Maximization of Hemisphere Inscribed Cylinder Volume

Let the radius and height of the cylinder be $r$ and $h$ respectively. We note that $r^2 = R^2 -h^2$ and the volume of the cylinder is:

$\begin{aligned} V & = \pi r^2 h \\ V(h) & = \pi h(R^2 - h^2) & \small \blue{\text{Differentiate both sides w.r.t. }h} \\ \frac {dV(h)}{dh} & = \pi (R^2 - 3h^2) & \small \blue{\text{Equating it to zero}} \\ \implies h & = \frac R{\sqrt 3} & \small \blue{\text{Since }h > 0} \end{aligned}$

Since $\dfrac {d^2 V(h)}{dh^2} < 0$ for $h > 0$ , $V\left(\dfrac R{\sqrt 3}\right) = \dfrac {2R^3}{3\sqrt 3}$ is the maximum cylinder volume. The volume of the hemisphere is $\dfrac {2\pi R^3}3$ . Therefore the required ratio is $\dfrac {2\pi R^3}{3\sqrt 3} \times \dfrac 3{2\pi R^3} = \dfrac 1{\sqrt 3}$ , $\implies n = \boxed 3$ .