The problem asks for the maximum volume of a polyhedron with a total edge length (perimeter) of 1. Through analysis of different polyhedral structures, including prisms, pyramids, deltahedrons, and Archimedean solids, it is found that prisms are particularly efficient for maximizing volume. Among prisms, the volume is maximized when the base is a regular polygon.

For a prism with an n-sided regu...

Short answer: What you conjectured is almost certainly true, but, to the best of current knowledge, there is no published complete proof that the regular triangular prism maximizes the volume among all polyhedra with fixed total edge length. It is, however, optimal among prisms (Cartesian products of a polygon with an interval), and beats all the usual suspects (tetrahedron, cube, octahedron, squa...

This is a fascinating optimization problem that combines geometry, calculus of variations, and discrete optimization. Let me analyze the current state of knowledge and provide some insights.

Current Results Summary

From your post, we have:

• Regular tetrahedron (edge length 1/6): Volume ≈ 0.0005456
• Cube (edge length 1/12): Volume ≈ 0.0005787
• **Triangular prism (all edges 1/9): Volume ≈ 0....

Short answer
Nobody knows the exact maximum. All that is known at the moment is

1. a concrete lower bound, obtained from the regular triangular prism whose nine edges all have length (1/9):   (V_{\text{prism}}=\dfrac{\sqrt 3}{2916}\approx 5.94\times 10^{-4};)

2. some rather crude general upper bounds – the best that has been written down so far seems to be the one you already gave, ...

Let P be any (convex) polyhedron whose total edge‐length (“perimeter”)

L(P)=∑_{e∈Edges(P)}|e|

is fixed to 1. We ask

Maximize V(P) subject to L(P)=1.

1. Easy lower‐bounds by the Platonic solids (all of whose edges we rescale so that their total edge‐length is 1) give

– Regular tetrahedron: 6 edges ⇒ a=1/6 ⇒
V= a³/(6√2)=√2/2592≈0.0005456
– Cube: 12 edges ⇒ a=1/...

First, I should recall that for a given surface area, the sphere has the maximum volume, but here it's ab...

Excellent question! This is a classic and surprisingly difficult unsolved problem in geometry. Your work so far is a great exploration of the problem, and your candidate shape is indeed a very strong one.

Let's break down the problem and what is known.

Summary of Your Findings

First, let's validate your calculations for the dimensionless ratio , where is the volume and is the...

Polyhedron with Maximum Volume for a Fixed Perimeter

This question asks which polyhedron has the maximum volume when the sum of all edge lengths (the perimeter) equals 1.

Analysis of Known Results

Let me verify the volumes of the candidates mentioned:

1. Regular tetrahedron: With 6 edges of length 1/6, the volume is: $V = \frac{\sqrt{2}}{12}(1/6)^3 = \frac{\sqrt{2}}{2592} \approx 0...

Let

  L(P)=∑e∈EP|e|

be the total length of the edges of a (convex) polyhedron P and let V(P) be its volume.
Because of the homogeneity of the two quantities we can, without loss of generality, fix L(P)=1 and try to maximise V(P).

1. What is known

• Tetrahedra. If a tetrahedron has fixed total edge length the regular one is volume -maximal (Fejes–Tóth, 1949). For L=1 the regular tetra...