Bowl Drinking
Have you ever drunk coffee in a bowl? When I did, I realized that the same amount of coffee somehow looks less plentiful in the bowl, but it works just as well as a cup for drinking. This got me thinking: suppose you are given 1 square meter of material which you can arrange into any geometric surface you desire. It could be a bowl, a cup, a fractal, or any other shape that you could potentially physically realize. The problem is simple: how should you shape the square meter of material to hold the most coffee? How much coffee would this hold?
Approach
I’ll state some sensible assumptions: the sheet is impermeable and has negligible thickness. The container can not self-intersect or stretch, and a constant force of gravity is present (which makes coffee flow over the sides of the container and limits the maximum volume of coffee that you can hold).
Ideally the problem would be rigorously defined mathematically, but the volume of “coffee” that a surface can “hold” is difficult for me to rigorously define. You’ll have to rely on intuition to interpret what this means for various surfaces. I can only tell you that the maximum volume of coffee that the surface $S$ can hold is the functional $V[S]$ (the volume depends only on the surface). When it comes to computing the volume of coffee that some particular surface can hold, it will have to be done by intuitively reasoning about how to do so.
Here, to model the real world, we are working in 3-dimensional Euclidean space. I will use the orthonormal coordinate frame such that the $xy$ plane is perpendicular to gravity, and gravity points in the $-z$ direction.
First, notice that volume scales like $V\propto s^3$ for some uniform scaling factor $s$. You can think of the volume of an object as being some proportion of its bounding box, and scaling the object (and its bounding box by extension) does not affect this proportion. Second, notice that surface area scales like $A\propto s^2$. Thus, what we can first do is create some sort of geometry with a surface area $A>0$ which contains a volume of coffee $V>0$, and then we can scale the geometry exactly such that its surface area is 1. Then, we want to maximize the resulting scaled volume. Let’s call the scaled surface area and scaled volume $A_2$ and $V_2$. Then
$$A_{2}=s^{2}A=1\implies s=\frac{1}{\sqrt{A}},$$ $$V_{2}=\sqrt{V_{2}^{2}}=\sqrt{s^{6}V^{2}}=\sqrt{A^{-3}V^{2}}$$
Thus, instead of requiring that we construct some geometry out of a certain surface area of material, we can simply construct any geometry and maximize $A^{-3}V^{2}$, which is an invariant quantity under uniform scaling. The maximum volume of coffee that will be held, then, will be $\sqrt{A^{-3}V^{2}}$.
Bound
There are two potential solutions: a closed container and an open container. The closed container is already a well-known problem, called the isoperimetric problem, first posed around the 2nd century B.C. by Zenodorus in a writing that was lost. It wasn’t until the 20th century that it was completely rigorously proven that the optimal closed container that solves this problem in 3D is a sphere. In particular, the final piece of the puzzle was in 1992 when Bruce Kleiner proved the Cartan–Hadamard conjecture from 1926. Because of this, we can provide a definitive lower bound on the maximum coffee that can be held by the container.
The unit sphere has volume $\frac{4}{3}\pi$ and surface area $4\pi$, so the lower bound on the maximum volume of coffee that we can hold is $V_2=\frac{1}{6\sqrt{\pi}}\approx0.0940316;m^3$.
Possible Solution
As I will show, the optimal geometry is an open container, and I hypothesize that it has a radial symmetry (but I have no satisfactory reason explaining why). My first step to improving the lower bound is to consider various types of shapes and determine the optimal geometry from those general types of shapes.
First, I think that an optimal solution may be a spherical cap, which is one of the portions of a sphere that has been cut by a plane. If we cut the unit sphere a distance $h$ from the top, then the portion above the cutting plane will have a volume $V=\frac{\pi}{3}\left(3-h\right)h^{2}$ and surface area $A=2\pi h$. Thus, the maximum volume of coffee that can be held shrinking this to a surface area of 1 is $V_2=\frac{\left(3-h\right)\sqrt{h}}{6\sqrt{2\pi}}$. Because $V_{2}’’(h)<0$ for $x\in(0,2)$, $V_2$ is maximized when $0=V_{2}’\left(h\right)=\frac{1-h}{4\sqrt{2\pi h}}$. This occurs when $h=1$, meaning the optimal sliced sphere is a hemisphere, which holds a maximum volume of $V_2=\frac{1}{3\sqrt{2\pi}}\approx0.13298;m^3$.
I will now show that two other geometries are not optimal because they are worse than this.
A cylinder with a height $h$ with the unit circle as its base has a volume $V=\pi h$ and surface area $A=2\pi\left(h+1\right)$. Thus $V_2=\frac{h\left(1+h\right)^{-\frac{3}{2}}}{2\sqrt{2\pi}}$. $V_{2}’’\left(h\right)<0$ for $h\in\left(0,\infty\right)$, so the optimal cylinder occurs when $0=V_{2}’\left(h\right)=\frac{\left(h-2\right)\left(1+h\right)^{-\frac{5}{2}}}{4\sqrt{2\pi}}$. This occurs when $h=2$, so $V_2=\frac{1}{3\sqrt{6\pi}}\approx0.076776$. That’s worse than the closed sphere!
Finally, consider a rectangular prism with a side length 1, and two other side lengths $a$ and $b$. It has volume $V=ab$ and surface area $A=2\left(a+b+ab\right)$. Thus $V_2=\frac{ab}{2\sqrt{2}}\left(ab+a+b\right)^{-\frac{3}{2}}$. For positive $a$ and $b$, the determinant of the Hessian is negative, so the local extremum that we find will be a global maximum. The extremum occurs when the gradient of $V_2$ is 0. That is: $$0=\frac{d}{da}V_{2}=\frac{\left(\left(2-a\right)b+a\right)b}{4\sqrt{2}\left(ab+a+b\right)^{5/2}}$$ $$0=\frac{d}{db}V_{2}=\frac{\left(\left(2-b\right)a+b\right)a}{4\sqrt{2}\left(ab+a+b\right)^{5/2}}$$ So $$\left(\left(2-b\right)a+b\right)a=0$$ $$\left(\left(2-a\right)b+a\right)b=0$$ Recall that $a$ and $b$ are positive. This this system is satisfied when $a=b=1$, in which case, we have the unit cube. Then $V_2=\frac{1}{6\sqrt{6}}\approx0.0704138;m^2$: the worst one yet!
Conclusion
With this, I hypothesize that the geometry that solves this problem is a hemisphere. However, I have no good reason to believe that this is the single optimal geometry, and I do not know how to prove it. If anyone discovers any updates/improvements/solutions to my problem, please let me know.