= Solution
\b[The minimum-volume aspect ratio is $H/R\simeq1.847$.] Keep the material parameter $B_m^2=(\eta-1)\Sigma_a/D$ fixed and positive. Write $a=2.405$ and $x=H/R$. The constraint from part (c) gives
$$
\frac{a^2}{R^2}+\frac{\pi^2}{H^2}=B_m^2,
\qquad
R^2=\frac{a^2+\pi^2/x^2}{B_m^2}.
$$
The <cylinder> <volume> is consequently
$$
V(x)=\frac{\pi x}{B_m^3}
\left(a^2+\frac{\pi^2}{x^2}\right)^{3/2},
\qquad x>0.
$$
Its logarithmic derivative is
$$
\frac{d\ln V}{dx}
=\frac1x-\frac{3\pi^2}{x(a^2x^2+\pi^2)}
=\frac{a^2x^2-2\pi^2}{x(a^2x^2+\pi^2)}.
$$
The denominator is positive. The derivative changes from negative to positive exactly once, and $V$ diverges as $x\to0$ or $x\to\infty$. Hence this stationary point is the global minimum:
$$
\boxed{\frac HR=x_*=\frac{\sqrt2\,\pi}{2.405}=1.84735.}
$$
Equivalently $H/(2R)\simeq0.924$. This is an aspect ratio in the supplied idealized <boundary value problem>; it does not determine an absolute size without a specified material parameter.
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