# Geometric Maximization Exercise

What is the height of a circular cone with surface line ("Mantellinie") s, which has the maximal volumina? My problem here is, that I am not really sure what surface line is..is this the same thing as just to say surface?

The "Mantellinie" $s$ is what you think it is: It's the distance from the tip of the cone to the rim at the bottom. Given $s$, you are free to choose the height $h$ in the interval $[0,s]$, and the radius of the bottom circle will then be $r=\sqrt{s^2-h^2}$. Therefore we have to maximize the function $$V(h)={\pi\over3} r^2 h={\pi\over3}(s^2 h-h^3)\qquad(0\leq h\leq s)\ .$$ As $V(0)=V(s)=0$ this maximum is taken for some $h\in\ ]0,s[\$, which will come to the fore by solving $V'(h)={\pi\over3}(s^2-3h^2)=0$ for $h$. There is only one solution in the interval $\ ]0,s[\$, namely $$h={s\over\sqrt{3}}\ ,$$ and this is the height you wanted to know.

The surface line should be a line on the surface which, when rotated, will form the surface in question. Usually I'd expect such surface line to lie in the same plane as the axis of rotation. So in the case of the cone, this would be a line segment starting at the origin, assuming your axis of rotation passes through the origin as well.

I'd think about this as an angle problem. Let's use $\alpha$ to denote the angle between surface line and axis of rotation. Then the height of the cone will be $s\cos\alpha$, the radius of its circular base will be $s\sin\alpha$, the area of the base will be $\pi s^2\sin^2\alpha$ and the volume will be $\frac13s^3\pi\cos\alpha\sin^2\alpha$. As you can see, the actual length of $s$ is irrelevant, you can choose your unit of length arbitrarily. All that matters is $\alpha$. You're looking for the value which maximizes $\cos\alpha\cdot\sin^2\alpha$. Simply typing this into Wolfram Alpha will, among others, tell you:

$$\max\left(\cos(\alpha) \sin^2(\alpha)\right) = \frac2{3\sqrt3} \quad\text{at}\quad \alpha = 2 n \pi\pm2 \tan^{-1}{\sqrt{2-\sqrt3}} \quad\text{for }\pi\in\mathbb Z$$

So your best solution is something close to $\alpha\approx54.73561°$. And since you can construct square roots using compass and ruler, you could even construct this angle using the classic set of construction tools. You can even Google for the above decimal representation. If you do, you will find various situations where that angle plays a certain role, and some documents will even give a nicer formula for the optimal angle:

$$\alpha=\arccos\frac{1}{\sqrt3}$$

All the above computation assumes that the base cap of the cone is not counted as part of the surface, since I did not count its radius towards the length of the surface line. If you do count that as well, things will become a lot more difficult. In particular, you'd have

\begin{align*} r &= h\tan\alpha \\ s &= h+r \end{align*}

From these two you could compute $h$ and $r$ given $s$ and $\alpha$, and then again maximize $V=\frac13\pi h r^2$. But since the equations are very non-linear, the result will likely look even uglier than the previous one. But the term “Mantellinie” in particular does usually not include the cap, so you probably won't have to worry about this.