Possible mistake finding the maximum volume of a box with the AM-GM inequality?

Question

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.

Answer 1

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.

Answer 2

Let $x=$ side of small square $=$ depth of the box;

then $2a-2x=$ side of the square forming bottom of the box, and the volume is $V=(2a-2x)^2x;$ which is the function to be made a maximum by varying $x$.

Applying the rule $$\dfrac{dV}{dx}=12x^2-16ax+4a^2$$

Solving $12x^2-16ax+4a^2=0$ gives the critical values $x=a,\dfrac a3$.

It is evident that $x=a$ must give a minimum, for then all the tin would be cut away, leaving no material out of which to make a box. By the usual test, $x=\dfrac a3$ is found to give a maximum volume of $V=\left(2a-2\left(\dfrac a3\right)\right)^2\dfrac a3=\dfrac{16a^3}{27}$