11. If real numbers $a, b$ satisfy the conditions $\left\{\begin{array}{l}a+b-2 \geq 0 \\ b-a-1 \leq 0 \\ a \leq 1\end{array}\right.$, then the maximum value of $\frac{a+2 b}{2 a+b}$ is equal to
Answer $\frac{7}{5}$.
Solution Represent the feasible region indicated by the constraint conditions, as shown in the figure, which is the interior and boundary of $\triangle ABC$, where $A\left(\frac{1}{2}, \frac{3}{2}\right), B(1,1), C(1,2)$. 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 $k_{O B}=1$. Therefore, $\frac{a+2 b}{2 a+b}=2-\frac{3 a}{2 a+b}=2-\frac{3}{2+\frac{b}{a}}$, which means when $\frac{b}{a}$ takes its maximum value of 3, $\frac{a+2 b}{2 a+b}$ also reaches its maximum value of $\frac{7}{5}$.