Question

The slant height of a right circular cone is $3cm$. The height of the cone for maximum volume is

• A. $5cm$
• B. $\sqrt{3}cm$
• C. $3cm$
• D. $\sqrt{5}cm$

Answer 1

Answer: (B)

Let height of a right circular cone $=hcm$ and $OA=rcm$

Given, slant height of a right circular cone $=3cm$ In $△OAB$,
$\angle BOA=90^{\circ }$
$(OB{)}^2+(OA{)}^2=(AB{)}^2$
[apply pythogoras theorem]
$(h{)}^2+(r{)}^2=(3{)}^2$
$r=\sqrt{9-h^2}\dots \text{ (i) }$
We know that, Volume of cone
$=\frac{\pi }{3}\pi r^{2}h$
$V=\frac{\pi }{3}\left(9-h^2\right)\times h$
$\left[\text{ from Eq. (i), }r=\sqrt{9-h^2}\right]$
$V=\frac{\pi }{3}\left(9h-h^3\right)$
$\frac{dV}{dh}=\frac{\pi }{3}\left(9-3h^2\right)$
$\therefore \frac{dV}{dh}=0\Rightarrow \frac{\pi }{3}\left(9-3h^2\right)=0$
$\Rightarrow \frac{h^2}{}=\frac{9}{3}=3\Rightarrow h=\sqrt{3}$
$\frac{d^{2}V}{d{h}^2}=\frac{-6\times h}{3}<0$
So, $h=\sqrt{3}$ of the cone for maximum volume