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

Simon Fraser
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Jack D'Aurizio
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    This is a very nice problem! – Dr. Sonnhard Graubner Jul 18 '18 at 20:36
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    It seems a very hard problem to me, related to geometric calculus of variations. Maybe it deserves a big bounty! – user Jul 18 '18 at 20:38
  • +1, nice question. Could you clarify whether "taping at the base" as in AppoopanThaadi's solution is allowed? That is, can we consider the paper as an abstract two-dimensional surface, and water is "contained" wherever this surface is; or can water "flow through the cracks" if the surface is folded onto itself? (Your two examples don't require such glueing.) – joriki Jul 18 '18 at 21:09
  • @orlp: it does not *falls flat* if you fold the corners: https://ibb.co/hNMbpJ – Jack D'Aurizio Jul 18 '18 at 21:11
  • @JackD'Aurizio Right, but that wasn't in your original design. So are we're supposed to take that into account or not? What about the finger pinching the paper? I don't believe the question is answerable without a clear concise model of what we're trying to solve here. – orlp Jul 18 '18 at 21:12
  • I have added some clarifications to the main question. – Jack D'Aurizio Jul 18 '18 at 21:17
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    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? – joriki Jul 18 '18 at 21:18
  • @joriki: you are right, if cut&paste is allowed the problem is trivial. – Jack D'Aurizio Jul 18 '18 at 21:20
  • You could approximate the volume of each of the "sides" by cones, assuming you could pinch it perfectly, and hold water, etc. The area of these cones should give you a better idea of if it *seems* larger or if it *is* larger. – Chickenmancer Jul 18 '18 at 22:30
  • A half-cylinder with the axis along the length of the paper can have radius $\frac x\pi$ and length $y-2r$ (it would require some intricate crumpling at the two semicircular walls), and thus volume $$ \frac\pi2r^2l=\frac{x^2}\pi\left(\frac y2-\frac x\pi\right)\approx1.146228\,l\;, $$ very slightly greater than the box and AppoopanThaadi's full cylinder along the other axis. – joriki Jul 18 '18 at 22:36
  • With an ideal piece of paper, can the butterfly shape even be formed? I feel like it requires slightly stretching the sheet of paper at the creases/corners. – BallpointBen Jul 19 '18 at 05:35
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    Let us enjoy [maximum-volume tacos](https://math.stackexchange.com/q/91937/856) while we ponder the optimal container to hold our salsa :) I look forward to the forthcoming question on mathematically ideal nachos. –  Jul 19 '18 at 05:43
  • @user202729 I’m indeed referring to the answer! – user Jul 19 '18 at 07:20
  • Is a flat torus an allowed shape here? ( http://www.science4all.org/article/flat-torus/ ) – asmaier Jul 19 '18 at 07:47
  • @asmaier Given that it's not convex, I don't think it will be able to hold more water than a normal torus (with the internal part "shrunken"). – user202729 Jul 19 '18 at 08:06
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    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*!" – Ister Jul 19 '18 at 11:48
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    If the A4+A4 bag uses two pieces of paper, does it not stand the reason that the optimal would probably be somewhere near the half of its volume? – jpmc26 Jul 19 '18 at 22:08
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    @jpmc26: of course, without any doubt. If we are able to inflate $\approx 3l$ inside such IV-bag, then the maximum capacity of a A4 sheet is around $1.5l$. – Jack D'Aurizio Jul 20 '18 at 01:00
  • So, the basic box achieves $72\%$ of the optimum, and @joriki's "slanted" box attains $84\%\,$. This is a good case for whatever the antonym to "diminishing returns" may be, when putting heavy machinery to work in what (deceivingly) looks like a simple problem ends up helping a great $16\%$ lot. My +1 to all. – dxiv Jul 20 '18 at 03:42
  • @Ister: That reminds me of the first few paragraphs of Ursula K. Le Guin's *The Dispossessed*. She makes a surprisingly convincing case for the mathematician's claim. :-) – ruakh Jul 21 '18 at 02:53
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    @JackD'Aurizio Why is your phone bragging on those photos? Is that desirable? – wizzwizz4 Jul 22 '18 at 11:41
  • I want MSE to make me available a feature called favorite [question](https://math.stackexchange.com/questions/2855975/what-is-the-maximum-volume-that-can-be-contained-by-a-sheet-of-paper) –  Oct 09 '21 at 18:23

