11. Given real numbers $a, b$ satisfy
$$
\left\{\begin{array}{l}
a+b-2 \geqslant 0, \\
b-a-1 \leqslant 0, \\
a \leqslant 1 .
\end{array}\right.
$$

Then the maximum value of $\frac{a+2 b}{2 a+b}$ is $\qquad$
11. $\frac{7}{5}$.

Draw the feasible region represented by the constraint conditions, as shown in Figure 2, which is the interior and boundary of $\triangle ABC$, where $A\left(\frac{1}{2}, \frac{3}{2}\right)$, $B(1,1)$, and $C(1,2)$.
Let $P(a, b)$ be any point in the feasible region.
Then $k_{O P}=\frac{b}{a}$ has a maximum value of $k_{O A}=3$ and a minimum value of $k_{O B}=1$.
$$
\text { Therefore, } \frac{a+2 b}{2 a+b}=2-\frac{3}{2+\frac{b}{a}} \text {. }
$$

Thus, when $\frac{b}{a}$ takes its maximum value of 3, $\frac{a+2 b}{2 a+b}$ achieves its maximum value of $\frac{7}{5}$.