Maximum volume enclosed by a piece of cloth in the shape of a unit circle

Question

I have a piece of cloth in the shape of a unit circle (or disk), a needle and thread. I want to stitch the cloth together so that it can contain stuff (say, lots of small beans) and the stuff cannot escape. What is the maximum volume that the cloth can contain?

I guess the optimal shape will be made by folding the cloth in half into a double-layered semicircle, then stitching along the arc (sort of like a dumpling, but with an infinitesimally narrow sealed strip).

I suppose we could physically make such a shape, stuff it with small beans, then pour the beans into another container that can measure volume. But is there a mathematical way to answer this question?

A crude upper bound for the maximum volume can be obtained by (erroneously) assuming that the cloth can be shaped into a sphere without wasting any of the cloth. Then the radius of the sphere, $r$, would satisfy $4\pi r^2=\pi$, so the volume of the sphere, i.e. the crude upper bound, would be $\pi/6\approx0.524$.

I did an internet search and found how to maximize the volume enclosed by a square net (here, here and a mathoverflow question), and an article called Profiles of inflated surfaces. From the comments, we have the paper bag problem. But I haven't seen the shape in this question. It seems that my question may be quite difficult; perhaps we can only find lower and upper bounds.

Answer 1

Obviously unanswered as of yet (w/o segmenting the cloth), I dare to suggest the simplest to calculate. Because the only tools are needle and thread, but no scissors, a change of the flat shape to "get" volume is possible by tucks.

Sewing infinitesimal many tucks in the border I may shape a becher (cupule, mug, beaker, paddling pool, youknowit), a circle as bottom with radius $r_b=r-h$ (with $r$ the radius of the given cloth in the shape of a unit circle), and a border height of $h$. Thus the volume is

$\pi(r-h)^2h$

derived once for $h$ is

$\pi\left(3h^2-4hr+r^2\right)$

set to null and solved for $h$ results

{$\displaystyle h_1=\frac{r}{3}$, $h_2=r$}

While the second solution is for sure a minimum, the first is a maximum.
The volume equation derived twice for $h$ at $h_1$

$2\pi\left(3h-2r\right)$ is $\lt 0$.

The volume of that resulting becher is $\displaystyle 4\pi\left (\frac{r}{3}\right )^3$

Remains the question, how about other shapes, a bowl, a hemisphere, something in between like a calotte?

Addendum: Shaping the given cloth with many tuks into a hemisphere, the diameter of the cloth $2r$ will be the half of the circumference of the resulting sphere. So its radius is $\displaystyle r_s=\frac{2r_c}{\pi}$ with $r_c=1$ the radius of the cloth. The volume of the resulting hemisphere is

$\displaystyle \frac{2}{3}\pi r_s^3=\frac{2}{3}\pi\left (\frac{2r_c}{\pi}\right )^3$ what is a bit more than the volume of the a. m. becher.

Addendum 2: Following the references in OP I found a paper describing also "The Mylar Balloon", chapter 4, p.58 ff. This triggered me to test an approximation of it, consisting of a cylinder and a quarter torus. So my ansatz is, to use $r_c-x$ for the bottom of the cylinder and height $h=f(x)$. $r_c$ is the radius of the given cloth, while with part $x$ of it I will form the torus. Thus this will be the circumference of the quarter torus attached to the outside of the cylinder. From $2\pi h=4x$ the resulting little radius of the torus and the height of the cylinder is $\displaystyle\frac{2x}{\pi}$.
The volume then is

$\displaystyle\pi(r-x)^2h+\frac{\pi h^2\left (\pi(r-x)+\displaystyle\frac{4h}{3}\right )}{2}$   Note: attached to the cylinder is the outer part of the torus, but only half of it as it's open to the top.
Replacing $h$ by a. m. $f(x)$

$\displaystyle\frac{2x\left(3\pi^2 r^2-3\pi^2 rx+8x^2\right)}{3\pi^2}$   A short check with $x=0$ (no cylinder, torus only) results a hemisphere, thus I'm convinced my ansatz could be ok.
The now following procedure is well known, derive once for $x$, solve for $x$, one result is not possible since $x\gt r_c$ (there is not more cloth available than $\displaystyle r_c^2\pi$), check for minimum with 2nd result

$\displaystyle x_2=\frac{-\left(\sqrt{\pi^2-8}\mathrm{abs}\left(r\right)-\pi r\right)\pi}{8}$   set $r=r_c=1$ and substitue with it $x$ in ansatz equation. Result:

$\displaystyle \frac{\left(\sqrt{\pi^2-8}\pi-\pi^2+\mathrm{16}\right)\left(\pi -\sqrt{\pi^2-8}\right)\pi}{\mathrm{96}}\approx \mathrm{0.60536}$   what is about 99,36% compared with a mylar balloon in a. m. paper.

Conclution: I give up. Why? Testing different shapes may only show "better than all before", what is no proof there could not exist another one even better but, alas, not found so far. An ansatz may find min and max of it, but (AFAIK) there is no method which may present a formula as result.

Answer 2

I am assuming that the given cloth can be arbitrarily cut and re-glued to enclose volume.

Using Variational calculus it can be established that:

Given area A and volume V, all minimal area surfaces of constant mean curvature H (instead of a single radius R) are satisfied by DeLaunay Unduloids and their combinations

$$ H =\frac{2A}{3V}=\frac{1}{r} = 1~ $$

being equivalent to a sphere, that cannot be exceeded.

A roughly sketched example of meridians of maximum 3D volume for given surface area.

EDIT1:

Alternately the warp and weft fibers should be rotated in order to make an iso-lineal mapping where fiber length is conserved for maximum enclosed volume.