10. In a rectangular parallelepiped $A B C D A_{1} B_{1} C_{1} D_{1}$ with edges $A B=4, A D=2$, $A A_{1}=3 \sqrt{2}$, a plane is drawn through the diagonal $B D_{1}$, intersecting the edge $A A_{1}$ such that the section of the parallelepiped by this plane has the smallest perimeter. Find the area of this section.
# Solution:

![](https://cdn.mathpix.com/cropped/2024_05_06_450936e33dcef50d2d99g-10.jpg?height=888&width=1282&top_left_y=575&top_left_x=453)

Extend the edge $AD$ and mark $AL = AB$; similarly, extend the edge $CD$ and mark $CK = BC$. Clearly, $B \in KL$. For any position of point $M$ on the edge $AA_1$, $BM = LM$, so the minimum value of the sum $MD_1 + MB$ will be when $M = AA_1 \cap LD_1$. Similarly, $N = CC_1 \cap KD_1$. The parallelogram $BM D_1 N$ is the desired section.

Let $AB = a$, $AD = b$, and $AA_1 = c$.

Draw $DE \perp KL$, where $E \in KL$, and connect $D_1$ and $E$.

\[
\begin{aligned}
& DE = \frac{DL}{\sqrt{2}} = \frac{AD + AL}{\sqrt{2}} = \frac{a + b}{\sqrt{2}}. \\
& ED_1 = \sqrt{DD_1^2 + DE^2} = \sqrt{c^2 + \frac{(a + b)^2}{2}} = \frac{\sqrt{2c^2 + (a + b)^2}}{\sqrt{2}}. \\
& \cos \angle DED_1 = \frac{DE}{D_1E} = \frac{a + b}{\sqrt{2c^2 + (a + b)^2}} = \frac{1}{\sqrt{1 + 2\left(\frac{c}{a + b}\right)^2}}. \\
& S_{BMD_1N} = \frac{S_{BADC}}{\cos \angle DED_1} = ab \sqrt{1 + 2\left(\frac{c}{a + b}\right)^2}.
\end{aligned}
\]

For $a = 4$, $b = 2$, and $c = 3\sqrt{2}$, $S_{BDD_1N} = 8\sqrt{2}$. Answer: $8\sqrt{2}$.