Question:

Solve the following linear programming problem graphically: Maximise \( Z = 2x + 3y \), subject to the constraints: \( x + y \leq 6, \, x \geq 2, \, y \geq 3, \, x \geq 0, \, y \geq 0. \)

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In linear programming, always evaluate the objective function at the corner points of the feasible region.
Updated On: Feb 19, 2025
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Solution and Explanation

Graph


Step 1: Plot the constraints
Graph the constraints \( x+y \leq 6, \, x \geq 2, \, y \geq 3, \, x \geq 0, \, y \geq 0 \) on the Cartesian plane. The feasible region is the shaded region bounded by these lines. 
Step 2: Find corner points
The corner points of the feasible region are: \[ A(2, 3), \, B(3, 3), \, C(6, 0), \, D(2, 0). \] 
Step 3: Evaluate \( Z = 2x + 3y \) at each point
\[ Z(2, 3) = 13, \quad Z(3, 3) = 15, \quad Z(6, 0) = 12, \quad Z(2, 0) = 4. \] 
Conclusion: The maximum value of \( Z \) is \( 15 \), which occurs at \( (3, 3) \). 
 

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