# What is the shape of the perfect coffee cup for heat retention assuming coffee is being drunk at a constant rate?

Note: I asked this on Mathematics SE and even though @TheSimpliFire offered a bounty on it, no-one had a good answer

Find the optimal shape of a coffee cup for heat retention. Assuming

1. A constant coffee flow rate out of the cup.
2. All surfaces radiate heat equally, i.e. liquid surface, bottom of cup and sides of cup.
3. The coffee is drunk quickly enough that the temperature differential between the coffee and the environment can be ignored/assumed constant.

So we just need to minimise the average surface area as the liquid drains

I have worked out the following 2 alternative equations for the average surface area over the lifetime of the liquid in the cup (see below for derivations):

$$S_{ave} =\pi r_0^2+ \frac{\pi^2}{V}\int_{0}^{h}{r^4ds}+\frac{2\pi^2}{V}\int_{0}^{h}{\int_{0}^{s}{r\sqrt{1+(\frac{dr}{dz})^2}\ dz\ }{r}^2ds\ } \tag{1}$$

$$S_{ave}=\pi r_0^2+\frac{\pi^2}{V}\int_{0}^{h}r^4ds+\frac{2\pi^2}{V}\int_{0}^{h}r{\underbrace{\int_{s}^{h}{{r}^2dz\ }}_{\text{Volume Drunk}}}\sqrt{1+(\frac{dr}{ds})^2}ds \tag{2}$$

If the volume of the cup is constant

$$V=\pi\int_{0}^{h}{r^2dz\ }$$

Can the function, $$r(z)$$, be found that minimises the average surface area $$S_{ave}$$?

If r is expressed as a parametric equation in the form $$r=f(t), z=g(t)$$ and $$f,g$$ are polynomials then a genetic search found the best function of parametric polynomials to be: $$r\left(z\right)=\sqrt{\frac{3}{2}}z^\frac{1}{2}-\frac{\sqrt6}{9}z^\frac{3}{2}, f\left(t\right)=\sqrt{\frac{3}{2}}t-\sqrt{\frac{3}{2}}t^3, g\left(t\right)=\frac{9}{2}t^2$$

This parametric shape has a maximum radius of 1, height of 4.5, starting volume of $$\frac{3^4}{2^5}\pi$$ and is shown here:

I can't prove that there is (or is not) a better $$r(z)$$ but...

the average surface area of this surface turns out to be $$12.723452r^2$$ or $$4.05\pi r_{max}^2$$. I suspect that the optimal surface will have the same surface area as a sphere, i.e. $$4\pi r_{max}^2$$ $$(12.5664)$$

Conjecture: The optimally shaped coffee cup has the same average surface area as a sphere of the same maximum radius.

Derivation of Surface Area Formula:

Surface area when surface of liquid is at level s is the sum of the areas of the top disc, bottom disc and the sides.

$$S(s)=\pi r_0^2+\pi r_s^2+2\pi\int_{0}^{s}{r(z)dldz}$$

$$S(s)=\pi r_0^2+\pi r_s^2+2\pi\int_{0}^{s}{r(z)\sqrt{1+\left(\frac{dr}{dz}\right)^2}\ dz\ }$$

The average surface area will be the sum of all the As’s times the time spent at each surface area.

$$S_{ave}=\frac{1}{T}\int_{t_0}^{t_h}{S(s)dt\ }$$

In order to have the drain rate constant we need to set the flow rate Q to be constant i.e. the rate of change volume is constant and $$Q=dV/dt =V/T$$

