= Solution
Put $q=2.4048$ and let the fixed constraint value be $C>0$. For $V=\pi R^2H$, the <Lagrange multiplier> equations give
$$
2\pi RH=\lambda\frac{-2q^2}{R^3},\qquad
\pi R^2=\lambda\frac{-2\pi^2}{H^3}.
$$
Dividing removes $\lambda$ and yields
$$
\frac{2H}{R}=\frac{q^2H^3}{\pi^2R^3},\qquad
\boxed{\frac HR=\frac{\sqrt2\pi}{q}=\frac{\sqrt2\pi}{2.4048}}.
$$
To establish a minimum, rather than merely a stationary ratio, set $u=q^2/R^2$ and $v=\pi^2/H^2$. Then $u+v=C$ and $V=\pi^2q^2/(u\sqrt v)$. The denominator squared is $u^2(C-u)$; its <derivative> is $u(2C-3u)$, giving a unique maximum at $u=2C/3$, $v=C/3$. The <volume> tends to infinity at either endpoint. This proves the stated ratio is the unique <global minimum>, as in the <optimal cylinder for a reciprocal-square constraint>.
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