# What is the smallest possible value of the sum of their squares if the sum of two positive numbers is 16?

Feb 16, 2015

If the sum of two positive integers, $x$ and $y$ is $16$
$x + y = 16$
$y = \left(16 - x\right)$

The sum of their squares is
${x}^{2} + {\left(16 - x\right)}^{2}$
$= 2 {x}^{2} - 32 x + 256$

The minimum will occur when the derivative $= 0$
i.e. when $4 x - 32 = 0$
That is the minimum occurs at $\left(x , y\right) = \left(8 , 8\right)$
and the minimum possible value of the sum of the squares is
${8}^{2} + {8}^{2}$
$= 128$