Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2016 / Paper 2 / 19 / b

Put $q=2.4048$ and let the fixed constraint value be $C>0$. For $V=\pi R^{2}H$, the Lagrange multiplier equations giveDividing removes $\lambda$ and yieldsTo establish a minimum, rather than merely a stationary ratio, set $u=q^2/R^2$ and $v={\pi }^2/H^2$. Then $u+v=C$ and $V={\pi }^2{q}^2/(u\sqrt{v})$. The denominator squared is $u^2(C-u)$; its derivative is $u(2C-3u)$, giving a unique maximum at $u=2C/3$, $v=C/3$. The volume tends to infinity at either endpoint. This proves the stated ratio is the unique global minimum, as in the optimal cylinder for a reciprocal-square constraint.