ve the linear programing problem using Simplex Method. \[ \begin{array}{l}\max Z=6 X_{1}+5 X_{2}+8 x_{3}\end{array} \] \[ \begin{array}{c}x_{1}+x_{2}+x_{3} \leq 12 \\ 2 x_{1}+x_{2}+2 x_{3} \leq 2 \\ x_{1}, x_{2}, x_{3} \quad \geq 0\end{array} \]

To solve the given linear programming problem using the Simplex Method, we will follow these steps: 1. **Identify the objective function and constraints.** - Objective function: \[ \max Z = 6X_1 + 5X_2 + 8X_3 \] - Constraints: \[ \begin{align*} 1. & \quad X_1 + X_2 + X_3 \leq 12 \\ 2. & \quad 2X_1 + X_2 + 2X_3 \leq 2 \\ 3. & \quad X_1, X_2, X_3 \geq 0 \end{align*} \] 2. **Convert inequalities to equalities by adding slack variables.** - Let \( S_1 \) and \( S_2 \) be the slack variables for the first and second constraints, respectively. - The constraints become: \[ \begin{align*} 1. & \quad X_1 + X_2 + X_3 + S_1 = 12 \\ 2. & \quad 2X_1 + X_2 + 2X_3 + S_2 = 2 \\ \end{align*} \] 3. **Set up the initial Simplex tableau.** The tableau will include the coefficients of the objective function and the constraints. \[ \begin{array}{c|cccccc|c} \text{Basis} & X_1 & X_2 & X_3 & S_1 & S_2 & Z & \text{RHS} \\ \hline S_1 & 1 & 1 & 1 & 1 & 0 & 0 & 12 \\ S_2 & 2 & 1 & 2 & 0 & 1 & 0 & 2 \\ \hline -Z & -6 & -5 & -8 & 0 & 0 & 1 & 0 \\ \end{array} \] 4. **Perform the Simplex iterations.** - Identify the pivot column (most negative in the Z-row). - Identify the pivot row (minimum ratio of RHS to pivot column). - Perform row operations to make the pivot column a unit column. Let's perform the calculations to find the optimal solution. ### Step 1: Identify the pivot column and row - The most negative value in the Z-row is -8 (for \( X_3 \)). - The ratios for the pivot row are: - For \( S_1 \): \( \frac{12}{1} = 12 \) - For \( S_2 \): \( \frac{2}{2} = 1 \) (minimum) Thus, \( S_2 \) will be the pivot row, and \( X_3 \) will be the pivot column. ### Step 2: Perform row operations We will pivot on the element in the second row and third column (2). 1. **Row 2 (new)**: Divide by 2: \[ R_2 = \frac{1}{2} R_2 \Rightarrow (1, 0.5, 1, 0, 0.5, 0, 1) \] 2. **Row 1 (new)**: Subtract \( R_2 \) from \( R_1 \): \[ R_1 = R_1 - R_2 \Rightarrow (0, 0.5, 0, 1, -0.5, 0, 11) \] 3. **Row 3 (new)**: Add 8 times \( R_2 \) to \( R_3 \): \[ R_3 = R_3 + 8R_2 \Rightarrow (0, 0, 0, 0, 4, 1, 8) \] The new tableau is: \[ \begin{array}{c|cccccc|c} \text{Basis} & X_1 & X_2 & X_3 & S_1 & S_2 & Z & \text{RHS} \\ \hline S_1 & 0 & 0.5 & 0 & 1 & -0.5 & 0 & 11 \\ X_3 & 1 & 0.5 & 1 & 0 & 0.5 & 0 & 1 \\ \hline -Z & 0 & -4 & 0 & 0 & 4 & 1 & 8 \\ \end{array} \] ### Step 3: Repeat the process - The most negative value in the Z-row is -4 (for \( X_2 \)). - The ratios for the pivot row are: - For \( S_1 \): \( \frac{11}{0.5} = 22 \) - For \( X_3 \): \( \frac{1}{0.5} = 2 \) (minimum) Thus, \( X_3 \) will be the pivot row, and \( X_2 \) will be the pivot column. ### Step 4: Perform row operations again 1. **Row 1 (new)**: Divide by 0.5: \[ R_1 = \frac{1}{0.5} R_1 \Rightarrow (0, 1, 0, 2, -1, 0, 22) \] 2. **Row 2 (new)**: Subtract \( 0.5R_1 \) from \( R_2 \): \[ R_2 = R_2 - 0.5R_1 \Rightarrow (1, 0, 1, -1, 0, 0, 0) \] 3. **Row 3 (new)**: Add 4 times \( R_1 \) to \( R_3 \): \[ R_3 = R_3 + 4R_1 \Rightarrow (0, 0, 0, 0, 0, 1, 0) \] The new tableau is: \[ \begin{array}{c|cccccc|c} \text{Basis} & X_1 & X_2 & X_3 & S_1 & S_2 & Z & \text{RHS} \\ \hline X_2 & 0 & 1 & 0 & 2 & -1 & 0 & 22 \\ X_3 & 1 & 0 & 1 & -1 & 0 & 0 & 0 \\ \hline -Z & 0 & 0 & 0 & 8 & 0 & 1 & 88