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Question
Find two positive numbers whose sum is 16 and the sum of whose cubes is minimum.
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Solution
Let the numbers be x and 16 - x and let
S = x3 + (16 - x)3
⇒ S = x3 + (16 - x)3
⇒ `(dS)/dx = 3x^2 + 3 (16 - x)^2 (-1)`
For minimum S, let `(dS)/dx = 0`
⇒ 3x2 - 3 (16 - x)2 = 0
⇒ x2 - (256 + x2 - 32x) = 0
⇒ 32x = 256
⇒ x = 8
`((d^2S)/dx^2) = 6x + 16 (16 - x) `
`((d^2S)/dx^2)_(x = 8) = 96 > 0`
∴ S has a minimum at x = 8
∴ The required numbers are 8 and 8.
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