Answer
The maximum volume$=V=\frac{2}{9\sqrt 3}\pi R^3$
Work Step by Step
As we know by Pythagoras Theorem $R^2=h^2+r^2$
$r^2=R^2-h^2$
The volume of the cone $=V=\frac{\pi}{3}r^2h$
$V=\frac{\pi}{3}(R^2-h^2)h$
$V=\frac{\pi}{3}(R^2h-h^3)$
thaking differentiate on both sides;
$V'(h)=\frac{\pi}{3}\frac{d}{dh}(R^2h-h^3)$
$=\frac{\pi}{3}(\frac{d}{dh}R^2h-\frac{d}{dh}h^3)$
$V'(h)=\frac{\pi}{3}(R^2-3h^2)$
Now,
$V'(x)=0$
so,
$\frac{\pi}{3}(R^2-3h^2)=0$
when $h=\frac{1}{\sqrt 3}R$
It gives absolute maximum, because,
$V'(h)>0$ for $0\frac{1}{\sqrt 3}R$
The maximum volume$=V=\frac{\pi}{3}(\frac{1}{\sqrt 3}R^3+\frac{1}{3\sqrt3}R^3)= \frac{2}{9\sqrt 3}\pi R^3$
