Question

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

Answer 1

We have to maximise $x^3+y^3$ and we have $x+y=16$

Put $x=16-y$ in $x^3+y^3$

we get $f\left(y\right)=48y^2-768y+4096$

First derivative is given by
$f^{{}^{\prime }}\left(y\right)=96y-768$

Putting this to zero we get, $y=8$

$f^{{}^{\prime \prime }}\left(y\right)=96$
Given that second derivative is positive
So given function will have minima when $y=8$

So $x=16-8=8$