Five. (20 points) Given a tetrahedron with four edges of length 1. Find the maximum volume of this tetrahedron.


(1) If the edges of length $x, y$ are opposite edges, as shown in Figure 2.
Let $P A=P B=C A=C B=1$,
$A B=x, P C=y$.
Let the midpoint of $A B$ be $M$, and the midpoint of $P C$ be $N$. Connect
$P M, C M, M N$.
Then $P M \perp A B$,
$C M \perp A B$,
$P M=C M$
$=\sqrt{1-\frac{x^{2}}{4}}$.

Therefore, $M N \perp P C, A B \perp$ plane $P C M$, and
$M N=\sqrt{1-\frac{x^{2}}{4}-\frac{y^{2}}{4}}$.
Thus, $V_{P-A E C}=\frac{1}{3} S_{\triangle S M C} \cdot A B=\frac{1}{3} \times \frac{1}{2} P C \cdot M N \cdot A B$
$=\frac{1}{6} x y \sqrt{1-\frac{x^{2}}{4}-\frac{y^{2}}{4}}=\frac{1}{12} \sqrt{x^{2} y^{2}\left(4-x^{2}-y^{2}\right)}$
$\leqslant \frac{1}{12} \sqrt{\left(\frac{x^{2}+y^{2}+4-x^{2}-y^{2}}{3}\right)^{3}}=\frac{2 \sqrt{3}}{27}$.
The equality holds when $x^{2}=y^{2}=4-x^{2}-y^{2}$, i.e., $x=y=\frac{2 \sqrt{3}}{3}$.
(2) If the edges of length $x, y$ are adjacent edges, as shown in Figure 3.

Let $P A=P B=P C=B C=1$,
$A B=x, A C=y$.
Let the projection of $P$ on the base be $O$, then $O$ is the circumcenter of
$\triangle A B C$.
Let $\angle B A C=\alpha$.
In $\triangle A B C$, by the Law of Sines we get
$$
\begin{array}{l}
=\frac{1}{12} \sqrt{-\left(x^{2}-\frac{y^{2}+2}{2}\right)^{2}-\frac{3}{4}\left(y^{2}-2\right)^{2}+3} \\
\leqslant \frac{\sqrt{3}}{12} \text {. } \\
\end{array}
$$

The equality holds when $x^{2}=\frac{y^{2}+2}{2}, y^{2}=2$, i.e.,
$$
x=y=\sqrt{2} \text {. }
$$
Since $\frac{\sqrt{3}}{12}>\frac{2 \sqrt{3}}{27}$, the maximum volume of the tetrahedron is $\sqrt{3}$.