Find two positive numbers whose sum is $16$ and the sum of whose cubes is minimum.

Let the numbers be $x$ and $16 - x$ and let $S = {x^3} + {(16 - x)^3}$
$\Rightarrow S = {x^3} + {(16 - x)^3} \Rightarrow \cfrac{{dS}}{{dx}} = 3{x^2} + 3{(16 - x)^2}( - 1)$

For minimum $S,\cfrac{{dS}}{{dx}} = 0$
$\Rightarrow 3{x^2} - 3{(16 - x)^2} = 0 \Rightarrow {x^2} - (256 + {x^2} - 32x) = 0$
$\Rightarrow 32x = 256 \Rightarrow x = 8$

$\cfrac{{{d^2}S}}{{d{x^2}}} = 6x + 6(16 - x)$ and ${\left( {\cfrac{{{d^2}S}}{{d{x^2}}}} \right)_{x = 8}} = 96 > 0$

Therefore, $S$ has a minimum value at $x = 8$.

Hence, the required numbers are $8{\rm{ and }}8$ .