Past exam of the mathematics course of the University of Cambridge / 2026 / ib / Paper 2 / 13B (Variational Principles)

With $V=\pi a^{2}l$ 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 $l/a=\sqrt{5}-1$, hence $A=2\pi \sqrt{5},a^2$ and $C=5$.

Substitute $y=-x-z$. The stationary equations are $-6x^2+2(x+z)=0$ and $2x+8z=0$, giving $(x,y,z)=(0,0,0)$ and $(1/4,-3/16,-1/16)$.