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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?


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 $$\Bigg(\sqrt{\frac{1600}{5}}\, \Bigg)^3 = 5724.334022\text{cm}^3.$$

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    $\begingroup$ You seem to have answered your own question: because the cube encloses less volume (as you have shown by direct computation). If you repeat the work, but with a lid, i.e. $1600 = 2x^2 + 4x h$, what shape do you get? $\endgroup$ – Eric Towers Nov 4 '20 at 20:41
  • $\begingroup$ You want to maximize $xyz$ given that $x,y,z$ are positive numbers satisfying $2xy + 2xz + yz = 1600 $. Hint: Try Weighted AM-GM inequality. $\endgroup$ – GohP.iHan Nov 4 '20 at 20:51
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    $\begingroup$ Because you don't need material for the top. If you have six sides, the answer is a cube. But think about how you would redistribute the material from the top to best effect if you don't need the top. That may help your intuition for why it comes out the way it does. $\endgroup$ – Mark Bennet Nov 4 '20 at 21:10
  • $\begingroup$ @EricTowers , you get half of a cube? The length of the square base is $\frac{20\sqrt{3}}{3}$ and the height is $\frac{10\sqrt{3}}{3}$. $\endgroup$ – Gurjinder Nov 4 '20 at 21:23
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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.

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