<s>[INST] 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!
 [/INST] 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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