= Solution
Along the stipulated fixed-aspect-ratio pathway, substitute $h=d/k$ before differentiating:
$$
W_n(d)=\frac{\pi\Delta G}{4k}d^3+\pi\left(\frac{\gamma_f}{2}+\frac{\gamma_e}{k}\right)d^2,
$$
$$
\frac{dW_n}{dd}=\frac{3\pi\Delta G}{4k}d^2+\pi\left(\gamma_f+\frac{2\gamma_e}{k}\right)d.
$$
The nonzero stationary <diameter> is
$$
d^*=-\frac4{3\Delta G}(k\gamma_f+2\gamma_e).
$$
Comparison with the stated <diameter> gives $k\gamma_f+2\gamma_e=6\gamma_f+2\gamma_e$, hence \b[$k=d/h=6$], provided $\gamma_f$ is nonzero. This is a plate-like shape, qualitatively consistent with the cheaper broad coherent faces. For negative $\Delta G$, substitution shows the stationary point is a maximum along this size pathway, as required for a <nucleation barrier>.
A useful distinction is that this calculation holds shape fixed. Minimizing the interface energy at fixed <volume> separately gives $d/h=\gamma_e/\gamma_f$ for the cylindrical model; if the selected aspect ratio is also an equilibrium shape, it would require $\gamma_e=6\gamma_f$. The given critical-diameter relation alone does not otherwise determine that energy ratio. In the degenerate zero-face-energy limit the coefficient comparison cannot determine $k$.
Back to article page
