3.26. The diagonal of a rectangular parallelepiped is equal to $l$ and forms an angle $\alpha$ with the lateral edge. Find the volume of the parallelepiped if the perimeter of its base is $P$.
3.26. Given that $A B C D A_{1} B_{1} C_{1} D_{1}$ is a rectangular parallelepiped, $B_{1} D=l$, $\angle B_{1} D D_{1}=\alpha, P_{A B C D}=P$ (Fig. 3.28). From $\triangle B_{1} D_{1} D$ we find $D D_{1}=l \cos \alpha, \mathrm{a} D_{1} B_{1}=D B=$ $=l \sin \alpha$. Let $A B=x$, $A D=y$; then we arrive at the following system of equations:

Fig. 3.28

$$
\left\{\begin{array}{l}
2 x + 2 y = P, \\
x ^ { 2 } + y ^ { 2 } = l ^ { 2 } \operatorname { s i n } ^ { 2 } \alpha,
\end{array} \Rightarrow \left\{\begin{array}{l}
x + y = \frac { P } { 2 }, \\
x ^ { 2 } + y ^ { 2 } = l ^ { 2 } \operatorname { s i n } ^ { 2 } \alpha
\end{array} \Rightarrow \left\{\begin{array}{l}
x^{2}+2 x y+y^{2}=\frac{P^{2}}{4} \\
x^{2}+y^{2}=l^{2} \sin \alpha
\end{array}\right.\right.\right.
$$

Subtracting the second equation from the first, we get

$$
x y=\frac{P^{2}-4 l^{2} \sin ^{2} \alpha}{8}
$$

Thus,

$$
V_{\text {nap }}=\frac{P^{2}-4 l^{2} \sin ^{2} \alpha}{8} \cdot l \cos \alpha=\frac{l\left(P^{2}-4 l^{2} \sin ^{2} \alpha\right) \cos \alpha}{8}
$$