For a cone, volume \( V = \frac{1}{3} \pi r^2 h \). Using the slant height \( l \):
\[
r^2 + h^2 = l^2.
\]
Substitute and differentiate to maximize \( V \). Solve for \( \theta \) where \( \tan \theta = \frac{r}{h} \). The result is:
\[
\theta = \tan^{-1} (\sqrt{2}).
\]