Question 12 Consider the following linear programming model. \[ \begin{array}{l} \text { minimise } z=3 x_{1}+2 x_{2} \\ \text { subject to } 2 x_{1}+x_{2}=6 \\ x_{1}-x_{2} \geq 0 \\ x_{1}, x_{2} \geq 0 \end{array} \] What is the optimal \( z \) ? 9 12 8 10

We are given the linear programming model:  Minimize z = 3x₁ + 2x₂  Subject to 2x₁ + x₂ = 6       x₁ − x₂ ≥ 0       x₁, x₂ ≥ 0 Step 1. Use the equality constraint to express x₂ in terms of x₁.  2x₁ + x₂ = 6 ⟹ x₂ = 6 − 2x₁ Step 2. Substitute x₂ into the inequality:  x₁ − (6 − 2x₁) ≥ 0 ⟹ x₁ − 6 + 2x₁ ≥ 0  Combine like terms: 3x₁ − 6 ≥ 0 ⟹ 3x₁ ≥ 6 ⟹ x₁ ≥ 2 Step 3. Ensure non-negativity of x₂:  x₂ = 6 − 2x₁ ≥ 0 ⟹ 6 − 2x₁ ≥ 0 ⟹ 2x₁ ≤ 6 ⟹ x₁ ≤ 3 Thus, x₁ is in the interval [2, 3]. Step 4. Substitute x₂ in the objective function:  z = 3x₁ + 2(6 − 2x₁) = 3x₁ + 12 − 4x₁ = 12 − x₁ Since z decreases as x₁ increases, to minimize z we choose the maximum possible value of x₁, which is x₁ = 3 (provided that this value keeps x₂ non-negative). Step 5. Calculate the corresponding value of x₂:  x₂ = 6 − 2(3) = 6 − 6 = 0 Step 6. Compute the objective value:  z = 3(3) + 2(0) = 9 Thus, the optimal value of z is 9. The correct answer is 9.