Suppose you are given a single unit square, and you are permitted to cut it into $k$ (connected) pieces (where $k=1$ means just the square). Your task is to construct the largest volume convex body in $\mathbb{R}^3$ by pasting the $k$ pieces together as its complete surface; so its surface area is $\le 1$. Denote its volume by $V_\max(k)$. (This is a more general version of my earlier question, Covering a Cube with a Square.)

The best one could hope for is to create a sphere with unit surface area, when the radius satisfies $4 \pi r^2 =1$ and so $r=\frac{1}{2\sqrt{\pi}} \approx 0.28$, and the volume is $V_\max(\infty)= V_\max = \frac{4}{3} \pi r^3 = \frac{1}{6 \sqrt{\pi}} \approx 0.094$.

In response to a question of Joe Malkevitch, with students I computed the optimum for $k=1$, with no cuts and no overlap, and found a maximum volume of $V_\max(1)\approx 0.056$, which is about 60% of the sphere volume. [p.418 in Geometric Folding Algorithms: Linkages, Origami, Polyhedra]:
            Max Volume Polyhedron
The seeming randomness of this shape discouraged my further pursuit.

The case $k=2$ resembles the famous tea-bag problem, but here I am restricting attention to convex bodies.

As $k \to \infty$, one should be able to approach $V_\max$ by, for example, cutting the square into many nearly equilateral triangles and constructing a geodesic dome.

I am wondering if anyone has heard of this problem in any guise before? And in any case, can anyone see any clear hypothesis, for any particular $k$? Presumably one could establish that $V_\max(k+1) > V_\max(k)$, but beyond that, I do not see how to make inroads. Is there an obvious candidate for $k=2$?

Update. Here is my interpretation of Gerhard's suggestion for $k=2$:
           Square to Box

  • $\begingroup$ Why is the $k=1$ case "no cuts" and not "one connected piece"? For the cases where cuts are allowed, is it clear that the supremum is realizable? On the side of the convex shape, Blaschke selection says the limit exists, but on the side of the cut up square, the limit could in theory involve thinner and thinner connected spiral strips. $\endgroup$ – Yoav Kallus May 6 '12 at 2:20
  • $\begingroup$ I thinl V_max(k) is unbounded as k goes to infiinity. I assume I am not supposed to cover the entire convex body, but just arrange a connected shape and then take the connvex hull. In that case, cut k many strips to form unit length edges of your favorite polyhedron. Gerhard "Also, I Could Have Misunderstood" Paseman, 2012.05.05 $\endgroup$ – Gerhard Paseman May 6 '12 at 3:43
  • $\begingroup$ For k=2 there is the 1 by 2 by 2 brick with volume 1/16. It is hard to see how to do better. Gerhard "Correct Up To Scaling Factor" Paseman, 2012.05.05 $\endgroup$ – Gerhard Paseman May 6 '12 at 6:08
  • $\begingroup$ @Gerhard: I clarified that I meant to cover the entire surface. Your $k=2$ idea is quite plausible! $\endgroup$ – Joseph O'Rourke May 6 '12 at 11:33
  • $\begingroup$ @Yoav: You are right, there could be a difference between one piece without cuts, and one piece with cuts. And I think it likely that $V_\max$ is unachievable for a finite $k$. $\endgroup$ – Joseph O'Rourke May 6 '12 at 11:36

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