= Solution
A central cross-section of the <sphere> gives the constraint $r^2+h^2/4=R^2$, and the cylinder's <volume> is $V=\pi r^2h$. At an interior optimum with $r,h>0$, the <Lagrange multiplier> equations for this <constrained optimization> problem are
$$
2\pi rh=2\lambda r,\qquad
\pi r^2=\lambda h/2,\qquad
r^2+h^2/4=R^2.
$$
The first gives $\lambda=\pi h$; substitution into the second gives $r^2=h^2/2$. The constraint then gives
$$
\boxed{h=\frac{2R}{\sqrt3},\qquad r=R\sqrt{\frac23}}.
$$
The feasible nonnegative $(r,h)$ form a compact arc, whose endpoints have zero <volume>. Since this is the only positive stationary candidate, it is the <global maximum>. Its <volume> and fraction of the sphere's <volume> are
$$
\boxed{V_{\max}=\frac{4\pi R^3}{3\sqrt3},\qquad
\frac{V_{\max}}{V_{\mathrm{sphere}}}=\frac1{\sqrt3}}.
$$
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