Question

Solve the following linear programming problem graphically: Maximize and minimize \( Z = 4x + 2y - 7 \) subject to the constraints \( x + 3y \le 60, x + y \ge 10, x - y \le 0, x \ge 0, y \ge 0 \).
Correct Answer: Max Value \( Z = 73 \) at \( (15, 15) \); Min Value \( Z = 13 \) at \( (5, 5) \). (Based on standard corner point analysis)

Hint:

Always double-check the intersection points by solving the equations of the boundary lines. This prevents errors in reading the graph.

Answer 1

Step 1: Understanding the Concept:
To solve an LPP graphically, we plot the linear inequalities to find the feasible region. The optimal value of the objective function occurs at the corner points of this region.
Step 2: Detailed Explanation:
1. Plot Constraints:
- \( x + 3y = 60 \): Points \( (60, 0), (0, 20) \).
- \( x + y = 10 \): Points \( (10, 0), (0, 10) \).
- \( x - y = 0 \): Line \( y = x \).
2. Feasible Region:
The region is bounded by the intersections of these lines.
3. Corner Points:
- Intersection of \( x+y=10 \) and \( y=x \): \( 2x=10 \implies (5, 5) \).
- Intersection of \( x+3y=60 \) and \( y=x \): \( 4x=60 \implies (15, 15) \).
- Intersection of \( x+3y=60 \) and \( x=0 \): \( (0, 20) \).
- Intersection of \( x+y=10 \) and \( x=0 \): \( (0, 10) \).
4. Evaluate Z at corner points:
- At \( (5, 5) \): \( Z = 4(5) + 2(5) - 7 = 20 + 10 - 7 = 23 \).
- At \( (15, 15) \): \( Z = 4(15) + 2(15) - 7 = 60 + 30 - 7 = 83 \).
- At \( (0, 20) \): \( Z = 4(0) + 2(20) - 7 = 33 \).
- At \( (0, 10) \): \( Z = 4(0) + 2(10) - 7 = 13 \).
Step 3: Final Answer:
Maximum value is 83 at \( (15, 15) \) and Minimum value is 13 at \( (0, 10) \).