Question

Solve the following linear programming problem graphically: \[ {Maximise } z = 4x + 3y, \quad {subject to the constraints:} \] \[ x + y \leq 800, \quad 2x + y \leq 1000, \quad x \leq 400, \quad x, y \geq 0. \]

Hint:

For graphical LPP, solve the intersection points of constraints accurately and evaluate the objective function at each vertex of the feasible region.

Answer 1

1. Plot the constraints: - Line 1: \( x + y = 800 \). - Line 2: \( 2x + y = 1000 \). - Line 3: \( x = 400 \). 2. Identify the feasible region: The feasible region is the area in the first quadrant where all constraints intersect. 3. Determine the vertices of the feasible region: - Intersection of \( x + y = 800 \) and \( 2x + y = 1000 \): Subtracting the first equation from the second yields \( x = 200 \) and \( y = 600 \). The vertex is \( (200, 600) \). - Intersection of \( x = 400 \) and \( 2x + y = 1000 \): Substituting \( x = 400 \) into the second equation gives \( 2(400) + y = 1000 \), resulting in \( y = 200 \). The vertex is \( (400, 200) \). - Intersection of \( x = 400 \) and \( x + y = 800 \): Substituting \( x = 400 \) into the first equation gives \( 400 + y = 800 \), resulting in \( y = 400 \). The vertex is \( (400, 400) \). 4. Evaluate the objective function \( z = 4x + 3y \): - At \( (200, 600) \): \( z = 4(200) + 3(600) = 800 + 1800 = 2600 \). - At \( (400, 200) \): \( z = 4(400) + 3(200) = 1600 + 600 = 2200 \). - At \( (400, 400) \): \( z = 4(400) + 3(400) = 1600 + 1200 = 2800 \). 5. Conclusion: The maximum value of \( z \) is \( \mathbf{2600} \) which occurs at \( \mathbf{(200, 600)} \). Final Answer: The maximum value of \( z \) is \( 2600 \) at \( (200, 600) \).