Question

Show that semi-vertical angle of right circular cone of given surface area and maximum volume a given slant height $\ell$ is ${\tan }^{-1}\sqrt{2}$.

Answer 1

The surface area of the cone will be $S=\pi r^2+\pi rl$, where $r$ is the radius and $l$ is the slant height of the cone.

$\Rightarrow l=\frac{S-\pi r^2}{\pi r}$

Also, we know that $h^2+r^2=l^2$, where $h$ is the verical height of the cone.

$\Rightarrow h^2+r^2={\left(\frac{S-\pi r^2}{\pi r}\right)}^2$

$\Rightarrow h^2$ $=\frac{S^2+{\pi }^2{r}^4-2\pi S{r}^2}{{\pi }^2{r}^2}-r^2=\frac{S^2+{\pi }^2{r}^4-2\pi S{r}^2-{\pi }^2{r}^4}{{\pi }^2{r}^2}=\frac{S^2-2\pi S{r}^2}{{\pi }^2{r}^2}$

$\Rightarrow h=\frac{\sqrt{S^2-2\pi S{r}^2}}{\pi r}$

Now, volume of the cone $V$ $=\frac{1}{3}\pi r^{2}h=\frac{1}{3}\pi r^2\frac{\sqrt{S^2-2\pi S{r}^2}}{\pi r}=\frac{r}{3}\sqrt{S^2-2\pi S{r}^2}$

For maximum volume, $\frac{dV}{dr}=0$

$\Rightarrow \frac{\sqrt{S^2-2\pi S{r}^2}}{3}+\frac{r}{3}\frac{-4\pi Sr}{2\sqrt{S^2-2\pi S{r}^2}}=0$

$\Rightarrow \frac{S^2-2\pi S{r}^2}{3}=\frac{2\pi S{r}^2}{3}$

$\Rightarrow S^2=4\pi S{r}^2$

Since, $S\ne 0$, we have

$S=4\pi r^2$

Now, $l=\frac{S-\pi r^2}{\pi r}=\frac{4\pi r^2-\pi r^2}{\pi r}=3r$

Let $\alpha$ be the semi-vertical angle of the cone.

Then $\sin \alpha =\frac{r}{l}=\frac{r}{3r}=\frac{1}{3}$

$\therefore \alpha ={\sin }^{-1}\left(\frac{1}{3}\right)$