= Optimal aspect ratios of closed cylinders
{title2=$\ell/a=2\text{ at fixed area};\quad\ell/a=\sqrt5-1\text{ inside a sphere}$}
A closed <circular cylinder> of radius $a>0$ and length $\ell>0$ has <volume> $\pi a^2\ell$ and total <surface area> $2\pi a(a+\ell)$, including both end disks. At fixed area $A$, <Lagrange multipliers> for maximizing the volume give $\ell=2a$, hence
$$
a=\sqrt{\frac A{6\pi}},\qquad V_{\max}=\frac13\sqrt{\frac{A^3}{6\pi}}.
$$
Equivalently, eliminate $\ell=A/(2\pi a)-a$ and maximize $(A/2)a-\pi a^3$ on $0<a<\sqrt{A/(2\pi)}$. The derivative changes from positive to negative at the displayed radius, and the volume tends to zero at both ends.
For a cylinder inscribed in a <sphere> of radius $R$, the constraint is $a^2+\ell^2/4=R^2$. Maximizing $a^2+a\ell$ by <Lagrange multipliers> gives $\ell^2+2a\ell=4a^2$, so the positive aspect ratio is $\ell/a=\sqrt5-1$. It is independent of $R$. With $q=\ell/a$, the area is $2\pi R^2(1+q)/(1+q^2/4)$; differentiating this expression confirms the unique maximum on $q>0$.
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