Find two positive numbers whose sum is 16 and the sum of whose cubes is minimum.
Let the two positive numbers be x and y.
Given: $x+y=16$
$\implies$ $y = 16-x$ $\;\; \cdots$ (1)
Sum of cubes of the numbers $= S = x^3 + y^3$
i.e. $S = x^3 + \left(16-x\right)^3$ [From equation (1)]
For minimum sum, $\dfrac{dS}{dx}=0$
Now, $\dfrac{dS}{dx}= 3x^2 - 3 \left(16-x\right)^2$ $\;\; \cdots$ (2)
$\therefore$ $\dfrac{dS}{dx}=0$ $\implies$ $3x^2 - 3 \left(16-x\right)^2 = 0$
i.e. $x^2 = \left(16-x\right)^2$
i.e. $x^2 = 256 -32x + x^2$
i.e. $32x = 256$ $\implies$ $x = 8$
From equation (2), $\dfrac{d^2 S}{dx^2} = 6x +6 \left(16-x\right) = 96 > 0 \; \forall \; x$
$\implies$ $x = 8$ gives a minimum for the sum of cubes of the numbers.
From equation (1), when $x = 8$, $y=16-8=8$
$\therefore$ The two numbers are 8, 8.