11. (20 points) Given the ellipse $\Gamma: \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$. Find the maximum perimeter of the inscribed parallelogram $A B C D$ in the ellipse.
11. The perimeter of $\square A B C D$ is a bounded continuous number in the closed interval of the positions of points $A, B, C, D$, and it must have a maximum value. Taking a set of points $A, B, C, D$ that maximize this perimeter, the ellipse with foci at $A, C$ and passing through $B, D$ is tangent to the ellipse $\Gamma$ at point $B$. Thus, the two curves have a common tangent line at point $B$.

Therefore, $\angle A B F_{1}=\angle F_{2} B C$, where $F_{1}, F_{2}$ are the two foci of the ellipse $\Gamma$.

Construct the point $F^{\prime}$, the reflection of $F_{2}$ about the midpoint of $B C$. Then the quadrilateral $B F_{2} C F^{\prime}$ is a parallelogram.

Since $F_{1} F_{2}$ is symmetric about the center of the parallelogram $A B C D$, we have $B F^{\prime}=F_{2} C=F_{1} A$.
Thus, the quadrilateral $B F^{\prime} F_{1} A$ is a parallelogram.
Connect $F^{\prime} F_{1}$. Then
$\angle B F_{1} F^{\prime}=\angle A B F_{1}=\angle C B F_{2}=\angle B C F^{\prime}$
$\Rightarrow B, F_{1}, C, F^{\prime}$ are concyclic
$\Rightarrow B C \cdot F^{\prime} F_{1}=B F^{\prime} \cdot C F_{1}+B F_{1} \cdot C F^{\prime}$.
Since $B F^{\prime}=A F_{1}, C F^{\prime}=D F_{1}, F^{\prime} F_{1}=B A$, we have
$A B \cdot B C=B F_{1} \cdot F_{1} D+A F_{1} \cdot F_{1} C$,
$A F_{1}+F_{1} C=B F_{1}+F_{1} D=A F_{1}+A F_{2}=2 a$.
By the median length formula, we get
$A F_{1}^{2}+C F_{1}^{2}=\frac{A C^{2}}{2}+\frac{F_{1} F_{2}^{2}}{2}=\frac{A C^{2}}{2}+2 c^{2}$,
$B F_{1}^{2}+D F_{1}^{2}=\frac{B D^{2}}{2}+2 c^{2}$.
And $A B^{2}+B C^{2}=\frac{A C^{2}}{2}+\frac{B D^{2}}{2}$,

Then by (3) + (4) - (5) we get
$$
\begin{array}{l}
A F_{1}^{2}+C F_{1}^{2}+B F_{1}^{2}+D F_{1}^{2} \\
=A B^{2}+B C^{2}+4 c^{2} .
\end{array}
$$

By (6) - 2 × (1) we get
$$
\begin{array}{l}
\left(A F_{1}+C F_{1}\right)^{2}+\left(B F_{1}+D F_{1}\right)^{2} \\
=(A B+B C)^{2}+4 c^{2} .
\end{array}
$$

Substituting equation (2) into equation (7) we get
$$
(A B+C D)^{2}=8 a^{2}-4 c^{2}=4\left(a^{2}+b^{2}\right) \text {. }
$$

Thus, $A B+C D=2 \sqrt{a^{2}+b^{2}}$.
Therefore, the perimeter of the parallelogram is $4 \sqrt{a^{2}+b^{2}}$. Hence, the maximum value of the perimeter of the parallelogram is $4 \sqrt{a^{2}+b^{2}}$.