Question

If the sum of two numbers is 20 and the sum of their squares is minimum then thenumbers are
(a) 10,10
(b) 5,15
(c) 4,16
(d) 3,17​

Answer 1

$\large\underline{\sf{Solution-}}$

Let assume that

• First number be x

and

• Second number be y.

So, given that,

$\rm :\longmapsto\:x + y = 20$

$\rm \implies\:\boxed{ \tt{ \: y \: = \: 20 \: - \: x \: }} \: - - - (1)$

Now, we have to find the values of x and y, such that sum of their squares is minimum.

Let assume that

$\rm :\longmapsto\:S = {x}^{2} + {y}^{2}$

On substituting the value of y, we get

$\rm :\longmapsto\:S = {x}^{2} + {(20 - x)}^{2}$

$\rm :\longmapsto\:S = {x}^{2} + 400 + {x}^{2} - 40x$

$\rm :\longmapsto\:S = 2{x}^{2} + 400 - 40x$

On differentiating both sides w. r. t. x, we get

$\rm :\longmapsto\:\dfrac{d}{dx}S =\dfrac{d}{dx}( 2{x}^{2} + 400 - 40x )\:$

$\rm :\longmapsto\:\dfrac{dS}{dx}=4x + 0 - 40$

$\rm :\longmapsto\:\dfrac{dS}{dx}=4x- 40 - - - (2)$

For maxima or minima, we have

$\rm :\longmapsto\:\dfrac{dS}{dx}=0$

$\rm :\longmapsto\:4x - 40 = 0$

$\rm :\longmapsto\:4x = 40$

$\bf\implies \:\boxed{ \tt{ \: x \: = \: 10 \: }}$

From equation (2), we have

$\rm :\longmapsto\:\dfrac{dS}{dx}=4x- 40$

On differentiating both sides w. r. t. x, we get

$\rm :\longmapsto\:\dfrac{ {d}^{2} S}{d {x}^{2} }=4$

$\rm :\longmapsto\:\dfrac{ {d}^{2} S}{d {x}^{2} } > 0$

$\bf\implies \:S \: is \: minimum \: when \: x = 10$

On substituting x = 10, in equation (1), we have

$\rm :\longmapsto\:y = 20 - 10$

$\bf\implies \:\boxed{ \tt{ \: y \: = \: 10 \: }}$

Hence,

• Option (a) is correct.

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

Basic Concept Used :-

Let y = f(x) be a given function.

To find the maximum and minimum value, the following steps are follow :

1. Differentiate the given function.

2. For maxima or minima, put f'(x) = 0 and find critical points.

3. Then find the second derivative, i.e. f''(x).

4. Apply the critical points ( evaluated in second step ) in the second derivative.

5. Condition :-

The function f (x) is maximum when f''(x) < 0.

The function f (x) is minimum when f''(x) > 0.

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

Additional Information :-

$\red{\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \dfrac{d}{dx}f(x) \\ \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf 0 \\ \\ \sf sinx & \sf cosx \\ \\ \sf cosx & \sf - \: sinx \\ \\ \sf tanx & \sf {sec}^{2}x \\ \\ \sf cotx & \sf - {cosec}^{2}x \\ \\ \sf secx & \sf secx \: tanx\\ \\ \sf cosecx & \sf - \: cosecx \: cotx\\ \\ \sf \sqrt{x} & \sf \dfrac{1}{2 \sqrt{x} } \\ \\ \sf logx & \sf \dfrac{1}{x}\\ \\ \sf {e}^{x} & \sf {e}^{x} \end{array}} \\ \end{gathered}}$

Answer 2

Answer:

$\large\underline{\sf{Solution-}}$

Let assume that

First number be x

and

Second number be y.

So, given that,

$\rm :\longmapsto\:x + y = 20$

$\rm \implies\:\boxed{ \tt{ \: y \: = \: 20 \: - \: x \: }} \: - - - (1)$

Now, we have to find the values of x and y, such that sum of their squares is minimum.

Let assume that

$\rm :\longmapsto\:S = {x}^{2} + {y}^{2}$

On substituting the value of y, we get

$\rm :\longmapsto\:S = {x}^{2} + {(20 - x)}^{2}$

$\rm :\longmapsto\:S = {x}^{2} + 400 + {x}^{2} - 40x$

$\rm :\longmapsto\:S = 2{x}^{2} + 400 - 40x$

On differentiating both sides w. r. t. x, we get

$\rm :\longmapsto\:\dfrac{d}{dx}S =\dfrac{d}{dx}( 2{x}^{2} + 400 - 40x )\:$

$\rm :\longmapsto\:\dfrac{dS}{dx}=4x + 0 - 40$

$\rm :\longmapsto\:\dfrac{dS}{dx}=4x- 40 - - - (2)$

For maxima or minima, we have

$\rm :\longmapsto\:\dfrac{dS}{dx}=0$

$\rm :\longmapsto\:4x - 40 = 0$

$\rm :\longmapsto\:4x = 40$

$\bf\implies \:\boxed{ \tt{ \: x \: = \: 10 \: }}$

From equation (2), we have

$\rm :\longmapsto\:\dfrac{dS}{dx}=4x- 40$

On differentiating both sides w. r. t. x, we get

$\rm :\longmapsto\:\dfrac{ {d}^{2} S}{d {x}^{2} }=4$

$\rm :\longmapsto\:\dfrac{ {d}^{2} S}{d {x}^{2} } > 0$

$\bf\implies \:S \: is \: minimum \: when \: x = 10$

On substituting x = 10, in equation (1), we have

$\rm :\longmapsto\:y = 20 - 10$

$\bf\implies \:\boxed{ \tt{ \: y \: = \: 10 \: }}$

Hence,

Option (a) is correct.

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

Basic Concept Used :-

Let y = f(x) be a given function.

To find the maximum and minimum value, the following steps are follow :

1. Differentiate the given function.

2. For maxima or minima, put f'(x) = 0 and find critical points.

3. Then find the second derivative, i.e. f''(x).

4. Apply the critical points ( evaluated in second step ) in the second derivative.

5. Condition :-

The function f (x) is maximum when f''(x) < 0.

The function f (x) is minimum when f''(x) > 0.

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

Additional Information :-

$\red{\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \dfrac{d}{dx}f(x) \\ \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf 0 \\ \\ \sf sinx & \sf cosx \\ \\ \sf cosx & \sf - \: sinx \\ \\ \sf tanx & \sf {sec}^{2}x \\ \\ \sf cotx & \sf - {cosec}^{2}x \\ \\ \sf secx & \sf secx \: tanx\\ \\ \sf cosecx & \sf - \: cosecx \: cotx\\ \\ \sf \sqrt{x} & \sf \dfrac{1}{2 \sqrt{x} } \\ \\ \sf logx & \sf \dfrac{1}{x}\\ \\ \sf {e}^{x} & \sf {e}^{x} \end{array}} \\ \end{gathered}}$