= Solution
Divide the objective <area> by $2\pi$, which does not change its maximizer, and use $F=a^2+a\ell$ with constraint $G=a^2+\ell^2/4=1$. The <Lagrange multiplier> equations are
$$
2a+\ell=2\lambda a,\qquad a=\frac{\lambda\ell}{2}.
$$
At a nondegenerate <cylinder> $a,\ell>0$, the second gives $\lambda=2a/\ell$. Eliminating it gives $\ell^2+2a\ell=4a^2$. Hence $q=\ell/a$ satisfies $q^2+2q-4=0$. Select its positive root:
$$
\boxed{\frac\ell a=\sqrt5-1,\qquad D=-1,\qquad E=5}.
$$
This is the sphere-inscribed case of <optimal aspect ratios of closed cylinders>. One may check maximality from $A(q)=2\pi(1+q)/(1+q^2/4)$: its <derivative> has the sign of $4-2q-q^2$, so it changes from positive to negative at the selected root.
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