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Wed, 06 Mar 2024
Optimal boxes with and without lids
Sometime around 1986 or so I considered the question of the dimensions that a closed cuboidal box must have to enclose a given volume but use as little material as possible. (That is, if its surface area should be minimized.) It is an elementary calculus exercise and it is unsurprising that the optimal shape is a cube. Then I wondered: what if the box is open at the top, so that it has only five faces instead of six? What are the optimal dimensions then? I did the calculus, and it turned out that the optimal lidless box has a square base like the cube, but it should be exactly half as tall. For example the optimal box-with-lid enclosing a cubic meter is a 1×1×1 cube with a surface area of !!6!!. Obviously if you just cut off the lid of the cubical box and throw it away you have a one-cubic-meter lidless box with a surface area of !!5!!. But the optimal box-without-lid enclosing a cubic meter is shorter, with a larger base. It has dimensions $$2^{1/3} \cdot 2^{1/3} \cdot \frac{2^{1/3}}2$$ and a total surface area of only !!3\cdot2^{2/3} \approx 4.76!!. It is what you would get if you took an optimal complete box, a cube, that enclosed two cubic meters, cut it in half, and threw the top half away. I found it striking that the optimal lidless box was the same proportions as the optimal complete box, except half as tall. I asked Joe Keane if he could think of any reason why that should be obviously true, without requiring any calculus or computation. “Yes,” he said. I left it at that, imagining that at some point I would consider it at greater length and find the quick argument myself. Then I forgot about it for a while. Last week I remembered again and decided it was time to consider it at greater length and find the quick argument myself. Here's the explanation. Take the cube and saw it into two equal halves. Each of these is a lidless five-sided box like the one we are trying to construct. The original cube enclosed a certain volume with the minimum possible material. The two half-cubes each enclose half the volume with half the material. If there were a way to do better than that, you would be able to make a lidless box enclose half the volume with less than half the material. Then you could take two of those and glue them back together to get a complete box that enclosed the original volume with less than the original amount of material. But we already knew that the cube was optimal, so that is impossible. [Other articles in category /math] permanent link |