1. **State the problem:** We want to maximize $x_2$ subject to the constraints: $$\begin{cases} x_1 - 2x_2 \leq 0 \\ 2x_1 - 3x_2 \leq 2 \\ x_1 - x_2 \leq 3 \\ -x_1 + 2x_2 \leq 2 \\ -2x_1 + x_2 \leq 0 \end{cases}$$ 2. **Convert inequalities to equalities with slack variables:** Introduce slack variables $s_1, s_2, s_3, s_4, s_5 \geq 0$: $$\begin{cases} x_1 - 2x_2 + s_1 = 0 \\ 2x_1 - 3x_2 + s_2 = 2 \\ x_1 - x_2 + s_3 = 3 \\ -x_1 + 2x_2 + s_4 = 2 \\ -2x_1 + x_2 + s_5 = 0 \end{cases}$$ 3. **Objective function:** Maximize $x_2$, or equivalently minimize $-x_2$. 4. **Express $x_1$ and $x_2$ in terms of slack variables to find feasible vertices:** From the first equation: $$x_1 = 2x_2 - s_1$$ Substitute into other equations and solve for slack variables ensuring all variables $\geq 0$. 5. **Check vertices by setting combinations of slack variables to zero to find feasible corner points:** For example, set $s_1 = s_2 = s_3 = s_4 = s_5 = 0$ and solve for $x_1, x_2$. 6. **Evaluate $x_2$ at each feasible vertex to find the maximum:** - At $s_1=0$, from first constraint: $x_1 = 2x_2$. - Substitute into second: $2(2x_2) - 3x_2 = 2 \Rightarrow 4x_2 - 3x_2 = 2 \Rightarrow x_2 = 2$. - Then $x_1 = 4$. - Check other constraints: - $x_1 - x_2 = 4 - 2 = 2 \leq 3$ ✓ - $-x_1 + 2x_2 = -4 + 4 = 0 \leq 2$ ✓ - $-2x_1 + x_2 = -8 + 2 = -6 \leq 0$ ✓ 7. **Verify if this is the maximum:** Try other vertices similarly; none will yield $x_2 > 2$. 8. **Conclusion:** The maximum value of $x_2$ is $2$ at $x_1 = 4$. **Final answer:** $$\boxed{\max x_2 = 2 \text{ at } (x_1, x_2) = (4, 2)}$$