Example 6 - To make a cone with the same volume as a cube with an upper edge length of $a$, what dimensions should the cone have for its total surface area to be the smallest? What is the central angle of the sector in the lateral surface development of the cone at this time?
Slightly solved: Let the radius of the base of the cone be $r$, and the apex angle of the cross-section be $2 \theta$. According to the problem, $\frac{1}{3} \pi r^{3} \operatorname{ctg} \theta=a^{3}$, then $r^{3}=\frac{3 a^{3}}{\pi} \operatorname{tg} \theta$.
$\therefore S_{\text {cone total }}=\pi r^{2}+\frac{\pi r^{2}}{\sin \theta}=\pi r^{2}\left(1+\frac{1}{\sin \theta}\right)$ $=\pi \sqrt{\left(\frac{3 a^{3}}{\pi}\right)^{2} \operatorname{tg}^{2} \theta}$
$\cdot\left(1+\frac{1}{\sin \theta}\right)$ $=\sqrt[3]{9} \pi a^{6}\left(\operatorname{tg}^{\frac{-2}{3}} \theta+\right.$ $\left.\frac{1}{\sin ^{\frac{-2}{3}} \theta \cdot \cos ^{\frac{-2}{3}} \theta}\right)$.
$S_{\dot{\theta}}^{\dot{\theta}}=-\frac{\sqrt{9 \pi a^{8}}}{3}$
$$
\begin{array}{c}
\cdot\left(\frac{2 \cos ^{-\frac{1}{3}} \theta+2 \sin \theta \cos ^{\frac{1}{3}} \theta}{\sin ^{-\frac{1}{3}} \theta \cdot \cos ^{2} \theta}\right. \\
\left.-\frac{\sin ^{-1} \theta \cdot \cos ^{\frac{7}{3}} \theta}{\sin ^{\frac{-1}{3}} \theta \cdot \cos ^{2} \theta}\right)
\end{array}
$$

Let $S^{\prime}=0$, then $2+2 \sin \theta-\frac{\cos ^{2} \theta}{\sin \theta}=0$, which simplifies to $3 \sin ^{2} \theta+2 \sin \theta-1=0$.
$\therefore \sin \theta=\frac{1}{3}, \sin \theta=-1$ (discard).
Therefore, when the ratio of the base radius to the slant height of the cone is 1:3, the total surface area of the cone is minimized, at which time the central angle $\alpha=\frac{2 \pi r}{3 r}=\frac{2 \pi}{3}$ of the sector in the lateral surface development of the cone.