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.