Let's solve the given Linear Programming (LP) problem using the **Simplex Method** in two steps, with explanations after each step: --- ## **Step 1: Write the Problem in Standard Form and Set Up the Initial Simplex Tableau** **Given Problem:** Maximize \(Z = 2X_1 + 2X_2\) Subject to: \(X_1 + X_2 \leq 6\) \(2X_1 + X_2 \leq 13\) \(X_1, X_2 \geq 0\) **Add slack variables \(S_1\) and \(S_2\) to convert inequalities to equations:** \[ \begin{align*} X_1 + X_2 + S_1 &= 6 \\ 2X_1 + X_2 + S_2 &= 13 \\ X_1, X_2, S_1, S_2 &\geq 0 \end{align*} \] **Objective function in terms of all variables:** \[ \text{Maximize}\ Z = 2X_1 + 2X_2 + 0S_1 + 0S_2 \] **Initial Simplex Tableau:** \[ \begin{array}{c|cccc|c} \text{Basic} & X_1 & X_2 & S_1 & S_2 & \text{RHS} \\ \hline S_1 & 1 & 1 & 1 & 0 & 6 \\ S_2 & 2 & 1 & 0 & 1 & 13 \\ \hline Z & -2 & -2 & 0 & 0 & 0 \\ \end{array} \] **Explanation:** We have set up the problem in standard form and constructed the initial simplex tableau, which will help us proceed with the Simplex method. --- ## **Step 2: Perform the Simplex Iteration** **Identify entering variable:** Look for the most negative coefficient in the Z-row: both \(X_1\) and \(X_2\) have -2. Let's choose \(X_1\) (could also choose \(X_2\), result will be the same). **Identify leaving variable:** Calculate the ratios: - Row 1: \(6/1 = 6\) - Row 2: \(13/2 = 6.5\) Smallest non-negative ratio is 6, so **S₁ leaves and X₁ enters**. **Pivot on Row 1, Column \(X_1\):** - Make Row 1 pivot (already 1). - Make other entries in the \(X_1\) column zero by row operations. **Row Operations:** - New Row 2: Row 2 - 2 × Row 1 - \(2-2(1) = 0\) - \(1-2(1) = -1\) - \(0-2(1) = -2\) - \(1-2(0) = 1\) - \(13-2(6) = 1\) - New Z-row: Z + 2 × Row 1 - \(-2+2(1) = 0\) - \(-2+2(1) = 0\) - \(0+2(1) = 2\) - \(0+2(0) = 0\) - \(0+2(6) = 12\) **Updated Tableau:** \[ \begin{array}{c|cccc|c} \text{Basic} & X_1 & X_2 & S_1 & S_2 & \text{RHS} \\ \hline X_1 & 1 & 1 & 1 & 0 & 6 \\ S_2 & 0 & -1 & -2 & 1 & 1 \\ \hline Z & 0 & 0 & 2 & 0 & 12 \\ \end{array} \] Now, look at the Z-row: all coefficients of variables not in the basis (\(X_2, S_2\)) are zero or positive (none negative), so **this is optimal**. **Explanation:** We performed one pivot operation and found the optimal solution because all the coefficients in the Z-row are now zero or positive, indicating no further improvement is possible. --- ## **Final Answer** **The optimal solution is:** \[ X_1 = 6,\quad X_2 = 0 \] **Maximum Value:** \[ Z = 2X_1 + 2X_2 = 2(6) + 2(0) = 12 \] --- **Summary Table of Solution:** | Variable | Value | |----------|-------| | \(X_1\) | 6 | | \(X_2\) | 0 | | \(Z\) | 12 |