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Suppose I want an open-topped cup to have the capacity to hold a volume $V$ of liquid. I want to find a shape for my cup that minimises its surface area.

My attempt

I strongly suspect that the shape of this cup will be rotationally symmetric, prompting me to use surface area and volume of revolution.

Suppose my cup has shape $f(x)$ and height $H$, and then revolved around the x-axis. Then the volume of my cup is \begin{equation} \displaystyle\int_0^H \pi (f(x))^2 dx \end{equation} (by volume of revolution).

Now the surface area of my cup is \begin{equation} \displaystyle\int_0^H 2\pi f(x)\sqrt{1+(f^{\prime}(x))^2} dx \end{equation} (by surface area of revolution).

So I need to find a function $f$ and a height $H$ that minimises $\displaystyle\int_0^H 2\pi f(x)\sqrt{1+(f^{\prime}(x))^2} dx$, given the restriction $\displaystyle\int_0^H \pi (f(x))^2 dx = V$.

Edit:

Can we find a function $f$ if we are given a $H$? And if a closed form for $f$ does not exist, can we approximate $f$?

Ultra
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  • I'll just comment here in case my procedure is wrong. Using the Euler-Lagrange eq. for the lagrangian $\mathcal L(x,f(x),f'(x))=2\pi f(x)\sqrt{1+f'(x)^2}$ in order to minimize the surface integral, you get $$f(x)f''(x)-f'(x)^2-1=0$$ WA gives $$f(x)=\pm\dfrac{c_1\tanh(c_1(c_2+x))}{\sqrt{\tanh^2(c_1(c_2+x))-1}}$$ Through the boundary condition and the constraint you can find $c_1$ and $c_2$ I think. – Hug de Roda May 05 '24 at 15:49
  • @Conreu tanh is between -1 and 1 so you are square rooting a negative – Ultra May 05 '24 at 15:55
  • You're right, I'll look into it. – Hug de Roda May 05 '24 at 16:03
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    Where is the volume constraint? – Ted Shifrin Jul 21 '24 at 05:49
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    I think the answer will be a sphere with a pin-prick at the top. – Gerry Myerson Jul 21 '24 at 11:20
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    @GerryMyerson You do considerably better with an open hemisphere. If both have volume $1$, the sphere has surface area $\sqrt[3]{36\pi}$, while the open hemisphere has surface area $\sqrt[3]{18\pi}$. – Chris Lewis Jul 21 '24 at 14:08
  • @GerryMyerson Yes, I think ChrisLewis is correct. I believe that if I imposed the constraint $f(0) = f(H) = 0$ then the sphere would be the answer. However, this is not the case – Ultra Jul 21 '24 at 16:03
  • There's a slight issue with the phrasing of the question, and I'm not sure what the best fix is. Nothing at the moment specifies the $x$-coordinate of the bottom of the cup. If the profile of the cup is given by $y=f(x)$ between $x=0$ and $x=H$, the profile $y=g(x)$ between $x=a$ and $x=a+H$ gives the "same" solution (up to translation) where $g(x)=0$ for $0\le x<a$ and $g(x)=f(x-a)$ for $x\ge a$. (This caused a problem when I was attempting the question numerically.) Are there any ways to include a constraint that $f(x)>0$ for $x>0$? – Chris Lewis Jul 22 '24 at 10:56
  • One idea for the approximation: let $N$ be large, and identify how to choose values for $r_k=f(kH/N)$, with $0\le k\le N$, with just lines connecting those points. The surface area is then $2\pi\cdot\frac{H}{N}\sum_{k=1}^N\frac{r_{k-1}+r_k}{2}\sqrt{1+\left(\frac{r_k-r_{k-1}}{H/N}\right)^2}$, and the volume is $\frac{\pi}{3}\cdot\frac{H}{N}\sum_{k=1}^N\left(r_{k-1}^2+r_{k-1}r_k+r_k^2\right)$ – Varun Vejalla Jul 22 '24 at 22:42
  • @VarunVejalla That's exactly the numerical approach I tried :-). Again, the problem is that nothing "forces" the bottom of the cup to be at $x=0$, so numerical optimisation methods have difficulty converging as there's no unique solution. – Chris Lewis Jul 23 '24 at 08:00
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    @Cristof012 hang on. Are we all overthinking this? Join two of the cups together and solve the equivalent problem (minimise surface area fully enclosing a given volume). This is just the isoperimetric problem in 3D whose answer (eg here https://en.m.wikipedia.org/wiki/Isoperimetric_inequality ) is well known to be a sphere. So the optimal cup is a hemisphere. – Chris Lewis Jul 23 '24 at 14:32
  • @ChrisLewis I can't see any mistakes in your reasoning. Maybe it is just that simple! – Ultra Jul 23 '24 at 15:17
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    @Cristof012 both answers below are different. I've asked the two solvers to post the actual surface area corresponding to their solutions. If they don't beat a hemisphere I'll post this as a solution! (I can't believe how much time I and others have spent trying to do this with calculus of variations...it's got some really interesting rabbit-holes, as I've mentioned above, but perhaps they aren't really needed!) – Chris Lewis Jul 23 '24 at 15:22
  • @ChrisLewis I don't think this works when $H$ is given, since we are more limited in how "tall" the whole cup can be. On a separate note, if you don't want it to be open-ended on both ends, I think you need to add $\pi f(0)^2$ to the surface area formula you have. – Varun Vejalla Jul 23 '24 at 15:39
  • @VarunVejalla Yes that is true, I suppose finding a solution for a given $H$ would be better, and I should maybe have phrased it like that. However, I had in mind to minimise over all $H$ in the end, which ChrisLewis pointed out would be a hemisphere. – Ultra Jul 23 '24 at 16:01

