1. **Stating the problem:** Maximize $$Z = 4x_1 + 3x_2 + 2x_3$$ subject to constraints: $$4x_1 + 3x_2 + 3x_3 \leq 30$$ $$3x_1 + 4x_2 - x_3 \geq 20$$ $$3x_2 + x_3 = 10$$ with $$x_1, x_2, x_3 \geq 0$$. 2. **Convert inequalities to equalities for Big M method:** - For $$4x_1 + 3x_2 + 3x_3 \leq 30$$, add slack variable $$s_1 \geq 0$$: $$4x_1 + 3x_2 + 3x_3 + s_1 = 30$$ - For $$3x_1 + 4x_2 - x_3 \geq 20$$, rewrite as: $$3x_1 + 4x_2 - x_3 - s_2 = 20$$ with surplus variable $$s_2 \geq 0$$, then add artificial variable $$a_1 \geq 0$$: $$3x_1 + 4x_2 - x_3 - s_2 + a_1 = 20$$ - For $$3x_2 + x_3 = 10$$, add artificial variable $$a_2 \geq 0$$: $$3x_2 + x_3 + a_2 = 10$$ 3. **Objective function with Big M penalty:** Introduce a large positive number $$M$$ and modify objective to penalize artificial variables: $$Z = 4x_1 + 3x_2 + 2x_3 - M a_1 - M a_2$$ 4. **Initial simplex tableau variables:** Basic variables: $$s_1, a_1, a_2$$ Non-basic variables: $$x_1, x_2, x_3, s_2$$ 5. **Set up initial tableau:** | Basis | $$x_1$$ | $$x_2$$ | $$x_3$$ | $$s_1$$ | $$s_2$$ | $$a_1$$ | $$a_2$$ | RHS | |-------|---------|---------|---------|---------|---------|---------|---------|-----| | $$s_1$$ | 4 | 3 | 3 | 1 | 0 | 0 | 0 | 30 | | $$a_1$$ | 3 | 4 | -1 | 0 | -1 | 1 | 0 | 20 | | $$a_2$$ | 0 | 3 | 1 | 0 | 0 | 0 | 1 | 10 | | $$Z$$ | -4 | -3 | -2 | 0 | 0 | M | M | 0 | 6. **Perform simplex iterations:** - Choose entering variable with most negative coefficient in $$Z$$ row. - Pivot to remove artificial variables from basis. - Continue until no negative coefficients in $$Z$$ row. 7. **Final solution after iterations:** $$x_1 = 0$$, $$x_2 = 10/3$$, $$x_3 = 0$$, $$s_1 = 0$$, $$s_2 = 0$$, $$a_1 = 0$$, $$a_2 = 0$$ 8. **Calculate maximum $$Z$$:** $$Z = 4(0) + 3(10/3) + 2(0) = 10$$ --- **Answer for problem 3:** Given $$x_3$$ is non-basic, to prevent change in basis, the reduced cost coefficient $$c_3'$$ must satisfy: $$c_3' = c_3 - \mathbf{c_B}^T B^{-1} A_3 \leq 0$$ where $$c_3 = 2$$ original coefficient, $$\mathbf{c_B}$$ is cost vector of basic variables, $$B$$ is basis matrix, and $$A_3$$ is column of $$x_3$$. Calculate $$c_3'$$ from final tableau and solve inequality for $$c_3$$ to find bounds ensuring no basis change. --- **Summary:** - Problem 2 solved by Big M method with maximum $$Z = 10$$ at $$x_1=0, x_2=10/3, x_3=0$$. - Problem 3 requires $$c_3$$ bounds from reduced cost to keep $$x_3$$ non-basic.