The sum of two +ve numbers is 20. If the sum of their squares is minimum then one of the number is

Given $x+y=20$,

Now $x^2+y^2=x^2+{(20-x)}^2$  

let  $f\left(x\right)=x^2+(20-x{)}^2$ , 

 $$\begin{gathered} diff w.r.to x on both sides \\ f\text{'}\left(x\right)=2x-2(20-x) \\ for max or min f\text{'}\left(x\right)=0 \\ {f}^1\left(x\right)=0\Rightarrow x=10 \end{gathered}$$