Find two positive numbers whose sum is 16 and | vedclass

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Question

(A) Let one number be $x$. Then,the other number is $(16-x)$.Let the sum of the cubes of these numbers be denoted by $S(x)$. Then,$S(x) = x^3 + (16-x)

Vedclass · Accepted Answer

$8, 8$

Vedclass · Answer

(A) Let one number be $x$. Then,the other number is $(16-x)$.Let the sum of the cubes of these numbers be denoted by $S(x)$. Then,$S(x) = x^3 + (16-x)^3$To find the minimum,we find the first derivative:$S'(x) = 3x^2 - 3(16-x)^2$Setting $S'(x) = 0$:$3x^2 - 3(16-x)^2 = 0$$x^2 - (256 - 32x + x^2) = 0$$32x - 256 = 0$$x = 8$Now,we find the second derivative to check for minima:$S''(x) = 6x + 6(16-x)$$S''(8) = 6(8) + 6(16-8) = 48 + 48 = 96$Since $S''(8) > 0$,the function has a local minimum at $x = 8$.Thus,the two numbers are $8$ and $16-8 = 8$.

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

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