Does a cube give the maximum volume for a fixed surface area?

Question

If I am making a fish tank with an open lid and $1600 \,\rm{cm}^2$ of material, why is the maximum volume that the tank can occupy not given by a cube?

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My work

Suppose the base of the fish tank is a square, with length $x$cm and height $h$cm.
Then the volume $V$ of the tank will be \begin{align} V = x^2h. \tag{1} \end{align} Given the constraint of the material, \begin{align} 1600 = x^2 + 4xh. \tag{2} \end{align} Eliminating $h$: $$V = 400x - \frac{1}{4}x^3.$$ The maximum volume will be given by: $$\frac{dV}{dx} = 400 - \frac{3}{4}x^2 =0.$$ Hence $x = +\frac{40\sqrt{3}}{3}.$

Then using $(2)$, we get $h = \frac{20\sqrt{3}}{3}.$

Therefore (using $(1)$) the volume of the tank will be about $6158$cm$^3$.

But if the tank is a cube, then the volume will be $$\Bigg(\sqrt{\frac{1600}{5}}\, \Bigg)^3 = 5724.334022\text{cm}^3.$$

Answer 1

You have already solved the problem, so you know the numerical reason.

The intuition is that if you put two of these open boxes together across the open face, you get a closed box with twice the surface area ($3200$) and twice the volume.

We know that the volume of the closed box is maximized when it's a cube (with side length $\sqrt{\frac{3200}{6}} = \frac{40\sqrt3}{3}$). So the optimal open box should be half of that cube.

That's exactly the solution you've found.