# 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 a 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 than 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. Mar 1, 2021 at 17:29
• You can get a better upper bound by using the theorem 2 from this paper. From that together with polyhedron cannot enclose more volume than a sphere, you get the upper bound of $\frac{1}{36\sqrt{6}\pi^2}\approx 0.00115$ May 14, 2021 at 18:59
• @TymaGaidash: What do you mean by a "volume formula"? The volume of a polyhedron is not uniquely determined by the multiset of its edge lengths, even given the associated edge graph (consider the family of parallelepipeds with unit edges), so there can be no deterministic mapping from a polyhedron's edge lengths to its volume. May 18, 2021 at 2:03
• This is essentially Melzak's Conjecture. ijgeometry.com/wp-content/uploads/2021/09/4.-63-73.pdf and scholar.rose-hulman.edu/cgi/…
– Dan
Dec 4, 2021 at 10:32
• Digging a bit on the Melzak conjecture literature, I found this paper: londmathsoc.onlinelibrary.wiley.com/doi/10.1112/jlms/s2-6.2.382 (unfortunately not freely available), which gives the upper bound $V<\frac{1}{432\pi}\approx 0.00073682$, which is pretty close to $0.00059$ Sep 27, 2022 at 10:21