The height of the cone of maximum volume which can be inscribed in a sphere of radius r is

Let R be the base radius, h be the height of the cone.

$R^2+{(h-r)}^2=r^2\Rightarrow R^2=r^2-{(h-r)}^2\Rightarrow R^2=r^2-h^2-r^2+2hr=2hr-h^2$

$\therefore$ Volume of the cone, $V=\frac{1}{3}\pi R^{2}h=\frac{\pi h}{3}\left[2hr-h^2\right]=\frac{2\pi r{h}^2}{3}-\frac{\pi h^3}{3}$

$\frac{dV}{dh}=\frac{4\pi rh}{3}-\pi h^2\Rightarrow \frac{d^{2}V}{d{h}^2}=\frac{4\pi r}{3}-2\pi h.$

$\frac{dV}{dh}=0\Rightarrow \frac{4\pi rh}{3}-\pi h^2=0\Rightarrow \frac{4r}{3}-h=0\Rightarrow h=\frac{4r}{3}$

${\left[\frac{{\displaystyle d^{2}V}}{{\displaystyle d{h}^2}}\right]}_{h=4r/3}=\frac{4\pi rh}{3}-\frac{8\pi r}{3}=-\frac{4\pi r}{3}<0.$ 

V has maximum at $h=4r/3.$