Example 3 (Question from the 11th "Hope Cup" Invitational Competition) From a circular iron sheet with a radius of 1 decimeter, a sector with a central angle of $x$ radians is cut out. The remaining part is rolled into a cone (ignoring the material used for the connection). When the volume of the cone is maximized, the value of $x$ is ( ).
A. $\frac{\pi}{2}$
B. $\frac{\pi}{3}$
C. $(3-\sqrt{2}) \pi$
D. $\frac{6-2 \sqrt{6}}{3} \pi$
Solution: Choose D. Reason: Let the volume of the cone be $V$, the radius of the base be $r$, and the height be $h$. According to the problem, we have
$$
\begin{aligned}
V & =\frac{1}{3} \pi r^{2} h=\frac{\pi}{3}\left(\frac{2 \pi-x}{2 \pi}\right)^{2} \cdot \sqrt{1-\left(\frac{2 \pi-x}{2 \pi}\right)^{2}} \\
& =\frac{1}{24 \pi^{2}}(2 \pi-x)^{2} \sqrt{4 \pi x-x^{2}}=\frac{1}{24 \pi^{2}} \sqrt{(2 \pi-x)^{4}\left(4 \pi x-x^{2}\right)} \\
& =\frac{1}{24 \pi^{2}} \sqrt{\left(4 \pi^{2}-4 \pi x+x^{2}\right)^{2}\left(8 \pi x-2 x^{2}\right) \cdot \frac{1}{2}} .
\end{aligned}
$$

where $\left(4 \pi^{2}-4 \pi x+x^{2}\right)+\left(4 \pi^{2}-4 \pi x+x^{2}\right)+\left(8 \pi x-2 x^{2}\right)=8 \pi^{2}$ (a constant).
Therefore, when $4 \pi^{2}-4 \pi x+x^{2}=8 \pi x-2 x^{2}$, i.e., $3 x^{2}-12 \pi x+4 \pi^{2}=0$, i.e., $x=\frac{6-2 \sqrt{6}}{3} \pi$, $V$ reaches its maximum value $\frac{1}{24 \pi^{2}} \sqrt{\frac{1}{2}\left(\frac{8 \pi^{2}}{3}\right)^{3}}=\frac{2 \sqrt{3}}{27} \pi$.