Let's solve the problem in **2 steps** as requested, with explanations at each step and a final answer at the end. --- ### **Step 1: Update the Coefficient in the Objective Function** **Change:** Change the coefficient of \( x_3 \) in the objective function to \( c_3 = 7 \). **Original Z Row (from tableau):** Z-row coefficients: \( [0,\ 0,\ 1,\ 0,\ 5,\ 0] \) - The "1" under \( x_3 \) means the current coefficient (reduced cost) of \( x_3 \) is 1. - We need to update the objective function coefficient of \( x_3 \) from its original value (which was 13) to the new value (which is 7). **We use the formula for the new Z-row:** \[ \text{New } Z\text{-row} = \text{Old } Z\text{-row} + (c_3^{\text{new}} - c_3^{\text{old}}) \times \text{Column of } x_3 \] - \( c_3^{\text{new}} = 7 \) - \( c_3^{\text{old}} = 13 \) - Difference: \( 7 - 13 = -6 \) - Column of \( x_3 \) from the tableau: \( [3, 10] \) (from the constraints), and for Z-row, the coefficient is 1. Apply the change across the tableau Z-row using the \( x_3 \) column: | | \( x_1 \) | \( x_2 \) | \( x_3 \) | \( x_4 \) | \( x_5 \) | RHS | |------|-----------|-----------|-----------|-----------|-----------|-----| | Z | 0 | 0 | 1 | 0 | 5 | 100 | | Row1 | -1 | 1 | 3 | 1 | 0 | 20 | | Row2 | 16 | 4 | 10 | -4 | 1 | 90 | But in tableau format, the Z-row coefficients are calculated as: \[ Z_j - c_j = (\sum c_B \cdot a_{ij}) - c_j \] Here, the change in \( c_3 \) only affects the Z-row. Specifically, we subtract 6 times the column of \( x_3 \) from the Z-row. - Z-row: \( [0,\ 0,\ 1,\ 0,\ 5] \) - Subtract \( 6 \) times the column of \( x_3 \) (from each row): | | \( x_1 \) | \( x_2 \) | \( x_3 \) | \( x_4 \) | \( x_5 \) | RHS | |------|-----------|-----------|-----------|-----------|-----------|-----| | Z | 0 | 0 | 1 | 0 | 5 | 100 | | | | | | | | | | 6 × x_3 column: | 6×(-1)=-6 | 6×1=6 | 6×3=18 | 6×1=6 | 6×0=0 | 6×20=120 | | | 6×16=96 | 6×4=24 | 6×10=60 | 6×(-4)=-24| 6×1=6 | 6×90=540 | But that's not the correct approach for the Z-row. Instead, **we only update the Z-row's \( x_3 \) coefficient**, reducing it by 6 (the change in \( c_3 \)), because only the coefficient for \( x_3 \) in the objective function is changing. So, the **updated Z-row** is: - Old row: \( [0,\ 0,\ 1,\ 0,\ 5,\ 100] \) - New row: \( [0,\ 0,\ 1 - 6,\ 0,\ 5,\ 100] = [0,\ 0,\ -5,\ 0,\ 5,\ 100] \) **Explanation:** Changing the coefficient of \( x_3 \) in the objective function only affects the Z-row in the simplex tableau, specifically the coefficient for \( x_3 \). We subtract 6 from the current Z-row coefficient for \( x_3 \). --- ### **Step 2: Test Feasibility and Optimality of the Current Solution** - The **current basic solution** is given by the values of the basic variables at the bottom right of the tableau: - \( x_2 = 20 \) - \( x_5 = 10 \) - Non-basic variables (\( x_1, x_3, x_4 \)) = 0 **Feasibility:** - All basic variables (including slack) are non-negative: \( x_2 = 20 \), \( x_5 = 10 \), so the solution is still feasible. **Optimality:** - In the simplex method for maximization: - The solution is optimal if all the coefficients in the Z-row (other than basic variables) are **non-negative**. - The new Z-row is \( [0,\ 0,\ -5,\ 0,\ 5] \) (excluding the Z and RHS columns). - **The coefficient of \( x_3 \) is now -5**, which is **negative**. **Explanation:** Since the Z-row for \( x_3 \) is negative (-5), the current solution is **not optimal** anymore. This means that increasing \( x_3 \) would improve the objective function, so the optimal solution will change if we continue simplex iterations. --- ## **Final Answer** ### **Step 1:** Change the coefficient of \( x_3 \) in the objective function to 7. The Z-row in the tableau becomes: \[ [0,\ 0,\ -5,\ 0,\ 5,\ 100] \] **Explanation:** Only the coefficient for \( x_3 \) in the Z-row is updated by subtracting 6 from its previous value. ### **Step 2:** Test the current basic solution for feasibility and optimality: - The current solution remains **feasible** since all basic variables are non-negative. - The solution is **not optimal** because the Z-row coefficient for \( x_3 \) is now negative (-5). --- **Final Answer:** After changing the coefficient of \( x_3 \) in the objective function to 7, the current basic solution is still feasible, but it is **not optimal** because the reduced cost for \( x_3 \) is negative, indicating that further improvement is possible.

**Step 1:** Update the coefficient of \( x_3 \) in the objective function to 7. This affects the Z-row in the tableau, which was previously \( [0,\ 0,\ 1,\ 0,\ 5,\ 100] \). Since the old coefficient was 13 and now it's 7, the change is \(-6\). We subtract 6 times the \( x_3 \) column from the Z-row: \[ \text{New Z-row} = \text{Old Z-row} - 6 \times \text{Column of } x_3 \] Resulting in the updated Z-row: \[ [0,\ 0,\ -5,\ 0,\ 5,\ 100] \] **Explanation:** Only the coefficient for \( x_3 \) in the Z-row is adjusted, reflecting the change in its objective function coefficient. --- **Step 2:** Evaluate the current basic solution for feasibility and optimality: - Basic variables \( x_2 = 20 \) and \( x_5 = 10 \) are non-negative, so the solution remains feasible. - To check optimality, look at the Z-row coefficients for non-basic variables: \( x_1, x_3, x_4 \). - The coefficient for \( x_3 \) is now \(-5\), which is negative. - Since there is a negative coefficient in the Z-row (excluding the RHS), the current solution is **not optimal**. Increasing \( x_3 \) could improve the objective value. **Conclusion:** After changing the coefficient to 7, the current solution remains feasible but is **not optimal** because the reduced cost for \( x_3 \) is negative.