I have found the following problem:
What is the box (without a top) of largest volume which can be constructed from
a square piece of tin of edge length $2a$ by cutting a square from each corner
and folding up the edges
I have tried to solve it with the AM-GM inequality: I have the sides $(2a-2h), (2a-2h),h$ and then:
$$\frac{(2a-2h)+ (2a-2h) + h}{3}\geq \sqrt[3]{(2a-2h)^2h}$$
Equality holds when $(2a-2h)=(2a-2h)=h$, when I try to solve, I find:
$$h=\frac{2a}{3}$$
When I try to do the same with derivatives, I find that the roots of the derivative of $(2a-2h)^2h$ are $a$ and $\frac{a}{3}$. I may be doing something extremely silly but I can't figure out what is wrong.
Best Answer
The problem is, that your bound on the left in AM/GM depends on $h$. Try $$\sqrt[3]{(2a-2h)^2h}=2^{1/3}\sqrt[3]{(a-h)^22h} \le2^{1/3}\frac{(a-h)+(a-h)+2h}{3}$$ instead. You get equality when $a-h=2h$ etc.