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

let x an d y are positive two numbers
It is given that
x + y = 16 , y = 16 - x
and  $x^3 + y^3$ is minimum
$f(x) = x^3 + (16-x)^3$
Now,
$f^{'}(x) = 3x^2 + 3(16-x)^2(-1)$
$f^{'}(x) = 0\\ 3x^2 - 3(16-x)^2 =0\\ 3x^2-3(256+x^2-32x) = 0\\ 3x^2 -3x^2+96x-768= 0\\ 96x = 768\\ x = 8\\$
Hence, x = 8 is the only critical point
Now,
$f^{''}(x) = 6x - 6(16-x)(-1) = 6x + 96 - 6x = 96\\ f^{''}(x) = 96$
$f^{''}(8) = 96 > 0$
Hence, x = 8 is the point of minima
y = 16 - x
= 16 - 8 = 8
Hence, values of x and y are 8  and 8 respectively

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