= Solution
For a cylinder of radius $r$ and length $L$, the <mean curvature> is $H=1/(2r)$ and the area is $2\pi rL$. Its <Helfrich membrane energy> per unit length is therefore
$$
\frac{E_m}{L}=2\pi r\left(\sigma+\frac{k_c}{2r^2}\right)
=2\pi\sigma r+\frac{\pi k_c}{r}.
$$
The first term favors a narrow tube and the second penalizes its curvature. Setting the <derivative> with respect to $r$ to zero gives
$$
2\pi\sigma-\frac{\pi k_c}{r^2}=0,
\qquad
\boxed{r_0=\sqrt{\frac{k_c}{2\sigma}}}.
$$
The positive second derivative confirms a minimum.
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