= Solution
Fix the total <area> at $A_0>0$. Apply <Lagrange multipliers> to $V=\pi a^2\ell$ with constraint $G=2\pi a(a+\ell)=A_0$. The two <derivative> equations are
$$
2\pi a\ell=\lambda(4\pi a+2\pi\ell),\qquad
\pi a^2=\lambda\,2\pi a.
$$
Since $a>0$, the second gives $\lambda=a/2$. Substitution in the first gives $\ell=2a$. The constraint then yields $a^2=A_0/(6\pi)$. Therefore
$$
\boxed{V_{\max}=2\pi\left(\frac{A_0}{6\pi}\right)^{3/2}
=\frac13\sqrt{\frac{A_0^3}{6\pi}},\qquad C=6}.
$$
For completeness, eliminating $\ell$ makes $V=(A_0/2)a-\pi a^3$ on $0<a<\sqrt{A_0/(2\pi)}$. Its <derivative> changes from positive to negative at the calculated radius and its <volume> tends to zero at both domain endpoints. Thus this stationary <cylinder> is indeed the maximum. It is the fixed-area case of <optimal aspect ratios of closed cylinders>.
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