Linear Programming

Plot the lines:
1. : Intercepts (3, 0), (0, 2). Shade below ( ).
2. : Intercepts (2, 0), (0, 2). Shade above ( ).
3. , : First quadrant.
Intersection points:
- and : Solve: Point: (0, 2).
- and : . Point: (3, 0).
- and : . Point: (0, 2).
- and : . Point: (2, 0).
Feasible region corner points: (0, 2), (2, 0), (3, 0).

Evaluate at each corner point:
- At (0, 10): .
- At (2, 2): .
- At (4, 0): .
Minimum at (2, 2).

We need to solve the system of inequalities graphically. First, graph each inequality and find the feasible region. Step 1: Graph the inequality . The boundary line is , which can be rewritten as . Plot this line and shade the region below it. Step 2: Graph the inequality . The boundary line is , which can be rewritten as . Plot this line and shade the region above it. Step 3: Graph the inequalities and , which represent the first quadrant. Step 4: Find the corner points of the feasible region by solving the system of equations formed by the boundary lines. The intersection points are: - Intersection of and : - Intersection of and : - Intersection of and : The corner points of the feasible region are , .

Step 1: Understanding the problem.
The given linear objective function is . We need to find the minimum value of using the feasible corner points. Step 2: Checking each point.
- For , . - For , . - For , . The minimum value occurs at with . Step 3: Conclusion.
Thus, the point of minimum is , so the correct answer is (A).

To determine the maximum value of , we evaluate it at the vertices of the feasible region.

Objective function:

Evaluating at the given points:

Conclusion:

The maximum value of is 95 at .

Thus, is maximum when and .