With $V=\pi a^2l$ and $A=2\pi a^2+2\pi al$, eliminating $l$ and differentiating gives $a^3=V/(2\pi)$ and $A=3(2\pi)^{1/3}V^{2/3}$, so $B=3$. Under $a^2+l^2/4=R^2$, maximising gives $l/a=\sqrt5-1$, hence $A=2\pi\sqrt5,a^2$ and $C=5$.


Solution

Past exam of the mathematics course of the University of Cambridge / 2026 / ib / Paper 2 / 13B / a / Solution

 Solution (source code)