Solve the following Linear Programming Problem graphically :
Maximize \( Z = \frac{2x}{5} + \frac{3y}{10} \)
subject to constraints
\( 2x + y \leq 1000 \)
\( x + y \leq 800 \)
\( x, y \geq 0 \).
Show Hint
Always shade the feasible region towards the origin for \( \leq \) constraints when intercepts are positive.
Evaluation of corner points is best presented in a table for clarity.
Concept: • Graphing inequalities to find the feasible region (common intersection).
• Identifying corner points of the bounded feasible region.
• Corner Point Theorem: Optimal solution occurs at a corner point.
Step 1: Find the boundary points for each constraint
Line 1 (\( 2x + y = 1000 \)):
• If \( x = 0 \), \( y = 1000 \). Point: \( (0, 1000) \)
• If \( y = 0 \), \( 2x = 1000 \Rightarrow x = 500 \). Point: \( (500, 0) \)
Line 2 (\( x + y = 800 \)):
• If \( x = 0 \), \( y = 800 \). Point: \( (0, 800) \)
• If \( y = 0 \), \( x = 800 \). Point: \( (800, 0) \)
Step 2: Identify the intersection point of the two lines
Solve \( 2x + y = 1000 \) and \( x + y = 800 \) simultaneously.
Subtracting the second equation from the first:
\[ (2x + y) - (x + y) = 1000 - 800 \]
\[ x = 200 \]
Substitute \( x = 200 \) into \( x + y = 800 \):
\[ 200 + y = 800 \Rightarrow y = 600 \]
Intersection point is \( (200, 600) \).
Step 3: Identify the feasible region corner points
The feasible region is bounded by the axes and the innermost constraints (since all are \( \leq \)).
The corner points are:
\( O(0, 0) \), \( A(500, 0) \), \( B(200, 600) \), and \( C(0, 800) \).
Step 4: Evaluate \( Z \) at each corner point
Objective Function: \( Z = 0.4x + 0.3y \).
• At \( O(0, 0) \): \( Z = 0.4(0) + 0.3(0) = 0 \)
• At \( A(500, 0) \): \( Z = 0.4(500) + 0.3(0) = 200 \)
• At \( C(0, 800) \): \( Z = 0.4(0) + 0.3(800) = 240 \)
• At \( B(200, 600) \): \( Z = 0.4(200) + 0.3(600) = 80 + 180 = 260 \)
Step 5: Conclusion
\includegraphics[width=0.5\linewidth]{Q29_sol.png}
The maximum value of \( Z \) is \( 260 \), which occurs at the point \( (200, 600) \).
