1. **State the problem:** We want to maximize the objective function $$Z = 3x_1 - x_2$$ subject to the constraints: $$2x_1 + x_2 \leq 2$$ $$x_1 + 3x_2 \geq 3$$ $$x_2 \leq 4$$ with $$x_1, x_2 \geq 0$$. 2. **Convert inequalities for Big-M method:** - For $$2x_1 + x_2 \leq 2$$, add slack variable $$s_1 \geq 0$$: $$2x_1 + x_2 + s_1 = 2$$ - For $$x_1 + 3x_2 \geq 3$$, rewrite as: $$x_1 + 3x_2 - s_2 = 3$$ with surplus variable $$s_2 \geq 0$$. Add artificial variable $$a_1 \geq 0$$ to form: $$x_1 + 3x_2 - s_2 + a_1 = 3$$ - For $$x_2 \leq 4$$, add slack variable $$s_3 \geq 0$$: $$x_2 + s_3 = 4$$ 3. **Formulate the Big-M objective function:** We want to maximize $$Z = 3x_1 - x_2$$. Add penalty $$-M a_1$$ for artificial variable to force it out: $$Z = 3x_1 - x_2 - M a_1$$ 4. **Set up initial simplex tableau variables:** Variables: $$x_1, x_2, s_1, s_2, s_3, a_1$$ all $$\geq 0$$. 5. **Summary of equations:** $$2x_1 + x_2 + s_1 = 2$$ $$x_1 + 3x_2 - s_2 + a_1 = 3$$ $$x_2 + s_3 = 4$$ 6. **Solve using simplex method:** - Start with basic variables $$s_1, a_1, s_3$$. - Pivot to remove artificial variable $$a_1$$ from basis. - Iterate to maximize $$Z$$. 7. **Check corner points of feasible region (graphically):** - Intersection of $$2x_1 + x_2 = 2$$ and $$x_1 + 3x_2 = 3$$: Solve system: $$2x_1 + x_2 = 2$$ $$x_1 + 3x_2 = 3$$ Multiply second by 2: $$2x_1 + 6x_2 = 6$$ Subtract first: $$5x_2 = 4 \Rightarrow x_2 = \frac{4}{5} = 0.8$$ Then $$x_1 = 3 - 3(0.8) = 3 - 2.4 = 0.6$$ - Check if $$x_2 \leq 4$$ and $$x_1, x_2 \geq 0$$: yes. - Evaluate $$Z$$ at this point: $$Z = 3(0.6) - 0.8 = 1.8 - 0.8 = 1.0$$ - Check other vertices: At $$x_2 = 4$$ and $$2x_1 + 4 \leq 2$$ impossible since $$2x_1 \leq -2$$ no solution. At $$x_1 = 0$$: From $$2(0) + x_2 \leq 2 \Rightarrow x_2 \leq 2$$ From $$0 + 3x_2 \geq 3 \Rightarrow x_2 \geq 1$$ From $$x_2 \leq 4$$ So $$1 \leq x_2 \leq 2$$. Evaluate $$Z$$ at $$x_1=0, x_2=1$$: $$Z = 3(0) - 1 = -1$$ At $$x_2=2$$: $$Z = 3(0) - 2 = -2$$ At $$x_2=0$$: From $$2x_1 + 0 \leq 2 \Rightarrow x_1 \leq 1$$ From $$x_1 + 0 \geq 3 \Rightarrow x_1 \geq 3$$ no solution. 8. **Conclusion:** Maximum value of $$Z$$ is $$1.0$$ at $$x_1 = 0.6, x_2 = 0.8$$. **Final answer:** $$\boxed{\max Z = 1 \text{ at } (x_1, x_2) = (0.6, 0.8)}$$