Find the altitude

Consider the diagram. Let $s$ be the slant height, $r$ be the base radius and $h$ be the altitude of the right circular cone.

By pythagorean theorem, we have $s=\sqrt{h^2+r^2}$ . By similar triangles, we have $\dfrac{2}{h-2}=\dfrac{r}{s}$ , however, $s=\sqrt{h^2+r^2}$ , substituting, we have

$\dfrac{2}{h-2}=\dfrac{r}{\sqrt{h^2+r^2}}$

Squaring both sides, we have

$\dfrac{4}{h^2-4h+4}=\dfrac{r^2}{h^2+r^2}$

Cross-multiplying and simplifying, we have

$4h^2+4r^2=h^2r^2-4hr^2+4r^2$

$4h^2=r^2(h^2-4h)$

$r^2=\dfrac{4h^2}{h^2-4h}=\dfrac{4h}{h-4}$

The volume of a right circular cone is given by, $V=\dfrac{1}{3}\pi r^2h$ . Substituting, we have

$V=\dfrac{1}{3}\pi \left(\dfrac{4h}{h-4}\right)(h)=\dfrac{4}{3}\pi \left(\dfrac{h^2}{h-4}\right)$

Differentiating both sides with respect to $h$ using the quotient rule , we have

$\dfrac{dV}{dh}=\dfrac{4}{3}\pi \left[\dfrac{(h-4)(2h)-h^2(1)}{(h-4)^2}\right]$

For volume to be minimum, $\dfrac{dV}{dh}=0$ . So

$2h^2-8h-h^2=0$

$h^2=8h$

$h=\boxed{8}$