Problem \( 2-31 \) Consider the following linear program: Min \( 3 A+4 B \) s.t. \[ \begin{array}{l}A+3 B \geq 6 \\ A+B \quad A, B \geq 0\end{array} \] a. Select the correct graph that shows the feasible region and the optimal solution for the problem.

To solve the linear programming problem, we need to follow these steps: 1. **Identify the objective function and constraints:** - Objective function: Minimize \( Z = 3A + 4B \) - Constraints: 1. \( A + 3B \geq 6 \) 2. \( A + B \geq 0 \) 3. \( A \geq 0 \) 4. \( B \geq 0 \) 2. **Convert the inequalities into equalities to find the boundary lines:** - From \( A + 3B = 6 \) - From \( A + B = 0 \) (which is simply the line \( B = -A \)) 3. **Graph the constraints:** - For \( A + 3B = 6 \): - When \( A = 0 \), \( 3B = 6 \) → \( B = 2 \) (point (0, 2)) - When \( B = 0 \), \( A = 6 \) (point (6, 0)) - For \( A + B = 0 \): - This line passes through the origin and has a slope of -1. 4. **Determine the feasible region:** - The inequality \( A + 3B \geq 6 \) indicates that we are interested in the area above the line \( A + 3B = 6 \). - The inequality \( A + B \geq 0 \) indicates that we are interested in the area above the line \( A + B = 0 \). - Since \( A \) and \( B \) must be non-negative, we are limited to the first quadrant. 5. **Find the intersection points of the constraints:** - To find the intersection of \( A + 3B = 6 \) and \( A + B = 0 \), we can solve these equations together. Let's solve the system of equations: \[ \begin{align*} 1. & \quad A + 3B = 6 \\ 2. & \quad A + B = 0 \implies A = -B \end{align*} \] Substituting \( A = -B \) into the first equation: \[ -B + 3B = 6 \implies 2B = 6 \implies B = 3 \implies A = -3 \] Since \( A \) cannot be negative, we discard this intersection. 6. **Identify the vertices of the feasible region:** - The feasible region is bounded by the lines and the axes. The vertices will be: - (0, 2) from \( A + 3B = 6 \) - (6, 0) from \( A + 3B = 6 \) - (0, 0) from the axes. 7. **Evaluate the objective function at the vertices:** - At (0, 2): \( Z = 3(0) + 4(2) = 8 \) - At (6, 0): \( Z = 3(6) + 4(0) = 18 \) - At (0, 0): \( Z = 3(0) + 4(0) = 0 \) 8. **Determine the optimal solution:** - The minimum value of \( Z \) occurs at (0, 2) with \( Z = 8 \). Now, you can select the correct graph that shows the feasible region and the optimal solution at the point (0, 2). The feasible region should be shaded above the line \( A + 3B = 6 \) and above the line \( A + B = 0 \), while also being in the first quadrant.