Question
Find two positive numbers whose sum is 16 and sum of whose cubes is minimum.
Solution
Let two numbers be x and 16 – x
Let f (x) = x3 + (16 – x)3
∴ f (x) = 3 x2 + 3 (16 – x)2 (– 1) = 3 x2 – 3 (16 – x)2
= 3 x2 – 3 (256 + x2 – 32 x) = 96 x – 768 = 96 (x – 8)
f ' (x) = 0 ⇒ 96 (x – 8) = 0 ⇒ x = 8
f ' ' (x) = 96
At x = 8, f ' ' (x) = 96 > 0
∴ f (x) is minimum when x = 8 ⇒ numbers are 8, 8.
