prove : semi verticle angle of right circular cone of given volume and least curved surface area is cot^-1 (root2) .

Let the semi-vertical angle of the cone be $\mathrm{\theta }$.


$$\begin{gathered} \mathrm{Volume}, \mathrm{V}=\frac{1}{3}{\mathrm{\pi r}}^2\mathrm{h} .....\left(1\right) \\ \mathrm{Curved} \mathrm{surface} \mathrm{area}, \mathrm{A}=\mathrm{\pi r}\sqrt{{\mathrm{r}}^2+{\mathrm{h}}^2} ......\left(2\right) \end{gathered}$$

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

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

Using (3) in (2):-

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

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

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

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

$So, \frac{dA}{dr}=0 \mathrm{when} 4\mathrm{\pi }$²r³+9V²-2r3=0

$\Rightarrow \frac{dA}{dr}=0 \mathrm{when} 4\mathrm{\pi }$²r³=18V2r3

$\Rightarrow \frac{dA}{dr}=0 \mathrm{when} \frac{4}{18}\mathrm{\pi }$²r⁶=13πr2h²

$\Rightarrow \frac{dA}{dr}=0 \mathrm{when} 2\mathrm{r}$⁶=r⁴h²

$\Rightarrow \frac{dA}{dr}=0 \mathrm{when} 2\mathrm{r}$²=h² 

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

$\Rightarrow \frac{dA}{dr}=0 \mathrm{when} {\cot }^{-1}\mathrm{\theta }=\sqrt{2}$

Hence Proved.