11. Let the plane point sets be
$$
\begin{array}{l}
A=\left\{(x, y) \left\lvert\,(y-x)\left(y-\frac{18}{25 x}\right) \geqslant 0\right.\right\}, \\
B=\left\{(x, y) \mid(x-1)^{2}+(y-1)^{2} \leqslant 1\right\} .
\end{array}
$$

If $(x, y) \in A \cap B$, find the minimum value of $2 x-y$.
Let $z=2 x-y$. Then $y=2 x-z,-z$ represents the y-intercept of the line $y=2 x-z$.

It is easy to know that when the line $y=2 x-z$ passes through point $P$ in the shaded area, $z=2 x-y$ reaches its minimum value.

Point $P$ is on the circle $(x-1)^{2}+(y-1)^{2}=1$, with coordinates $(1+\cos \theta, 1+\sin \theta)$. From the graph, we know $\theta \in\left(\frac{\pi}{2}, \pi\right)$.
Since point $P$ is on the curve $y=\frac{8}{25 x}$, we have,
$$
\begin{array}{l}
(1+\cos \theta)(1+\sin \theta)=\frac{18}{25} \\
\Rightarrow \sin \theta \cdot \cos \theta+\sin \theta+\cos \theta+\frac{7}{25}=0 .
\end{array}
$$

Let $\sin \theta+\cos \theta=t$. Then
$$
\sin \theta \cdot \cos \theta=\frac{1}{2}\left(t^{2}-1\right) \text {. }
$$

Substituting into equation (1) we get
$$
\begin{array}{l}
\frac{1}{2}\left(t^{2}-1\right)+t+\frac{7}{25}=0 \\
\Rightarrow t=\frac{1}{5} \text { or } t=-\frac{11}{5}(\text { discard }) \\
\Rightarrow \sin \theta+\cos \theta=\frac{1}{5} \\
\Rightarrow \sin \theta=\frac{4}{5}, \cos \theta=-\frac{3}{5} \\
\Rightarrow P\left(\frac{2}{5}, \frac{9}{5}\right), z_{\min }=2 \times \frac{2}{5}-\frac{9}{5}=-1 .
\end{array}
$$