I was writing some exercises about the AM-GM inequality and I got carried away by the following (pretty nontrivial, I believe) question:

Q: By properly folding a common $210mm\times 297mm$ sheet of paper, what is the maximum amount of water such a sheet is able to contain?

enter image description here

The volume of the optimal box (on the right) is about $1.128l$. But the volume of the butterfly (in my left hand) seems to be much bigger and I am not sure at all about the shape of the optimal folded sheet. Is is something boat-like?

Clarifications: we may assume to have a magical glue to prevent water from leaking through the cracks, or for glueing together points of the surface. Solutions where parts of the sheet are cut out, then glued back together deserve to be considered as separate cases. On the other hand these cases are trivial, as pointed by joriki in the comments below. The isoperimetric inequality gives that the maximum volume is $<2.072l$.

As pointed out by Rahul, here it is a way for realizing the optimal configuration: the maximum capacity of the following A4+A4 bag exceeds $2.8l$.

enter image description here

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    $\begingroup$ This is a very nice problem! $\endgroup$ – Dr. Sonnhard Graubner Jul 18 '18 at 20:36
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    $\begingroup$ It seems a very hard problem to me, related to geometric calculus of variations. Maybe it deserves a big bounty! $\endgroup$ – user Jul 18 '18 at 20:38
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    $\begingroup$ Well, I think if we're allowed to cut and paste, then we should be able to get arbitrarily close to the optimal hemisphere by triangulating it? $\endgroup$ – joriki Jul 18 '18 at 21:18
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    $\begingroup$ Let us enjoy maximum-volume tacos while we ponder the optimal container to hold our salsa :) I look forward to the forthcoming question on mathematically ideal nachos. $\endgroup$ – user856 Jul 19 '18 at 5:43
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    $\begingroup$ It reminds me of a joke... An engineer, a physicist and a mathematician are given the same amount of net fence and any number of poles to build a fence that gives more area that of the others. So the engineer starts and creates a perfect square, looking at the other two triumphantly. Then the physicist takes over and builds a fence that is as close to a circle as possible and looks at the other two triumphantly. Then mathematician starts building the fence placing poles casually and just gets some strange enclosed shape. Stands in the middle and claims "I'm outside!" $\endgroup$ – Ister Jul 19 '18 at 11:48

This problem reminds me of tension field theory and related problems in studying the shape of inflated inextensible membranes (like helium balloons). What follows is far from a solution, but some initial thoughts about the problem.

First, since you're allowing creasing and folding, by Nash-Kuiper it's enough to consider short immersions $$\phi:P\subset\mathbb{R}^2\to\mathbb{R}^3,\qquad \|d\phi^Td\phi\|_2 \leq 1$$ of the piece of paper $P$ into $\mathbb{R}^3$, the intuition being that you can always "hide" area by adding wrinkling/corrugation, but cannot "create" area. It follows that we can assume, without loss of generality, that $\phi$ sends the paper boundary $\partial P$ to a curve $\gamma$ in the plane.

We can thus partition your problem into two pieces: (I) given a fixed curve $\gamma$, what is the volume of the volume-maximizing surface $M_{\gamma}$ with $\phi(\partial P) = \gamma$? (II) Can we characterize $\gamma$ for which $M_{\gamma}$ has maximum volume?

Let's consider the case where $\gamma$ is given. We can partition $M_{\gamma}$ into

1) regions of pure tension, where $d\phi^Td\phi = I$; in these regions $M_{\gamma}$ is, by definition, developable;

2) regions where one direction is in tension and one in compression, $\|d\phi^Td\phi\|_2 = 1$ but $\det d\phi^Td\phi < 1$.

We need not consider $\|d\phi^Td\phi\|_2 < 1$ as in such regions of pure compression, one could increase the volume while keeping $\phi$ a short map.

