Trying to find the right answer to this..

Let two numbers be x and 16 – x Let f (x) = x 3 + (16 – x) 3 ∴ f (x) = 3 x 2 + 3 (16 – x) 2 (– 1) = 3 x 2 – 3 (16 – x) 2 = 3 x 2 – 3 (256 + x 2 – 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.

Application of Derivatives

Question

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

Answer

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.