= Solution
Use a <Lagrange multiplier> to maximize $V=\pi r^2h$ subject to $2\pi r(r+h)=A$, with $r,h>0$. The equations $\nabla V=\lambda\nabla A$ are
$$
2\pi rh=\lambda(4\pi r+2\pi h),\qquad\pi r^2=2\lambda\pi r.
$$
The second equation gives $\lambda=r/2$. Substituting into the first gives $2rh=r(2r+h)$, hence $h=2r$. The <constraint> then gives $A=6\pi r^2$, so
$$
\boxed{r=\sqrt{\frac A{6\pi}},\qquad h=2\sqrt{\frac A{6\pi}},\qquad V_{\max}=2\pi\left(\frac A{6\pi}\right)^{3/2}=\frac13\sqrt{\frac{A^3}{6\pi}}.}
$$
This is the same aspect ratio as the <minimum-area closed cylinder at fixed volume>. To confirm maximality directly, eliminate $h$: $V(r)=Ar/2-\pi r^3$ on $0<r<\sqrt{A/(2\pi)}$. Its only stationary point is the value above, its second <derivative> there is negative, and the <volume> tends to zero at either end of the interval.
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