 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 areAt 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: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.