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#### Example – 01:

- If x + y = 3 show that the maximum value of x
^{2}y is 4 **Solution:**

Given x + y = 3

hence y = 3 – x

∴ x^{2} y = x^{2}(3 – x) = 3x^{2} – x^{3}

Let ƒ(x) = 3x^{2} – x^{3} ………….. (1)

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

ƒ’(x) = 6x – 3x^{2} ………….. (2)

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

ƒ’’(x) = 6 – 6x ………….. (3)

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

6x – 3x^{2} = 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) = x
^{x}is minimum when x = 1/e **Solution:**

Let ƒ(x) = x^{x} ………….. (1)

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

ƒ’(x) = x^{x}(1 + logx) ………….. (2)

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

ƒ’'(x) = x^{x}(0 + 1/x) + (1 + logx)x^{x}(1 + logx)

ƒ’'(x) = x^{x}/x + x^{x}(1 + logx)^{2}

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

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

x^{x}(1 + logx) = 0

∴ x^{x} = 0 or/and (1 + logx) = 0

x^{x} = 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)/x^{2} = 0

∴ (1 – logx) = 0

∴ log x = 1

∴ x = e^{1}

∴ 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 x
^{2}.e^{x}. **Solution:**

Let ƒ(x) = x^{2}.e^{x } ………….. (1)

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

ƒ’(x) = x^{2}.e^{x} + e^{x}. 2x

ƒ’(x) = x^{2}.e^{x} + 2 x e^{x}………….. (2)

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

ƒ’’(x) = x^{2}.e^{x} + e^{x}. 2x + 2(x e^{x} + e^{x}(1))

ƒ’’(x) = x^{2}.e^{x} + 2x e^{x} + 2x e^{x} + 2e^{x}

ƒ’’(x) = e^{x}(x^{2} + 2x + 2x + 2)

ƒ’’(x) = e^{x}(x^{2} + 4x + 2)………….. (3)

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

x^{2}.e^{x} + 2 x e^{x} = 0

∴ x.e^{x} ( x + 2) = 0

e^{x} ≠ 0

∴ x = 0 or/and x + 2 = 0

∴ x = 0 or/and x = – 2

Let us consider x = 0

ƒ’’(0) = e^{0}(0^{2} + 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) = 0^{2}.e^{0} = 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/e^{2} < 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/e^{2}

Thus point of maximum is (- 2, 4/e^{2})

**Ans:** The maximum value is 4/e^{2} 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

a^{2} cos^{2}x = b^{2} sin^{2} x

a^{2} (1 – sin^{2}x) = b^{2} sin^{2} x

a^{2 } – a^{2}sin^{2}x)= b^{2} sin^{2} x

a^{2 } = a^{2}sin^{2}x + b^{2} sin^{2} x

a^{2 } = (a^{2} + b^{2})sin^{2} x

Hence the function is increasing and has minimum value

Hence the function is increasing and has minimum value

Science > Mathematics > Applications of Derivatives > You are Here |

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