Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2010 / Paper 2 / 19

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 areSince $a>0$, the second gives $\lambda =a/2$. Substitution in the first gives $\ell =2a$. The constraint then yields $a^2=A_0/(6\pi )$. ThereforeFor 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.

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.