Question

If the maximum volume (in cu.m) of the right circular cone having slant height $3m$ is equal to $k\sqrt{3}\pi m^3$, then $k$ is equal to

Answer 1

Answer: 2

$h^2+r^2=3^2=9\dots (1)$
Volume of cone
$V=\frac{1}{3}\pi r^{2}h\dots (2)$
From (1) and (2),
$\Rightarrow V=\frac{1}{3}\pi (9-h^2)h$
$\Rightarrow V=\frac{1}{3}\pi (9h-h^3)$
$\Rightarrow \frac{dv}{dh}=\frac{1}{3}\pi (9-3h^2)$
For maxima/minima,
$\frac{dV}{dh}=0$
$\Rightarrow \frac{1}{3}\pi (9-3h^2)=0$
$\Rightarrow h=\pm \sqrt{3}$
$\Rightarrow h=\sqrt{3}\left(\because h>0\right)$
Now ; $\frac{d^{2}V}{d{h}^2}=\frac{1}{3}\pi \left(-6h\right)=-2\pi h=-2\sqrt{3}\pi$
Here, ${\left(\frac{d^{2}V}{d{h}^2}\right)}_{\text{at}h=\sqrt{3}}<0$
Then, $h=\sqrt{3}$ is point of maxima
Hence, the required maximum volume is,
$V=\frac{1}{3}\pi \left(9-3\right)\sqrt{3}=2\sqrt{3}\pi m^3$
$\Rightarrow k\sqrt{3}\pi =2\sqrt{3}\pi$
$\Rightarrow k=2$