Maximization of Inscribed Cylinder Volume

Let the radius of the cone circular base be $R$ . Then the height of the cone is $2R$ . Let the radius and height of the cylinder be $r$ and $h$ respectively. Then we note that $\dfrac {2R-h}r = 2 \implies h = 2R - 2r$ and the volume of the cylinder:

$\begin{aligned} V & = \pi r^2 h = 2\pi r^2(R-r) & \small \blue{\text{Differentiate both sides w.r.t. }r} \\ \frac {dV(r)}{dr} & = 2\pi \left(2Rr - 3r^2 \right) & \small \blue{\text{Equate to zero.}} \\ 2 \pi r(2R - 3r) & = 0 \\ \implies r & = \begin{cases} 0 \\ \frac 23 R \end{cases} \end{aligned}$

It is obvious that $V(0) = 0$ is the minimum, and $V\left(\dfrac 23 R\right) = \pi \left(\dfrac 23 R\right)^3$ is maximum since $\dfrac {d^2V(r)}{dr^2} < 0$ , when $r = \dfrac 23 R$ (Note that $r = h$ , when $V$ is maximum.)

Since the volume of the cone is $\dfrac 13 \pi R^2 (2R)$ , the required ratio is $\dfrac {\pi \left(\frac 23R\right)^3}{\frac 23 \pi R^2} = \left(\dfrac 23 \right)^2$ . Therefore $a+b = 2+3 = \boxed 5$ .