The maximum volume (in cubic meter) of the right circular cone having slant height 3m is.

 $V=\frac{1}{3}\pi r^{2}h\&r^2+h^2=l^2$ where $l=3$

$V=\frac{\pi }{3}\left(9-h^2\right)h\Rightarrow \frac{dv}{dh}=0$

$\Rightarrow h=\sqrt{3}\&\frac{d^{2}v}{d{h}^2}-6h<0$

 $\Rightarrow$ vis maximum at  $h=\sqrt{3}$
Maximum volume 
$=\frac{\pi }{3}\times \left(9-3\right)\sqrt{3}=2\sqrt{3}\pi$