The two positive numbers whose sum is 16 and the sum of whose squares is minimum then the positive numbers are:

$8, 8$

The correct answer is Option (3) → $8, 8$ **

Let the numbers be $x$ and $y$ with $x+y=16,\ x>0,\ y>0$.

Minimize $S=x^{2}+y^{2}$. Using $y=16-x$:

$S(x)=x^{2}+(16-x)^{2}=2x^{2}-32x+256$

$\frac{dS}{dx}=4x-32=0\Rightarrow x=8$

Then $y=16-8=8$.

Therefore the two positive numbers are $8$ and $8$.