3 Answers3

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Promoting my comment to an answer.

If two such optimal cups are joined together to form a single closed surface, that surface is the solution to the question "what is the surface with the least surface area that bounds a volume of $2V$?", which is well known to be a sphere (see, for example, here). So the answer to this question is a hemisphere.

This is a little unsatisfactory. Since it might be helpful, here's a summary of how a calculus of variations approach could work.

The interesting aspect of this is that $H$, the height of the cup, is to be found. This is not the case for the "usual" formulation of a calculus of variations problem. The difficulty with this is that an optimal cup can be translated along the $x$-axis without changing its volume or surface area. The translated cup has a different $H$ but is still a solution. This ambiguity means numerical methods don't consistently converge: my initial attempts would tend to hemispheres, but sometimes with a zero-thickness "stem" at the bottom of the cup. It's also the reason analytical approaches are so difficult.

With that in mind, a two-step approach may work better:

  1. Solve the problem "what shape of cup with height $h$ bounding a volume $V$ has the least surface area?" (This is a standard CoV problem; solutions will vary between elongated champagne flutes and flatter coupes)
  2. This defines a function $A(h)$ - the smallest surface area for a given height. The second step is to minimise this function for $h$.
Chris Lewis
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The mathematical set up here doesn't match the common sense interpretation of the question, because the OP doesn't include a term for the bottom of the cup. That is to say, the OP asks us to minimize $$\int_{x=0}^H 2 \pi f(x) \sqrt{1+f'(x)^2} dx \qquad (1)$$ for a fixed value of $$\int_{x=0}^H \pi f(x)^2 dx \qquad (2).$$ The integral (2) computes the volume of the cup, but (1) only computes the area of the side of the cup, not including the bottom.

The solution to this problem is that it is unbounded! For any $R>0$, consider the cylinder of radius $R$ and height $h=V/(\pi R^2)$. It has volume $V$, and surface area $2 \pi R h = 2 V/R$. As $R \to \infty$, we have $2 V/R \to 0$. So we can get an arbitrarily small surface area by making a platter of radius $R$ with a very short rim around it.

If we do the more physically reasonable thing and include a term for the surface area of the bottom of the cup (namely, $\pi f(0)^2$), then Chris Lewis is right -- the optimal shape is a hemisphere.

David E Speyer
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  • Yes, the context of 'open-topped cup' implies that either $f(0)=0$ or we need to add on the $\pi f(0)^2$ to our final result for the surface area. – Ultra Jul 26 '24 at 14:36
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Supportive answer.

By using Euler-Lagrange equation and a Lagrange multiplier, Conreu found that
$$\frac\lambda 2 y^2+c=\frac y{\sqrt{1+y'^2}}.$$ I obtained the same equation. Since $y(0)=0$, we have $c=0.$ Hence, $$y'=\frac{\sqrt{4-\lambda^2y^2}}{\lambda y}\tag1$$ which is separable. Using boundary conditions $y(0)=y(H)=0$ again, the solution curve is $$y=\sqrt H\sqrt{x-\frac{x^2}H}.$$ But then $V$ is irrelevant. I think we need another Lagrange multiplier. I can't see how to do that.

If we accept the natural symmetry with respect to $x=\frac H 2$ axis, since the solution of $(1)$ with $y(0)=0$ is $$y=f(x)=\sqrt{\frac4\lambda x-x^2}\tag2$$ the volume condition $\int_0^{\frac H2}y^2dx=\frac V2$ gives $\frac1\lambda=\frac V{\pi H^2}+\frac H{12}.$ Then the solution of the problem is $y=f(x)$ for $0\leq x\leq\frac H2$ and $y=f(H-x)$ for $\frac H2\leq x\leq H,$ allowing a singularity in the middle.

Bob Dobbs
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