= Solution
As in the diagram, an inscribed <cylinder> rests on $z=0$, is coaxial with the paraboloid, and has its top rim on the curved boundary. Thus
$$
h+\frac{Hr^2}{R^2}=H,\qquad V_c=\pi r^2h.
$$
Allowing contact with the boundary is the usual meaning of \i[inscribed]. Form the <Lagrange multiplier> <function>
$$
\mathcal L=\pi r^2h-\lambda\left(h+\frac{Hr^2}{R^2}-H\right).
$$
At a nondegenerate stationary <cylinder>, its two <derivative> equations are
$$
2\pi rh=\lambda\frac{2Hr}{R^2},\qquad \pi r^2=\lambda.
$$
Eliminating $\lambda$ gives $h=Hr^2/R^2$. The <constraint> then gives $h=H/2$ and $r=R/\sqrt2$, so
$$
\boxed{V_{c,\max}=\frac{\pi R^2H}{4}}.
$$
The degenerate endpoints $r=0$ or $h=0$ have zero <volume>, consistent with the positive stationary value being the <maximum>. If the strict inequalities defining the open interior were interpreted as forbidding boundary contact, the same value would instead be a supremum approached by smaller cylinders.
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