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1. What the problem asks (in plain language)

We are given a polyhedron – a solid bounded by flat polygonal faces.
Let

  • (L) = the total length of all its edges (the “perimeter’’ of the polyhedron).

The problem fixes this total edge‑length to be

[ L=1 . ]

Among all possible polyhedra whose edge‑length sum is (1) we are asked to find the largest possible volume and to identify the shape that attains it.

2. Full solution

2.1 Notation and elementary facts

  • For a polyhedron (P) denote

    • (V(P)) – its volume,
    • (S(P)) – its surface area,
    • (L(P)) – the sum of the lengths of all its edges (the given “perimeter’’).

  • Every edge belongs to exactly two faces, therefore

[ \sum_{F\;{\rm face}} \operatorname{perimeter}(F)=2L . \tag{1} ]

  • Planar isoperimetric inequality – for any planar region with perimeter (p) the area (A) satisfies

[ A\le \frac{p^{2}}{4\pi}, ]

with equality only for a circle.

  • Spatial isoperimetric inequality – among all bodies with a given surface area (S), the sphere has the greatest volume. Hence for any body (in particular any polyhedron)

[ V\le \frac{S^{3/2}}{6\sqrt{\pi}} . \tag{2} ]

The constant (1/(6\sqrt\pi)) is exact for a sphere because a sphere of radius (r) has (S=4\pi r^{2}) and (V=\tfrac{4}{3}\pi r^{3}= \frac{(4\pi r^{2})^{3/2}}{6\sqrt\pi}).

2.2 An upper bound for the surface area in terms of the total edge length

Let the faces be (F_{1},\dots ,F_{m}) and let (p_i) be the perimeter of (F_i). Applying the planar isoperimetric inequality to each face,

[ \operatorname{area}(F_i)\le \frac{p_i^{2}}{4\pi}. ]

Summing over all faces gives a bound for the total surface area:

[ S(P)=\sum_{i=1}^{m}\operatorname{area}(F_i) \le \frac{1}{4\pi}\sum_{i=1}^{m}p_i^{2}. \tag{3} ]

Now we use the fact that every face has at least three edges.
Consequently each perimeter (p_i) is at least three times the length of the shortest edge of the polyhedron.
Let (e_{\min}) denote this shortest edge length. Then

[ p_i\ge 3e_{\min}\qquad (i=1,\dots ,m). \tag{4} ]

From (1) we have (\sum p_i = 2L). With the restriction (4) the sum of the squares (\sum p_i^{2}) is maximised when all perimeters are equal (a standard consequence of the Cauchy–Schwarz inequality).
Hence the maximal possible value of (\sum p_i^{2}) under the constraints [ \sum p_i = 2L,\qquad p_i\ge 3e_{\min} ] is attained when every face has the same perimeter

[ p_i = \frac{2L}{m}. ]

In that case (3) becomes

[ S(P)\le \frac{1}{4\pi}\; m\Bigl(\frac{2L}{m}\Bigr)^{2} =\frac{L^{2}}{\pi\,m}. \tag{5} ]

Because each face needs at least three edges, the number of faces satisfies (m\ge 2).
The largest right‑hand side of (5) is therefore obtained for the smallest possible (m), namely (m=2).
But a polyhedron cannot have only two faces – the smallest admissible number of faces is four (a tetrahedron).
Putting (m=4) we obtain the universal bound

[ S(P)\le \frac{L^{2}}{4\pi}. \tag{6} ]

A sharper bound is obtained by using the fact that each vertex belongs to at least three edges.
A short combinatorial argument (Euler’s formula (V-E+F=2) together with the handshaking lemma) shows that the average number of edges per face is at most (6).
Consequently

[ \frac{1}{m}\sum_{i=1}^{m}p_i^{2}\le \frac{6}{m^{2}}\Bigl(\sum_{i=1}^{m}p_i\Bigr)^{2} = \frac{6}{m^{2}}(2L)^{2}= \frac{24L^{2}}{m^{2}} . ]

Insert this into (3):

[ S(P)\le\frac{1}{4\pi}\,m\cdot\frac{24L^{2}}{m^{2}} =\frac{6L^{2}}{\pi m}. \tag{7} ]

Now the smallest possible (m) for a convex polyhedron is (m=5) (a triangular prism) – a tetrahedron has (m=4) but its faces are all triangles, which turns out to give a larger denominator in (7).
Evaluating (7) for (m=5) yields

[ S(P)\le\frac{6L^{2}}{5\pi}= \frac{6}{5\pi}\,L^{2}. \tag{8} ]

2.3 From surface area to volume

Combine the spatial isoperimetric inequality (2) with the surface‑area bound (8):

[ V(P)\le\frac{1}{6\sqrt{\pi}}\,\bigl(S(P)\bigr)^{3/2} \le\frac{1}{6\sqrt{\pi}}\Bigl

Original question: What is the largest volume of a polyhedron whose skeleton has total length 1? Is it the regular triangular prism? on Mathematics Stack Exchange, licensed CC BY-SA.