Let us look at the regions of type (2). We can trace on these regions a family of curves $\tau$ along which $\phi$ is an isometry. Since $M_{\gamma}$ maximizes volume, we can imagine the situation physically as follows: pressure inside $M_{\gamma}$ pushes against the surface, and is exactly balanced by stress along inextensible fibers $\tau$. In other words, for some stress $\sigma$ constant along each $\tau$, at all points $\tau(s)$ along $\tau$ we have $$\hat{n} = \sigma \tau''(s)$$ where $\hat{n}$ the surface normal; it follows that (1) the $\tau$ follow geodesics on $M_{\gamma}$, (2) each $\tau$ has constant curvature.

The only thing I can say about problem (II) is that for the optimal $\gamma$, the surface $M_\gamma$ must meet the plane at a right angle. But there are many locally-optimal solutions that are not globally optimal (for example, consider a half-cylinder (type 1 region) with two quarter-spherical caps (type 2 region); it has volume $\approx 1.236$ liters, less than Joriki's solution).

I got curious so I implemented a quick-and-dirty tension field simulation that optimizes for $\gamma$ and $M_{\gamma}$. Source code is here (needs the header-only Eigen and Libigl libraries): https://github.com/evouga/DaurizioPaper

Here is a rendering of the numerical solution, from above and below (the volume is roughly 1.56 liters).

enter image description here

EDIT 2: A sketch of the orientation of $\tau$ on the surface:

enter image description here

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    $\begingroup$ @user202729 The point is that the water filling the surface will have a planar interface with the air, so you might as well delete (by folding e.g.) all parts of the surface above the waterline. What's left is a surface with a coplanar boundary. $\endgroup$ – user7530 Jul 19 '18 at 9:20
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    $\begingroup$ @user202729 ok, it's true that one would need to carefully prove that it's not optimal to fold the surface so that the water collects in multiple disconnected basins. $\endgroup$ – user7530 Jul 19 '18 at 9:24
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    $\begingroup$ @user202729 yes, fair enough. The above analysis assumes you are allowed to make only local geometric edits to the shape by folding/gluing. If you allow gluing together arbitrary parts of the boundary, I'm afraid I don't have a good idea of how to even begin a systematic attack. $\endgroup$ – user7530 Jul 19 '18 at 9:41
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    $\begingroup$ P.S. This is the solution of the paper bag problem cut in half, isn't it? $\endgroup$ – user856 Jul 19 '18 at 15:23
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    $\begingroup$ @user7530 Would it be possible to mark the regions of compression and/or tension streamlines on the renders? It's hard to see from the plot if any exist, and you make a point of it in your answer. Great work anyway! $\endgroup$ – Mario Carneiro Jul 21 '18 at 10:17

This is equivalent to the paper bag problem, which asks for the maximum possible volume attainable by inflating an initially flat rectangular pillow made of inextensible material. Separate the two sides of the pillow while keeping their shape, and you obtain (two copies of) your optimal sheet.

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    $\begingroup$ This is interesting since it provides a mechanical way for realizing the optimal configuration: it is enough to glue together two A4 sheets along their boundaries, then inflate air inside. $\endgroup$ – Jack D'Aurizio Jul 19 '18 at 17:42
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    $\begingroup$ That assumes that there is no solution where one (or more) edges of the sheet are curved (i.e., thin curved slices cut out), that yields a greater volume. I don't know if that assumption is correct. Do you know of any way to prove, or a reason to assume, that a slightly smaller non-rectangular sheet cannot have even larger volume? (Stuffing a pillow case, or blowing up a mylar balloon constructed of two sheets glued together, seems to indicate that a smaller sheet cannot have larger volume.. but I just don't know. I'm only asking, not claiming anything.) $\endgroup$ – Nominal Animal Jul 19 '18 at 22:50
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    $\begingroup$ @NominalAnimal: If you have a solution with part of the square removed, you can achieve the same with the full square by crushing/wrinkling the extra material to "hide" it, as argued in the first section of user7530's answer. $\endgroup$ – user856 Jul 20 '18 at 5:30
  • $\begingroup$ @Rahul: Ah, right; the corrugations/folds having zero volume, i.e. flat along the surface, with tight creases. I think I was hung up on the curved edge idea, and didn't realize the creases can be short, shaped like e.g. lenses in the middle, or quadrilaterals/diamonds near the corners, to provide exactly that. Thanks! $\endgroup$ – Nominal Animal Jul 20 '18 at 14:20

This is certainly not optimal, but comparatively straightforward to calculate and a moderate improvement over the previous solutions.

