Question

If the total surface area of a cone is given, its volume is maximum when the semi vertical angle is

• A. ${\sin }^{-1}\frac{1}{3}$
• B. ${\sin }^{-1}\frac{1}{\sqrt{3}}$
• C. ${\tan }^{-1}\frac{1}{3}$
• D. ${\tan }^{-1}\frac{1}{\sqrt{3}}$

Answer 1

The correct option is A ${\sin }^{-1}\frac{1}{3}$

Total surface area of the cone is given. Total surface area of a cone in terms of radius 'r' and slant height 'l' is
$A=\pi rl+\pi r^2$ .......(1)
$\Rightarrow {\pi }^2{r}^2{l}^2={(A-\pi r^2)}^2$
Volume of a cone is given by the formula $V=\frac{1}{3}\pi r^{2}h$
$=\frac{1}{3}\pi r^2\sqrt{l^2-r^2}=\frac{1}{3}r\sqrt{{\pi }^2{r}^2{l}^2-{\pi }^2{r}^4}=\frac{1}{3}r\sqrt{{(A-\pi r^2)}^2-{\pi }^2{r}^4}=\frac{1}{3}r\sqrt{A^2-2A\pi r^2}$
To find maximum or minimum volume we find derivative of V with respect to r:
$\frac{dV}{dr}=\frac{1}{3}[\frac{-2A\pi r^2}{\sqrt{A^2-2A\pi r^2}}+\sqrt{A^2-2A\pi r^2}]$
$\frac{dV}{dr}=\frac{1}{3}[\frac{-2A\pi r^2}{\sqrt{A^2-2A\pi r^2}}+\sqrt{A^2-2A\pi r^2}]\frac{dV}{dr}=0\Rightarrow 2A\pi r^2=A^2-2A\pi r^2\Rightarrow A=4\pi r^2$
Using this value of $A$ in (1) we get $\frac{r}{l}=\frac{1}{3}\Rightarrow \theta ={\sin }^{-1}\frac{1}{3}$
( where $\theta$ is semi vertical angle of the cone)