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

Let one number be x. Then, the other number is (16 − x).

Let the sum of the cubes of these numbers be denoted by S(x). Then,

$S(x)=x^{3}+(16-x)^{3}$

$\therefore S^{\prime}(x)=3 x^{2}-3(16-x)^{2}, S^{\prime \prime}(x)=6 x+6(16-x)$'

Now, $S^{\prime}(x)=0 \Rightarrow 3 x^{2}-3(16-x)^{2}=0$

$\Rightarrow x^{2}-(16-x)^{2}=0$

$\Rightarrow x^{2}-256-x^{2}+32 x=0$

$\Rightarrow x=\frac{256}{32}=8$

Now, $S^{\prime \prime}(8)=6(8)+6(16-8)=48+48=96>0$

∴ By second derivative test, x = 8 is the point of local minima of S.

Hence, the sum of the cubes of the numbers is the minimum when the numbers are 8 and 16 − 8 = 8.