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.

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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.
The optimal solution is at (0, 2) with a minimum value of 8.

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.

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