Answer
See proof
Work Step by Step
By similar triangles
$\frac{H}{R}$ = $\frac{H-h}{r}$
$h$ = $\frac{H}{R}(R-r)$
The volume of the inner cone is
$V$ = $\frac{1}{3}{\pi}r^{2}h$
$V(r)$ = $\frac{1}{3}{\pi}r^{2}\left[\frac{H}{R}(R-r)\right]$ = $\frac{\pi{H}}{3R}(Rr^{2}-r^{3})$
$V'(r)$ = $\frac{\pi{H}}{3R}(2Rr-3r^{2})$
$V'(r)$ = $0$
$\frac{\pi{H}}{3R}(2Rr-3r^{2})$ = $0$
$r$ = $0$ or $r$ = $\frac{2R}{3}$
From
$h$ = $\frac{H}{R}(R-r)$
we get
$h$ = $\frac{H}{R}\left(R-\frac{2R}{3}\right)$ = $\frac{H}{3}$
$V'(r)$ changes from positive to negative at $r$ = $\frac{2R}{3}$, so the inner cone has a maximum volume of
$V$ = $\frac{1}{3}{\pi}r^{2}h$ = $\frac{1}{3}{\pi}\left(\frac{2R}{3}\right)^{2}\left(\frac{H}{3}\right)$ = $\frac{4}{27}\left[\frac{1}{3}{\pi}R^{2}H\right]$
which is about $15$% of the volume of the larger cone
