Question

Solve the following Linear Programming Problem using graphical method : Maximize \( Z = 100x + 50y \) subject to the constraints \[ 3x + y \leq 600, \quad x + y \leq 300, \quad y \leq x + 200, \quad x \geq 0, \quad y \geq 0. \]

Hint:

In Linear Programming, the optimal solution is found at one of the corner points of the feasible region. Always evaluate the objective function at these points.

Answer 1

1. Plot the constraints: The constraints are plotted as lines: - \( 3x + y = 600 \) - \( x + y = 300 \) - \( y = x + 200 \) 2. Determine the feasible region: The feasible region is the area where all the inequalities are satisfied. This is the region bounded by the lines. 3. Find the corner points of the feasible region: - Intersection of \( 3x + y = 600 \) and \( x + y = 300 \): The corner point is \( (150, 150) \). - Intersection of \( x + y = 300 \) and \( y = x + 200 \): The corner point is \( (50, 250) \). - Intersection of \( 3x + y = 600 \) and \( y = x + 200 \): The corner point is \( (100, 300) \). 4. Evaluate the objective function at the corner points: - At \( (150, 150) \), \[ Z = 100(150) + 50(150) = 22500. \] - At \( (50, 250) \), \[ Z = 100(50) + 50(250) = 17500. \] - At \( (100, 300) \), \[ Z = 100(100) + 50(300) = 25000. \] The maximum value of \( Z \) is 25000 at the corner point \( (100, 300) \). 5. Final Answer: The optimal solution is \( x = 100 \), \( y = 300 \), and the maximum value of \( Z = 25000 \).