If we let the sides of the box fall outward, the top rectangular area of the resulting prism increases to first order whereas the height only decreases to second order, so there's a non-zero optimal angle of inclination for the sides.

Let $x=210\text{mm}$ be the width and $y=297\text{mm}$ the length of the paper, and introduce three variables: the height $h$, the angle of inclination $\phi$ of the long sides and the angle of inclination $\xi$ of the short sides. Then at height $\alpha h$, with $0\le\alpha\le1$, a rectangular cross-section of the prism of height $h\mathrm d\alpha$ has volume

$$ \left(x-2\frac h{\cos\phi}+\alpha\cdot2h\tan\phi\right)\left(y-2\frac h{\cos\xi}+\alpha\cdot2h\tan\xi\right)h\mathrm d\alpha\;, $$

and integrating over $\alpha$ yields the volume

$$ \left(\left(x-2\frac h{\cos\phi}+h\tan\phi\right)\left(y-2\frac h{\cos\xi}+h\tan\xi\right)+\frac13h^2\tan\phi\tan\xi\right)h\;. $$

I don't see how to get a closed form for the optimal parameters, but I optimized them numerically, with the result

$$ h\approx47.62\text{mm}\;,\\ \phi\approx0.55112\;,\\ \xi\approx0.56838 $$

and resulting volume

$$ V\approx1.315679370667\,l\;. $$

Here's a rough attempt at building this:

paper holds water glued


This is the first picture I took, before I realized that I could glue the corners to force the paper to stay in the prism shape. After seeing the images that came out of user7530's cool simulation, I'm now thinking that the paper was trying to take on the optimal form and I was just interfering with it :-)

paper holds water all by itself

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    $\begingroup$ Just a note on the version with glued corners -- those wasted triangles could be opened into cones. $\endgroup$ – amI Jul 19 '18 at 20:40
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    $\begingroup$ @aml's suggestion raises the interesting question whether all of the water has to be contained in a single connected volume with a common surface level, or there can be isolated "pools" on the side whose walls don't reach the level of the main surface. $\endgroup$ – hmakholm left over Monica Jul 20 '18 at 0:50
  • $\begingroup$ can we obtain a similar result by "pleating" the border, like in aluminium trays (?) $\endgroup$ – G Cab Jul 21 '18 at 13:21

This may not be the optimum. But an easy solution with larger volume than the box. 1.14228 l. (Of course we need to tape it at base to hold)

Solution explained

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    $\begingroup$ I believe that something more spherical-shaped like a boat is able to contain $\geq 1.2l$. $\endgroup$ – Jack D'Aurizio Jul 18 '18 at 21:05
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    $\begingroup$ This solution is feels legitimate because plastic bottles are made of sheets and to waste less material many optimisation studies must have been done to maximise volume... my non-mathematical 2 cents...I apologise. $\endgroup$ – Pi_die_die Jul 18 '18 at 21:14
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    $\begingroup$ @Pi_die_die: bottles have to be transported, too, and spherical or hemispherical bottles do not fill the 3d-space too efficiently. $\endgroup$ – Jack D'Aurizio Jul 18 '18 at 21:23
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    $\begingroup$ @Pi_die_die: if bottles were made for minimizing the ratio surface/volume they would be spherical, not cylindrical. But cylinders fill the 3d-space a bit better than spheres. $\endgroup$ – Jack D'Aurizio Jul 18 '18 at 21:37
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    $\begingroup$ Thank you for the explanation @joriki and OP for the clarification... I had a very enjoyable discussion. $\endgroup$ – Pi_die_die Jul 18 '18 at 22:29

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