Time spent at a particular liquid level $$dt\ =\frac{T}{V}dV$$ and $$dV={\pi r}^2ds$$

$$dt=\frac{T\pi r^2}{V}ds$$

$$S_{ave}=\int_{s=0}^{s=h}{S(s)\frac{T\pi{r(s)}^2}{V}ds\ }$$ $$S_{ave}=\frac{\pi}{V}\int_{s=0}^{s=h}{(r_0^2+r(s)^2+2\int_{z=0}^{z=s}{r(z)\sqrt{1+(\frac{dr(z)}{dz})^2}\ dz\ })\pi{r(s)}^2ds\ }$$

$$S_{ave}=\frac{\pi}{V}r_0^2\int_{0}^{h}{\pi{r(s)}^2ds}+\frac{\pi}{V}\int_{s=0}^{s=h}{\left(r\left(s\right)^2+2\int_{z=0}^{z=s}{r(z)\sqrt{1+(\frac{dr(z)}{dz})^2}\ dz\ }\right)\pi{r(s)}^2ds\ }$$

$$S_{ave}=\pi\ r_0^2+\frac{\pi^2}{V}\int_{s=0}^{s=h}{\left(r^2+2\int_{z=0}^{z=s}{r(z)\sqrt{1+\left(\frac{dr(z)}{dz}\right)^2}\ dz\ }\right)r^2ds\ }$$

Alternative Formula Derivation:

Surface area of highlighted ribbon in the diagram is:

$$S_{ribbon}=2\pi rdl$$

And the contribution towards the average surface area lasts for the ratio of volume of the liquid above the current level to the total volume.

$$S_{sides}=2\pi\frac{\pi}{V}{\underbrace{\int_{s}^{h}{{r\left(z\right)}^2dz\ }}_{\text{Volume Drunk}}}rdl$$

$$S_{sides}=\frac{2\pi^2}{V}{\underbrace{\int_{s}^{h}{{r\left(z\right)}^2dz\ }}_{\text{Volume Drunk}}}r\left(s\right)\sqrt{1+\left(\frac{dr}{ds}\right)^2}ds$$

Integrate the contribution of all such sections.

$$S_{sides}=\frac{2\pi^2}{V}\int_{0}^{h}{r\left(s\right){\underbrace{\int_{s}^{h}{{r\left(z\right)}^2dz\ }}_{\text{Volume Drunk}}}\sqrt{1+\left(\frac{dr}{ds}\right)^2}ds}$$

Contribution of top surfaces to average surface area is area of top by proportion of volume that area x dz is:

$$S_{tops}=\frac{1}{V}\int_{0}^{h}{\pi{r(s)}^2{\pi r(s)}^2}ds$$

Contribution of bottom surface is constant $$\pi r_0^2$$ so adding together all three gives:

$$S_{ave}=\pi r_0^2+\frac{\pi^2}{V}\int_{0}^{h}r\left(s\right)^4ds+\frac{2\pi^2}{V}\int_{0}^{h}r{\underbrace{\int_{s}^{h}{{r\left(z\right)}^2dz\ }}_{\text{Volume Drunk}}}\sqrt{1+\left(\frac{dr}{ds}\right)^2}ds$$

• The Italians solved this problem a long time ago. Commented May 20, 2023 at 11:24
• @DanFox: the Italians didn't worry specifically about heat retention, but about the total sweetness of life, which involves integrating many more terms. Commented May 20, 2023 at 11:57
• I don't want to post an answer duplicating my post on MSE so this will be a comment: if we let $Q(s)=\int_s^hr(z)^2\,dz$, then the equation for the minimum "simplifies" to the third-order ODE $$\frac{4Q}{Q''}R'+Q''R=1+\frac C{Q'^2}$$ subject to $Q(0)=V/\pi$ and $Q(h)=0$, where $Q''^2-4Q'=R^{-2}$ and $C$ is an arbitrary constant. Commented May 21, 2023 at 9:44
• A corollary is that if $r(h)=0$ then $C=0$. This reduces to $$4(QR)'+\frac1R=Q''.$$ Commented May 21, 2023 at 15:32
• @MichaelMcL1960 It seems it does work but I'm stuck on the global minimum issue. Commented May 30, 2023 at 23:17

A coffee cup, drained at a constant flow rate, whose volume is $$\frac{81\pi}{32}$$ and whose shape is the surface of revolution of $$r(z)=\sqrt\frac{3}{2}z^\frac{1}{2}-\frac{\sqrt{6}}{9}z^{\frac{3}{2}},\quad 0 has minimum average surface area.