Question

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

Answer 1

$h^{2}+r^{2} = 3^{2}=9 \,\dots (1)$
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 ; $\frac{d^{2}V}{dh^{2}}=\frac{1}{3} \pi \left(-6h\right)=-2\pi h =-2\sqrt{3}\pi$
Here, $\left(\frac{d^{2} V}{dh^{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 $
$\Rightarrow k=2$