 Solution

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 $C=5$.