Let's break down the problem step by step: ### **Given LP:** \[ \begin{align*} \text{max } & z = 4x_1 + x_2 \\ \text{s.t. } & 3x_1 + 2x_2 \leq 6 \\ & 6x_1 + 3x_2 \leq 10 \\ & x_1, x_2 \geq 0 \end{align*} \] Introduce slack variables \( s_1 \) and \( s_2 \): \[ \begin{align*} 3x_1 + 2x_2 + s_1 &= 6 \\ 6x_1 + 3x_2 + s_2 &= 10 \\ x_1, x_2, s_1, s_2 \geq 0 \end{align*} \] --- ## **Given:** - Betsy says the optimal tableau Row 0 (z-row) is: [Z X1 1 0 X2 2 S1 0 $2 rhs 20 3] (This seems like a typo, but let's try to interpret it) - Basic variables are \(\{s_1, x_1\}\). --- ## **Step 1: Read the optimal dual solution \( y = [y_1, y_2] \) from the tableau** Recall: - Dual variables correspond to constraints. - The optimal dual solution is \( y = c_B B^{-1} \), where \( c_B \) are the objective coefficients of the basic variables. - Here, the basics are \( s_1 \) (coefficient 0) and \( x_1 \) (coefficient 4). - \( B \) is the basis matrix (columns for \( s_1 \) and \( x_1 \)). Let's write the constraint matrix: \[ A = \begin{bmatrix} 3 & 2 \\ 6 & 3 \\ \end{bmatrix} \] Slack columns: - \( s_1 \): [1, 0]^T - \( s_2 \): [0, 1]^T So, the full constraints in variable order [\(x_1, x_2, s_1, s_2\)]: \[ \begin{align*} 3x_1 + 2x_2 + 1s_1 + 0 s_2 &= 6 \\ 6x_1 + 3x_2 + 0s_1 + 1 s_2 &= 10 \end{align*} \] **The basis is \([s_1, x_1]\):** - \( s_1 \) column: [1, 0]^T - \( x_1 \) column: [3, 6]^T So, \[ B = \begin{bmatrix} 1 & 3 \\ 0 & 6 \end{bmatrix} \] \( c_B = \begin{bmatrix} 0 \\ 4 \end{bmatrix} \) **Compute \( B^{-1} \):** \[ B = \begin{bmatrix} 1 & 3 \\ 0 & 6 \end{bmatrix} \] Inverse is: \[ B^{-1} = \begin{bmatrix} 1 & -0.5 \\ 0 & \frac{1}{6} \end{bmatrix} \] **So,** \[ y^T = c_B^T B^{-1} = [0, 4] \begin{bmatrix} 1 & -0.5 \\ 0 & \frac{1}{6} \end{bmatrix} = [0 \cdot 1 + 4 \cdot 0, \quad 0 \cdot (-0.5) + 4 \cdot \frac{1}{6}] = [0, \frac{4}{6}] = [0, \frac{2}{3}] \] So, \[ \boxed{y_1 = 0, \quad y_2 = \frac{2}{3}} \] --- ## **Step 2: Will the current optimal dual solution remain feasible if Change 4 is considered where \( c_2 = 1 \) and \( a_2 = ? \) (y/n)** The question is incomplete because it says \( a_2 = ? \). But let's clarify: dual feasibility requires that for all \(j\): \[ c_j - y^T a_j \leq 0 \quad \text{if } x_j = 0 \text{ in the basis} \] We need to check dual feasibility after changing \(c_2\) (the coefficient of \(x_2\)) to 1. But without knowing \(a_2\), we cannot answer. So, **cannot answer** (n/a). --- ## **Step 3: Will the current optimal dual solution remain feasible if \( c_2 = 1 \) and \( a_2 = 21 \)? (y/n)** Assume \( a_2 \) is the right-hand side for constraint 2 (\(6x_1 + 3x_2 \leq 21\), i.e., b2 is changed to 21). But the dual feasibility relates to cost coefficients and constraint matrix, not the RHS. The question is likely asking about primal feasibility (not dual feasibility). But let's check: changing \( b_2 \) to 21, does that affect dual feasibility? No, dual feasibility is only affected by changes in the costs or the constraint matrix, not the RHS. However, changing \( c_2 \) to 1 affects dual feasibility for variable \( x_2 \): The dual constraint for \( x_2 \): \[ c_2 - y^T a_2 \leq 0 \] \( a_2 \) is the column of \( x_2 \) in the constraint matrix: [2, 3]^T. With \( y = [0, 2/3] \): \[ c_2 - y^T a_2 = 1 - [0, 2/3] \cdot \begin{bmatrix}2 \\ 3\end{bmatrix} = 1 - (0 \cdot 2 + \frac{2}{3} \cdot 3) = 1 - 2 = -1 \] So \( -1 \leq 0 \), so dual feasibility **holds**. So, **YES**, the current optimal dual solution remains feasible. \[ \boxed{y} \] --- ## **Step 4: What is the new value of \(z\) if the change \(b_2 = 21\) is made?** When the RHS changes, the change in optimal \(z\) is given by: \[ z^*_{\text{new}} = z^*_{\text{old}} + y_2 \cdot \Delta b_2 \] Given that \( y_2 = \frac{2}{3} \), \( \Delta b_2 = 21 - 10 = 11 \): \[ z^*_{\text{new}} = z^*_{\text{old}} + \frac{2}{3} \times 11 = z^*_{\text{old}} + \frac{22}{3} \] But what was \( z^*_{\text{old}} \)? Let's compute it: From the original problem, with the optimal basis [\( s_1, x_1 \)], let's solve for the optimal solution. From constraints: \[ \begin{align*} 3x_1 + 2x_2 + s_1 &= 6 \\ 6x_1 + 3x_2 + s_2 &= 10 \\ \end{align*} \] In basis [\( s_1, x_1 \)], set \( x_2 = 0, s_2 = 0 \): \[ \begin{align*} 3x_1 + s_1 &= 6 \\ 6x_1 &= 10 \implies x_1 = \frac{10}{6} = \frac{5}{3} \\ s_1 = 6 - 3x_1 = 6 - 3 \times \frac{5}{3} = 6 - 5 = 1 \\ \end{align*} \] Thus, \[ x_1 = \frac{5}{3},\quad x_2 = 0 \] \[ z^*_{\text{old}} = 4 \times \frac{5}{3} + 1 \times 0 = \frac{20}{3} \] Now, with \( b_2 = 21 \): \[ z^*_{\text{new}} = \frac{20}{3} + \frac{22}{3} = \frac{42}{3} = 14 \] --- ### **Final Answers** **(1) Dual solution:** \[ \boxed{y = [0, \frac{2}{3}]} \] **(2) Will the current optimal dual