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 S(x) be the sum of these number. Then,


S(x) = x3 + (16-x)3


⇒ S’(x) = 3x2 -3(16-x)2


⇒ S’’(x) = 6x + 6(16-x)


Now, S’(x) =0


⇒ 3x2 -3(16-x)2 = 0


⇒ x2 -(16-x)2 = 0


⇒ x2 – 256 - x2 + 32x = 0


⇒ 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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