# What is the largest volume of a polyhedron whose skeleton has total length 1? Is it the regular triangular prism?

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 $$10$$.)

• 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. – Steven Stadnicki Mar 1 at 17:29
• @StevenStadnicki: Just an update that I did investigate something like this for some small polyhedra and didn't find any improvements (see the edit). – RavenclawPrefect Mar 9 at 21:24