For a circle inscribed in a triangle, its center is at the point of intersection of the angular bisector of the triangle called the

incenter (see figure).

For the problem:

$V = \frac{1}{3}\pi r^2 h$

From the figure:

$\tan \theta = a/r$

$\tan 2\theta = h/r$

$\dfrac{2\tan \theta}{1 - \tan^2 \theta} = \dfrac{h}{r}$

$\dfrac{2(a/r)}{1 - a^2/r^2} = \dfrac{h}{r}$

$\dfrac{2a)}{\dfrac{r^2 - a^2}{r^2}} = h$

$h = \dfrac{2ar^2)}{r^2 - a^2}$

Thus,

$V = \frac{1}{3}\pi r^2 \left( \dfrac{2ar^2)}{r^2 - a^2} \right)$

$V = \frac{2}{3}a \pi \left( \dfrac{r^4)}{r^2 - a^2} \right)$

$\dfrac{dV}{dr} = \frac{2}{3}a \pi \left[ \dfrac{(r^2 - a^2)(4r^3) - r^4(2r)}{(r^2 - a^2)^2} \right] = 0$

$4r^3(r^2 - a^2) - 2r^5 = 0$

$2(r^2 - a^2) - r^2 = 0$

$2r^2 - 2a^2 - r^2 = 0$

$r^2 = 2a^2$

$h = \dfrac{2ar^2}{r^2 - a^2}$

$h = \dfrac{2a(2a^2)}{2a^2 - a^2}$

$h = \dfrac{4a^3}{a^2}$

$h = 4a$ (*okay!*)