Optimization Problem

Let the volume of the cone ${\bf V_{cone} = \frac{1}{3} \pi R^2 H }$ and the volume of the cylinder ${\bf V = \pi r^2 h. }$

From the geometry of the problem we have two similar triangles whose proportion is: ${\bf \frac{R}{R - r} = \frac{H}{h} \implies }$ ${\bf h = \frac{H(R - r)}{R} \implies V = \frac{H \pi}{R} * (Rr^2 - r^3) \implies}$ ${\bf \frac{dV}{dr} = \frac{H \pi }{R} * r * (2R - 3r) = 0, r <> 0 \implies r = \frac{2R}{3}. }$

${\bf \frac{d^2V}{dr^2} = \frac{H \pi}{R} * (2R - 6r)}$ and ${\bf \frac{d^2V}{dr^2}|_{r = \frac{2R}{3}} = \frac{-2H \pi}{R} < 0 \implies }$ we have a maximum at ${\bf r = \frac{2R}{3} }$

${\bf r = \frac{2R}{3} \implies h = \frac{H}{3} \implies V = \pi * (\frac{2R}{3})^2 * (\frac{H}{3}) = \frac{4}{27} * \pi R^2 H = \frac{4}{9} * V_{cone}. }$

Since ${\bf V_{cone} = 1 \implies V = \frac{4}{9} = \frac{a}{b} \implies a + b = 13.}$

Way too much laziness to use LaTeX. So, handwritten solution:

Height of the cylinder should be 1/3 the height of the cone.