Compact Cap on Cone

The volume of a spherical sector of semi-vertical angle $\theta$ and radius $R$ is $V \; = \; \tfrac23\pi R^3(1 - \cos\theta)$ The surface area of the cap is $2\pi R^2(1 - \cos\theta)$ while the cone has base radius $R\sin\theta$ and slant length $R$ , and hence surface area $\pi R^2\sin\theta$ . Thus we are asked to minimise $A \; =\; 2\pi R^2(1 - \cos\theta) + \pi R^2 \sin\theta$ subject to the constraint $\tfrac23\pi R^3 (1 - \cos\theta) \; = \; 1$ Eliminating $R$ from these equations, we deduce that $A \; = \; \big(\tfrac{9\pi}{2}\big)^{\frac13} f\big(\tfrac12\theta\big)$ where $\begin{aligned} f(x) & = 2\big(\sin x)^{\frac23} + \big(\sin x\big)^{-\frac13} \cos x \\ f'(x) & = \tfrac43\big(\sin x\big)^{-\frac13}\cos x - \tfrac13\big(\sin x\big)^{-\frac43} \cos^2x - \big(\sin x\big)^{\frac23} \\ & = \tfrac13\big(\sin x\big)^{-\frac43}(3\sin x - \cos x)(\cos x - \sin x) \end{aligned}$ and hence turning points for the function $f$ in the range $0 < x \le \frac12\pi$ occur at $x = \tan^{-1}\tfrac13$ and $x = \tfrac14\pi$ . It is easy to check that the first of these is a local minimum, and the second is a local maximum.

When $\theta = 2\tan^{-1}\tfrac13 \approx 36.87^\circ$ , the surface area of the sector is $\boxed{5.61165}$ . This is a minimum surface area for $0 < \theta \le \tfrac12\pi$ , namely for acute semi-vertical angles.

However, if we are prepared to allow the semi-vertical angle $\theta$ to be obtuse, and take values $0 \le \theta \le \pi$ , then the minimum at $2\tan^{-1}\tfrac13$ is no longer a global minimum. The function $f(x)$ is minimized over the interval $0 < x \le \tfrac12\pi$ at $x = \tfrac12\pi$ , as the below graph of $f(x)$ against $x$ shows (the minimum value is $2$ ):

Thus, if we allowed obtuse semi-vertical angles, the minimum surface would be achieved at $\theta = \pi$ , and would therefore be the surface area of a sphere of unit volume, which would be $4.83598$ .