Q16 of 168 Page 231

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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