Sattelhofsk

2022-06-20

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?

Jaida Sanders

Beginner2022-06-21Added 18 answers

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

$V(h)=\frac{\pi}{3}{r}^{2}h=\frac{\pi}{3}({s}^{2}h-{h}^{3})\phantom{\rule{2em}{0ex}}(0\le h\le s)\text{}.$

As $V(0)=V(s)=0$ this maximum is taken for some $h\in \text{}]0,s[\text{}$ , which will come to the fore by solving ${V}^{\prime}(h)=\frac{\pi}{3}({s}^{2}-3{h}^{2})=0$ for $h$. There is only one solution in the interval $\text{}]0,s[\text{}$ , namely

$h=\frac{s}{\sqrt{3}}\text{},$

and this is the height you wanted to know.

$V(h)=\frac{\pi}{3}{r}^{2}h=\frac{\pi}{3}({s}^{2}h-{h}^{3})\phantom{\rule{2em}{0ex}}(0\le h\le s)\text{}.$

As $V(0)=V(s)=0$ this maximum is taken for some $h\in \text{}]0,s[\text{}$ , which will come to the fore by solving ${V}^{\prime}(h)=\frac{\pi}{3}({s}^{2}-3{h}^{2})=0$ for $h$. There is only one solution in the interval $\text{}]0,s[\text{}$ , namely

$h=\frac{s}{\sqrt{3}}\text{},$

and this is the height you wanted to know.