Question

Prove that the semi-vertical angle of the right circular cone of given volume and least curved surface area is ${\cot }^{-1}\sqrt{2}.$

Answer 1

Volume , V $=\frac{1}{3}\pi r^{2}h$ --- (1)

Surface area, A $=\pi r\sqrt{r^2+h^2}$ --- (2)

Since volume if constant, we can write h in terms of V from (1):

$h=\frac{3V}{\pi r^2}$ --- (3)

Using (3) in (2)

$A=\pi r\sqrt{r^2+\frac{9V^2}{{\pi }^2{r}^4}}$

$\Rightarrow A=\sqrt{{\pi }^2{r}^4+\frac{9V^2}{r^2}}$

$\Rightarrow \frac{dA}{dr}=\frac{4{\pi }^2{r}^3+9V^2\left(\frac{-2}{r^3}\right)}{2\sqrt{{\pi }^2{r}^4+\frac{9V^2}{r^2}}}$

Now A is maximum or minimum when $\frac{dA}{dr}=0$

So $\frac{dA}{dr}=0$ when $4{\pi }^2{r}^3=18V^2{r}^3$

$\Rightarrow \frac{dA}{dr}=0$ when $4{\pi }^2{r}^3=18V^2{r}^3$

$\Rightarrow \frac{dA}{dr}=0$ when $\frac{4}{18}{\pi }^2{r}^6=\frac{1}{3}\pi r^2{h}^2$

$\Rightarrow \frac{dA}{dr}=0$ when $2r^6=r^4{h}^2$

$\Rightarrow \frac{dA}{dr}=0$ when $2r^2=h^2$

$\Rightarrow \frac{dA}{dr}=0$ when $\frac{h}{r}=\sqrt{2}$

$\Rightarrow \frac{dA}{dr}=0$ when $co{t}^{-1}=\sqrt{2}$

Hence proved.