Finding maxima of a function $f(x) = 4x(16 - x)(6 - x)$ without calculcus My question is how to prove that $f(x) = 4x(16 - x)(6 - x)$ has its maximum at point $x_0 = \frac{8}{3}$.
I am trying to use arithmetic-geometric inequality, but I am unable to solve it.
The problem can be greatly simplified by considering $f(8/3+t)-f(8/3)$.
This is $4(t^3-14t^2)$. The graph of this cubic has a repeated root at $0$ and a single root at $14$ and is negative at $7$. It therefore clearly has a local maximum at $t=0$ i.e. $x=8/3$.
The same idea can be used for other polynomial functions.
$f(x)=4x(16-x)(6-x)=4x^3 - 88x^2 +384x$
Now calculate it's derivative
$f'(x)=12x^2 -176x + 384 = 12(x-6)(x - \frac{8}{3})$
Now just find increasing and decreasing intervals
$f(x)$ is increasing on $ x&lt; \frac{8}{3} $ and $ x&gt;6$
And decreasing on ($ \frac{8}{3} ; 6)$ so after that you can say $x_0 = \frac{8}{3} $ is a local maximum point