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 represented graphically as lines: - \( 3x + y = 600 \) - \( x + y = 300 \) - \( y = x + 200 \)2. Identify the feasible region: The feasible region is the area satisfying all inequalities, bounded by the constraint lines.3. Determine the vertices of the feasible region: - Intersection of \( 3x + y = 600 \) and \( x + y = 300 \): Vertex: \( (150, 150) \). - Intersection of \( x + y = 300 \) and \( y = x + 200 \): Vertex: \( (50, 250) \). - Intersection of \( 3x + y = 600 \) and \( y = x + 200 \): Vertex: \( (100, 300) \).4. Evaluate the objective function at each vertex: - 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 \( Z \) value is 25000, occurring at vertex \( (100, 300) \).5. Conclusion: The optimal solution is \( x = 100 \), \( y = 300 \), yielding a maximum \( Z \) of 25000.