4 Answers4


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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  • Can you explain what is a short immersion? I'm not very familiar with that, and [Wikipedia page](https://en.wikipedia.org/wiki/Immersion_(mathematics)) doesn't say anything about it. – user202729 Jul 19 '18 at 03:51
  • @user202729 Oh, I mean an immersion that does not increase distances: for any tangent vector $v$ on $P$, $\|d\phi v\| \leq \|v\|.$ – user7530 Jul 19 '18 at 03:54
  • I'm not sure about "$\phi$ sends the paper boundary $\partial P$ to a curve $\gamma$ in the plane.". For example, that's probably not true with the hemisphere-capped cylinder. – user202729 Jul 19 '18 at 04:03
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    @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. – user7530 Jul 19 '18 at 09:20
  • But the boundary can be put below the surface as well, you're allowed to glue them together. – user202729 Jul 19 '18 at 09:21
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    @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. – user7530 Jul 19 '18 at 09:24
  • Not disconnected, but like [this solution](https://math.stackexchange.com/a/2855998/261710). – user202729 Jul 19 '18 at 09:26
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    @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. – user7530 Jul 19 '18 at 09:41
  • I'd upvote this again if I could -- great answer! Interestingly, this looks a lot like what the paper kept trying to do while I was trying to fold it into my prism solution. Take a look at the P.S. with the second picture I added to my answer :-) – joriki Jul 19 '18 at 10:29
  • I love the way this answer illustrates how useful it is to construct a simple model, then numerically optimize it, to get a good, practical answer. It also shows how hard it is to prove the result is optimal; I think one needs to prove the initial model leads to an optimal result.. (While I personally love working out such practical results, I don't know enough about topology etc. to develop really good models -- and proving optimality is way, *way* outside my own math-fu!) – Nominal Animal Jul 19 '18 at 13:41
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    P.S. This is the solution of the [paper bag problem](https://en.wikipedia.org/wiki/Paper_bag_problem) cut in half, isn't it? –  Jul 19 '18 at 15:23
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    @Rahul Yes, nice find! Robin's formula gives the estimate 1.452 liters for the optimal volume. – user7530 Jul 19 '18 at 15:30
  • @user7530 How come that's so far from the simulated 1.56? – JollyJoker Jul 20 '18 at 09:00
  • @JollyJoker the formula is just an estimate. According the Wikipedia article there are more accurate ones in Robin’s paper but I haven’t looked into it. – user7530 Jul 20 '18 at 09:02
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    Could you comment on how your code attempts to optimize for $\gamma$? The only thing mentioned in your fantastic answer is that $M_\gamma$ must meet $\gamma$ at a right angle. But how did you compute various $\gamma$? – Greg Hurst Jul 21 '18 at 02:29
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    I did a small [comparison](http://folk.uio.no/hansw/comp.png) of your optimization code with the paper-bag approximate formula in Wiki. The results are qualitatively the same with the main difference being that you have a roughly 7% higher value (tested this for $w/h \in [0.4,2.0]$ and checked convergence wrt the mesh size). Assuming your code has no bugs a likely reason for the discrepancy is that the approximate formula is simply not that accurate, perhaps it was computed with a too low mesh resolution which would lead to a smaller volume. – Winther Jul 21 '18 at 02:45
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    @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! – Mario Carneiro Jul 21 '18 at 10:17
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    What happens if you run your simulation for a disk, instead of a rectangle? – Jack D'Aurizio Jul 23 '18 at 12:48
  • @MarioCarneiro tension streamlines would require some nontrivial coding, but I've added a figure sketching the direction of tension. Don't trust the hourglass region at the top---the surface is very close to being in pure tension there and there is a lot of numerical noise. – user7530 Jul 26 '18 at 19:18

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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    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. – Jack D'Aurizio Jul 19 '18 at 17:42
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    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.) – Nominal Animal Jul 19 '18 at 22:50
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    @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](https://math.stackexchange.com/a/2856108/856). –  Jul 20 '18 at 05:30
  • @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! – 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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    Just a note on the version with glued corners -- those wasted triangles could be opened into cones. – amI Jul 19 '18 at 20:40
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    @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. – hmakholm left over Monica Jul 20 '18 at 00:50
  • can we obtain a similar result by "pleating" the border, like in aluminium trays (?) – 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

Eric Duminil
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    I believe that something more spherical-shaped like a boat is able to contain $\geq 1.2l$. – Jack D'Aurizio Jul 18 '18 at 21:05
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    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. – Pi_die_die Jul 18 '18 at 21:14
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    @Pi_die_die: bottles have to be transported, too, and spherical or hemispherical bottles do not fill the 3d-space too efficiently. – Jack D'Aurizio Jul 18 '18 at 21:23
  • But isn't that because for the same surface area these shapes occupy least volumes? – Pi_die_die Jul 18 '18 at 21:26
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    @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. – Jack D'Aurizio Jul 18 '18 at 21:37
  • That is my point precisely-the ratio is inversely proportional to packing efficiency that is why spheres are not used to make bottles since for the same amt of material they can enclose less volume and as a consequence of the high area to volume ratio they reduce packing efficiency. I propose a prism solution....with longer side as axis. – Pi_die_die Jul 18 '18 at 22:16
  • @Pi_die_die: No, you're confusing packing efficiency and surface-to-volume ratio. Spheres have the minimal surface-to-volume ratio. It just happens to be wasteful to pack them. – joriki Jul 18 '18 at 22:23
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    Thank you for the explanation @joriki and OP for the clarification... I had a very enjoyable discussion. – Pi_die_die Jul 18 '18 at 22:29
  • @JackD'Aurizio Assuming in the cut/paste option (taken to an indiscriminate limit ! ) we cut the A4 paper into infinitely small spherical triangles and re-glue edges to form large $p-$ division Fullerene domes... we have $ 2.1*2.97 = 4 \pi R^2,\, R=0.704503 , \rightarrow \frac43 \pi R^3= 1.46466 $ litres as maximum volume by variational calculus. – Narasimham Jul 19 '18 at 05:56
  • @Narasimham: in such a case the optimal volume is enclosed by a hemisphere, it is $V=\frac{2\pi}{3}\left(\frac{S}{2\pi}\right)^{3/2}\approx 2.071l$. – Jack D'Aurizio Jul 19 '18 at 06:01
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    @Pi_die_die Bottles are usually made by [blow moulding](https://en.wikipedia.org/wiki/Blow_molding), not by gluing sheets of material. – David Richerby Jul 22 '18 at 18:34