Model Answer
0 min readIntroduction
Linear Programming (LP) is a mathematical technique used to optimize an objective function, subject to a set of constraints. It’s widely applied in resource allocation, production planning, and various other managerial decision-making processes. The graphical method is a visual approach to solving LP problems with two decision variables. This method allows for a clear understanding of the feasible region and the optimal solution. In this problem, we aim to minimize the cost function C = 5x1 + 6x2, given constraints related to production or resource limitations. The constraints define the boundaries within which the optimal solution must lie.
Problem Formulation and Graphical Representation
The given Linear Programming Problem is:
Minimize cost : C = 5x1 + 6x2
Subject to:
- x1 + x2 = 1000
- x1 ≤ 300
- x2 ≥ 150
- x1, x2 ≥ 0
Step 1: Convert Inequalities to Equalities (where applicable) and Plot the Constraints
The constraints are plotted on a graph with x1 on the x-axis and x2 on the y-axis.
- x1 + x2 = 1000: This is a straight line. When x1 = 0, x2 = 1000. When x2 = 0, x1 = 1000.
- x1 ≤ 300: This is a vertical line at x1 = 300. The feasible region lies to the left of this line.
- x2 ≥ 150: This is a horizontal line at x2 = 150. The feasible region lies above this line.
- x1, x2 ≥ 0: These constraints restrict the solution to the first quadrant.
Step 2: Identify the Feasible Region
The feasible region is the area on the graph that satisfies all the constraints simultaneously. It is the intersection of all the regions defined by the constraints. In this case, it's a polygon bounded by the lines x1 + x2 = 1000, x1 = 300, x2 = 150, x1 = 0, and x2 = 0.
Step 3: Determine the Corner Points of the Feasible Region
The corner points are the vertices of the feasible region. These are the points where the constraint lines intersect. We need to find the coordinates of these points.
- A: Intersection of x1 = 300 and x2 = 150. Coordinates: (300, 150)
- B: Intersection of x1 = 300 and x1 + x2 = 1000. Substituting x1 = 300 into the equation, we get 300 + x2 = 1000, so x2 = 700. Coordinates: (300, 700)
- C: Intersection of x2 = 150 and x1 + x2 = 1000. Substituting x2 = 150 into the equation, we get x1 + 150 = 1000, so x1 = 850. Coordinates: (850, 150)
Step 4: Evaluate the Objective Function at Each Corner Point
The objective function is C = 5x1 + 6x2. We evaluate C at each corner point:
- At A (300, 150): C = 5(300) + 6(150) = 1500 + 900 = 2400
- At B (300, 700): C = 5(300) + 6(700) = 1500 + 4200 = 5700
- At C (850, 150): C = 5(850) + 6(150) = 4250 + 900 = 5150
Step 5: Identify the Optimal Solution
Since we are minimizing the cost, we choose the corner point with the lowest value of C. In this case, the minimum cost is 2400, which occurs at the point A (300, 150).
Therefore, the optimal solution is x1 = 300 and x2 = 150, with a minimum cost of 2400.
Conclusion
In conclusion, by employing the graphical method, we have successfully minimized the cost function C = 5x1 + 6x2 subject to the given constraints. The optimal solution, x1 = 300 and x2 = 150, yields a minimum cost of 2400. This demonstrates the power of linear programming in optimizing resource allocation and decision-making. The graphical method, while effective for two variables, becomes less practical for problems with more variables, necessitating the use of more advanced techniques like the Simplex method.
Answer Length
This is a comprehensive model answer for learning purposes and may exceed the word limit. In the exam, always adhere to the prescribed word count.