 4

TThus the maximum value is 4 (Proved as required)

#### Example – 02:

• show that f(x) = xx is minimum when x = 1/e
• Solution:

Let ƒ(x) = xx  ………….. (1)

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

ƒ’(x) = xx(1 + logx) ………….. (2)

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

ƒ’'(x) = xx(0 + 1/x) + (1 + logx)xx(1 + logx)

ƒ’'(x) = xx/x + xx(1 + logx)2

ƒ’'(x) = (1/x + 1(1 + logx)2)  ………….. (3)

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

xx(1 + logx) = 0

∴  xx = 0 or/and (1 + logx) = 0

xx = 0 implies x = 0 makes term logx resundant

Hence x = 0 is not possible

∴  (1 + logx) = 0

∴   log x = -1

∴   x = e-1

∴   x = 1/e

ƒ’’(1/e) = (1/(1/e) + 1(1 + log(1/e))2)

ƒ’’(1/e) = (e + 1(1 – loge)2) = (e + 0) = e > 0

Hence the function is increasing at x = 1/e and has minimum value at x = 1/e (Proved as required)

#### Example – 03:

• show that f(x) = (logx)/x (x ≠ 0) is maximum at x = e
• Solution:

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

(1 – logx)/x2 = 0

∴  (1 – logx) = 0

∴   log x = 1

∴   x = e1

∴   x = e

Hence the function is decreasing at x = e and has maximum value at x = e (Proved as required)

#### Example – 04:

• Find the maximum and minimum values of x2.ex.
• Solution:

Let ƒ(x) = x2.e ………….. (1)

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

ƒ’(x) = x2.ex + ex. 2x

ƒ’(x) = x2.ex + 2 x ex………….. (2)

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

ƒ’’(x) = x2.ex + ex. 2x + 2(x ex + ex(1))

ƒ’’(x) = x2.ex + 2x ex + 2x ex + 2ex

ƒ’’(x) = ex(x2 + 2x + 2x + 2)

ƒ’’(x) = ex(x2 + 4x + 2)………….. (3)

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

x2.ex + 2 x ex = 0

∴  x.ex ( x + 2) = 0

ex ≠ 0

∴   x = 0 or/and x + 2 = 0

∴   x = 0 or/and x = – 2

Let us consider x = 0

ƒ’’(0) = e0(02 + 4(0) + 2) = 1(0 + 0 + 2) = 2 > 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(2) = 02.e0  = 0

Thus point of minimum is (0, 0)

Let us consider x = – 2

ƒ’’(-2) = e-2((-2)2 + 4(-2) + 2) = e-2(4 – 8 + 2) = – 2/e2 < 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.e-2  = 4/e2

Thus point of maximum is (- 2, 4/e2)

Ans: The maximum value is 4/e2 at x = -2 and minimum value is -0 at x = 0

#### Example – 05:

• Find extreme values of ƒ(x) = a sin x + b cos x
• Solution:

Let ƒ(x) = ƒ(x) = a sin x + b cos x  ………….. (1)

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

ƒ’(x) = ƒ(x) = a cosx – b sin x  ………….. (2)

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

ƒ’’(x) = – a sin x – b cos x = – ƒ(x)  ………….. (3)

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

a cosx – b sin x = 0

a cosx = b sin x

Squaring both sides

a2 cos2x = b2 sin2 x

a2 (1 – sin2x) = b2 sin2 x

a – a2sin2x)= b2 sin2 x

a = a2sin2x + b2 sin2 x

a = (a2 + b2)sin2 x

Hence the function is increasing and has minimum value

Hence the function is increasing and has minimum value

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