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

Answer

Let one number be $x$. Then, the other number is $(16 - x)$.
Let S(x) be the sum of these number. Then,
$S(x) = x^3 + (16 - x)^3$
$\Rightarrow S\ ' (x) = 3x^2 - 3(16 - x)^2$
$\Rightarrow S\ '' (x) = 6x + 6(16 - x)$
Now $, S\ ' (x) = 0$
$\Rightarrow 3x^2 - 3(16 - x)^2 = 0$
$\Rightarrow x^2 - (16 - x)^2 = 0$
$\Rightarrow x^2 - 256 - x^2 + 32x = 0$
$\Rightarrow x = 8$
Now $, S\ '' (8) = 6(8) + 6(16 - 8)$
$= 48 + 48 = 96 > 0$
Then, by second derivative test,$ x = 8$ is the point of local minima of $S.$
Therefore, the sum of the cubes of the numbers is the minimum when the numbers are $8$ and $16 - 8 = 8$.

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