Generalized optimisation question about open topped boxes made from regular polygons

Question

This is based on the question about maximizing the volume of an open box being formed from a square with corners cut.

Original question is here Optimisation question

edit: added above picture for extra clarity about the case of a square base.

I also solved this for the base being a triangle, pentagon and hexagon, the trig gets a little hectic. Spoiler alert - the answer in all cases ended up with $x$ being $1/6th$ the length of the original shape's side.

My question is how to prove if is this the case for all regular polygons.

In this low quality sketch, the cuts are along the green lines, making the rectangular flaps that fold up to make the triangular based box. Similar kite shapes need to be cut for the other shapes.

triangle base

Answer 1

Notice that the base of the box is shrinked version of the original polygon. Let the scale factor be $\alpha$. If $S$ is the area of the original polygon, then the area of shrinked polygon is $\alpha^2 S$. If $h$ is the length of a perpendicular from the center of the polygon to one of it's sides, then the length of the shrinked perpendicular is $\alpha h$. Hopefully, it's clear that the height of the box is $(h - \alpha h) = (1-\alpha)h$.

Now the volume of the box is $\alpha^2 S (1 - \alpha) h$. It reaches it's maximum when $\alpha = \frac23$.

If $L$ is the length of a side of the polygon, then it's shrinked length is $\alpha L$, which means $x = (L - \alpha L) / 2 = L / 6$.

Answer 2

Let the edge length of the original polygon be $1$. The base of the polygonal box will then have edge length $1-2x$.

The height of the polygonal box is proportional to $x$ (because the polygon angle is fixed, the edges of the kites you cut off are proportional to $x$).

Therefore, the volume is proportional to $x(1-2x)^2$, which reaches its maximum at $x=\frac{1}{6}$.