Question

If the slant height of a right circular cone is $3cm$ then the maximum volume of the cone is:

• A. $2\sqrt{3}\mathrm{\pi }{\mathrm{cm}}^3$
• B. $4\sqrt{3}\mathrm{\pi }{\mathrm{cm}}^3$
• C. $\left(2+\sqrt{3}\right)\mathrm{\pi }{\mathrm{cm}}^3$
• D. $\left(2-\sqrt{3}\right)\mathrm{\pi }{\mathrm{cm}}^3$

Answer 1

The correct option is A

$2\sqrt{3}\mathrm{\pi }{\mathrm{cm}}^3$

Explanation for the correct option.

Step 1: Find the expression for the volume

Given that, the slant height of the right circular cone $3cm$.

We know that in the right circular cone: $l^2=r^2+h^2$; $l,r,h$ are slant height, radius & height of cone.

$\begin{array}{rcl}{3}^2& =& r^2+h^2\\ r^2& =& 9-h^2....\left(i\right)\end{array}$

Now using the volume of cone$=\frac{1}{3}{\mathrm{\pi r}}^2\mathrm{h}$

$\begin{array}{rcl}V& =& \frac{1}{3}\mathrm{\pi }\left(9-{\mathrm{h}}^2\right)\mathrm{h}\\ V& =& \frac{1}{3}\mathrm{\pi }\left(9\mathrm{h}-{\mathrm{h}}^3\right)....\left(1\right)\end{array}$

Step 2: Find the maximum volume

To find the maximum volume differentiate equation (1) with respect to $h$ we get:

$\frac{dV}{dh}=\frac{1}{3}\mathrm{\pi }\left(9-3{\mathrm{h}}^2\right)$

For critical points put $\frac{dV}{dh}=0$ we get:

$$\begin{gathered} \Rightarrow \frac{1}{3}\mathrm{\pi }\left(9-3{\mathrm{h}}^2\right)=0 \\ \Rightarrow 9-3{\mathrm{h}}^2=0 \\ \Rightarrow \mathrm{h}=\sqrt{3} \end{gathered}$$

Put $h=\sqrt{3}$ in equation (i) we get:

$r^2=6$

So, the maximum volume of the cone is :

$\begin{array}{rcl}V& =& \frac{1}{3}\mathrm{\pi }\left(6\right)\times \sqrt{3}\\ & =& 2\sqrt{3}\mathrm{\pi }{\mathrm{cm}}^3\end{array}$

Hence, option A is correct.