9. Find the maximum value of the volume of a regular quadrilateral frustum with the upper base edge and side edge both equal to 1 (accurate to 0.001).

Note: If the upper base area of the regular quadrilateral frustum is $S_{1}$, the lower base area is $S_{2}$, and the height is $h$, then its volume
$$
V=\frac{1}{3} h\left(S_{1}+S_{2}+\sqrt{S_{1} S_{2}}\right) .
$$
II. 9. As shown in Figure 1, let the regular quadrilateral frustum be $A B C D-$ $A_{1} B_{1} C_{1} D_{1}, O$ and $O_{1}$ be the centers of the upper and lower bases, respectively, and let the angle between the lateral edge and the base be $\alpha$ $\left(\alpha \in\left(0, \frac{\pi}{2}\right)\right)$. Then
$$
O O_{1}=\sin \alpha, A_{1} O_{1}=\frac{\sqrt{2}}{2}+\cos \alpha \text {. }
$$

Thus, the area of the upper base $S_{1}=1$,
the area of the lower base $S_{2}=2\left(\frac{\sqrt{2}}{2}+\cos \alpha\right)^{2}$.
Therefore, the volume of the regular quadrilateral frustum is
$$
\begin{aligned}
V= & \frac{1}{3} h\left(S_{1}+S_{2}+\sqrt{S_{1} S_{2}}\right) \\
= & \frac{1}{3} \sin \alpha\left(1+1+2 \sqrt{2} \cos \alpha+2 \cos ^{2} \alpha+\right. \\
& 1+\sqrt{2} \cos \alpha) \\
= & \frac{1}{3} \sin \alpha\left(2 \cos ^{2} \alpha+3 \sqrt{2} \cos \alpha+3\right) .
\end{aligned}
$$

Using a graphing calculator, we get $V_{\operatorname{mix}}=1.674$.