Find two positive numbers whose sum is 15 and the sum of whose squares is minimum.

The correct answer is $x = \dfrac{15}{2}$ and $y = \dfrac{15}{2}$

Let $x$ and $y$ be two positive numbers.

Given that $x+y = 15$ ⇒ $y = 15 - x$


Let $S$ be the sum of squares of these two numbers.

∴ $S = x^2 + y^2$ = $x^2 + (15 - y)^2 $ = $2x^2 - 30x +225$


∴ $\dfrac{dS}{dx}$ = $4x - 30$ and $\dfrac{d^2S}{dx^2}$ = $4$


Now, $\dfrac{dS}{dx} = 0$ = $4x - 30$ ⇒ $4x = 30$ or $x = \dfrac{15}{2}$



At $x = \dfrac{15}{2}$, $ \left( \dfrac{d^2S}{dx^2} \right)_{x= \frac{15}{2}}$ = $4 > 0$

∴ By second derivative test, we have a local minima of $S$ at $x = \dfrac{15}{2}$

As $x+y=15$, $y= \dfrac{15}{2}$

Hence the sum of squares is minimum when the numbers are $\dfrac{15}{2}$ and $\dfrac{15}{2}$