Say that the perimeter of a polyhedron is the sum of its edge lengths. What is the maximum volume of a polyhedron with unit perimeter?

A reasonable first guess would be the regular tetrahedron of side length $1/6$, with volume $\left(\frac16\right)^3\cdot\frac1{6\sqrt{2}}=\frac{\sqrt{2}}{2592}\approx 0.0005456$. However, the cube fares slightly better, at $\frac{1}{1728}\approx 0.0005787$.

After some experimentation, it seems that the triangular prism with all edges of length $1/9$ is optimal, at a volume of $\frac{\sqrt{3}}{2916}\approx0.00059398$. I can prove that this is optimal among all prisms (the Cartesian product of any polygon with an interval), and that there is no way to cut off a small corner from the shape and improve it.

Is the triangular prism the largest polyhedron with a fixed perimeter?

I can prove a weak upper bound of $\frac1{12\pi^2\sqrt{2}}\approx 0.00597$ on the volume of such a polyhedron, by combining the isoperimetric inequalities in both $2$ and $3$ dimensions (i.e., the fact that polygons cannot enclose more surface area than a circle and that a polyhedron cannot enclose more volume that a sphere of the same surface area) along with the observation that a single face of a polyhedron cannot take up the majority of its surface area. Note the number of leading zeros - this upper bound is a bit over $10$ times my lower bound!

Edit: A friend of mine has confirmed with Mathematica that no polyhedron with $5$ or fewer vertices, or anything combinatorially equivalent to the triangular prism, improves on this bound. (With some work, this approach might be extended to all polyhedra on at most $k$ vertices, for $k$ on the order of $7$ to $10$.)

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    $\begingroup$ I wonder whether you could approach this by specific combinatorial counting; the number of polyhedra with low edge counts is relatively small. An orange-slice shape suggests that it might be hard to lower-bound the number of 'long' edges, though. This is a great question. $\endgroup$ – Steven Stadnicki Mar 1 at 17:29
  • $\begingroup$ @StevenStadnicki: Just an update that I did investigate something like this for some small polyhedra and didn't find any improvements (see the edit). $\endgroup$ – RavenclawPrefect Mar 9 at 21:24

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