# Calculus optimisation

This meant to be a relatively easy problem but I cannot get my head around it. It is from Burkill's "First course in Analysis", book $4$(f), $10$.

An open bowl is in the form of a segment of a sphere of metal of negligible thickness. Find the shape of the bowl if its volume is the greatest for a given area of metal. (Solution: Hemisphere)

Could anyone help me with the solution of the problem?

Here is one of my attempts. I assumed that the problem is circumferentially symmetric so I considered the planar problem instead. I took the area of the segment of a disk with radius R, central angle $\theta$, and area A which I calculated as follows:

$$A = \text{sector area} - \text{area of triangle} = \frac{R^2\pi}{2\pi} \theta - 2 \frac{1}{2} R \cos\left(\frac{\theta}{2}\right) R\sin\left(\frac{\theta}{2}\right) = \frac{R^2\theta}{2}-\frac{R^2}{2}\sin\theta.$$

This is constrained by the area that is the length of material we have say $L$:

$$L = R\theta.$$

Substituting in for $R$:

$$A = \frac{L^2}{2\theta} - \frac{L^2}{2\theta^2} \sin\theta.$$

Differentiate to find turning value:

$$\frac{dA}{d\theta} = \frac{L^2}{2}\left(-\frac{1}{\theta^2} + \frac{2}{\theta^3} \sin\theta + \frac{1}{\theta^2} \cos\theta\right) = 0.$$

I am bit stuck now how to get $\theta$ out of this and I am questioning whether my method is really correct. Could anyone help me out? Thank you!

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The sine and cosine functions are denoted `\sin` and `\cos` in $\TeX$, respectively. – Brett Frankel May 1 '12 at 20:30
Thanks for the edit! – adamG May 1 '12 at 21:23

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– copper.hat May 1 '12 at 21:42

## 2 Answers

I think the problem implies that the bowl is obtained by cutting a sphere with a plane, so that the bowl is indeed rotationally symmetric. Let $2 \theta$ be the aperture of the bowl, with $0 < \theta < \pi$. Then the area of the metal is $A(R,\theta) = 4 \pi R^2 \sin^2\left(\frac{\theta}{2}\right)$.

The volume of liquid such a bowl could hold is given by an integral, obtained by shell method: $$\begin{eqnarray} V_\theta &=& \int_{R \cos \theta}^R A(\rho, \arccos\left( \frac{R \cos\theta}{\rho} \right)) \mathrm{d} \rho = \int_{R \cos \theta}^R 4 \pi \rho^2 \sin^2\left(\frac{1}{2} \arccos\left( \frac{R \cos\theta}{\rho} \right)\right) \rho \mathrm{d} \rho \\ &=& \int_{R \cos\theta}^R 2 \pi \rho \left( \rho - R \cos(\theta) \right) \mathrm{d} \rho = \frac{4}{3} \pi R^3 \left( 3 - 2 \sin^2\left(\frac{\theta}{2}\right) \right) \sin^4\left(\frac{\theta}{2}\right) = \frac{4}{3} \pi R^3 \left( 3 - 2 \frac{A}{4 \pi R^2} \right) \frac{A^2}{16 \pi^2 R^4} \end{eqnarray}$$ The above expression is maximal for $A = 4 \pi R^2$, meaning $\theta = \pi$, i.e. exactly the hemisphere.

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Thank you Sasha! It is interesting how many ways one can interpret this problem! – adamG May 2 '12 at 21:34

It is useful to know the (curved) surface area and the volume of a spherical cap. These are obtainable by straightforward integration, or more nicely by arguments that go back to Archimedes.

If the radius of the sphere the cap was cut from is $r$, and the maximum depth of the cap is $h$, then the spherical cap has curved surface area $2\pi rh$ and volume $\frac{\pi h^2}{3}(3r-h)$.

Let $A=2\pi k$ be the surface area. We want to maximize $h^2(3r-h)$ given that $2\pi rh =A=2\pi k$. So $r=\frac{k}{h}$.

We therefore want to maximize $h^2\left(\frac{3k}{h}-h\right)$, which is $3kh-h^3$. This is a very easy calculus problem. The maximum is reached when $h=\sqrt{k}$. The corresponding $r$ is $\frac{k}{\sqrt{k}}=\sqrt{k}$.

So the maximum is reached when $r=h$, that is, when the bowl is a hemisphere.

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Thank you Andre! Simple solution! You interpret the shape as not a spherical segment but rather a spherical cap despite the wording of the problem, which is slightly confusing I must admit. Nevertheless, this is the closest to my approach to the problem. – adamG May 2 '12 at 21:52

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– Sasha May 3 '12 at 5:29
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