# Find two positive numbers whose sum is 14 and the sum of whose squares is minimum.

Last updated date: 21st Mar 2023

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Answer

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**Hint:**First of all, let the two numbers be x and y. Now, we will get an equation $x + y = 14$ and from this we can substitute $y = 14 - x$. Now, let the sum of the squares be S and form an equation. Now, to find the minimum value, we need to differentiate the equation twice and we will get our answer.

**Complete step-by-step solution:**

In this question, we are given that the sum of two positive numbers is 14 and their sum of squares is minimum. We need to find these numbers.

Let these two numbers be x and y.

So, therefore, we get an equation

$ \Rightarrow x + y = 14$ - - - - - - (1)

Therefore, we can say that y will be equal to

$ \Rightarrow y = 14 - x$

Now, let the sum of the squares of these two numbers be S. Therefore, we get an equation

$ \Rightarrow S = {x^2} + {y^2}$

But, $y = 14 - x$

$ \Rightarrow S = {x^2} + {\left( {14 - x} \right)^2}$ - - - - - - - (2)

Now, to find the minimum sum, we need to use calculus.

If $f\left( x \right)$ is any given function and $f''\left( x \right) > 0$ then the value of $f\left( x \right)$ is minimum and if $f''\left( x \right) < 0$, then the value of $f\left( x \right)$ is maximum.

Steps for finding the Max/Min value of $f\left( x \right)$:

Step 1: Differentiate the given function.

Therefore, differentiating equation (2) w.r.t x, we get

$\Rightarrow \dfrac{{dS}}{{dx}} = \dfrac{d}{{dx}}{x^2} + \dfrac{d}{{dx}}{\left( {14 - x} \right)^2} \\

\Rightarrow \dfrac{{dS}}{{dx}} = 2x + 2\left( {14 - x} \right)\left( { - 1} \right) \\

\Rightarrow \dfrac{{dS}}{{dx}} = 2x - 28 + 2x $

$ \Rightarrow \dfrac{{dS}}{{dx}} = 4x - 28$ - - - - - - (3)

Step 2: Let $f'\left( x \right) = 0$ and find the critical points.

Therefore,

$ \Rightarrow 4x - 28 = 0 \\

\Rightarrow 4x = 28 \\

\Rightarrow x = \dfrac{{28}}{4} \\

\Rightarrow x = 7 $

Step 3: Find second derivative.

Therefore, differentiating equation (3), we get

$ \Rightarrow \dfrac{{{d^2}S}}{{d{x^2}}} = \dfrac{d}{{dx}}\left( {4x - 28} \right) = 4$$ > 0$

Here, the second derivative is $f''\left( x \right) > 0$. Therefore, we can say that the value of

$f\left( x \right)$ is minimum.

Now, for $x = 7$, the value of $f\left( x \right)$ will be minimum. Therefore, for finding the value of y, substitute $x = 7$ in equation (1), we get

$ \Rightarrow x + y = 14 \\

\Rightarrow 7 + y = 14 \\

\Rightarrow y = 14 - 7 \\

\Rightarrow y = 7 $

**Therefore the numbers whose sum is 14 and the sum of their squares is minimum are 7 and 7.**

**Note:**Note that if we had got $\dfrac{{{d^2}S}}{{d{x^2}}}$ equal to with some variable, then to find the maximum or minimum value, we had to put $x = 7$ in that equation and the answer obtained would have been the minimum or the maximum value of the function.

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