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 operational decisions. In the context of university administration, LP can be effectively used to minimize costs associated with procuring necessary resources, such as chemicals for practical examinations. This problem demonstrates a classic application of LP, where the goal is to determine the optimal quantity of each type of box to purchase to meet the department’s chemical needs at the lowest possible cost.
Problem Formulation
Let:
- x = number of boxes of type A
- y = number of boxes of type B
Objective Function: Minimize the total cost (Z)
Z = 300x + 200y
Constraints:
- Chemical X: 3x + y ≥ 10
- Chemical Y: 2x + 2y ≥ 12
- Chemical Z: x + 2y ≥ 7
- Non-negativity: x ≥ 0, y ≥ 0
Graphical Solution
To solve this LPP graphically, we first convert the inequalities into equations and plot them on a graph.
Step 1: Plotting the Constraints
- 3x + y = 10 => y = 10 - 3x
- 2x + 2y = 12 => y = 6 - x
- x + 2y = 7 => y = (7 - x)/2
We need to determine the feasible region, which satisfies all the constraints. Since the constraints are "greater than or equal to," the feasible region will be above each line.
Step 2: Identifying the Feasible Region
The feasible region is the area where all constraints are simultaneously satisfied. This region is bounded by the lines and the axes (x ≥ 0, y ≥ 0). The corner points of the feasible region are crucial for finding the optimal solution.
Step 3: Finding the Corner Points
The corner points are the intersections of the constraint lines:
- Intersection of 3x + y = 10 and 2x + 2y = 12: Solving these equations, we get x = 2, y = 4. Point A (2, 4)
- Intersection of 3x + y = 10 and x + 2y = 7: Solving these equations, we get x = 1.8, y = 4.6. Point B (1.8, 4.6)
- Intersection of 2x + 2y = 12 and x + 2y = 7: Solving these equations, we get x = 5, y = 1.5. Point C (5, 1.5)
Step 4: Evaluating the Objective Function at Corner Points
We evaluate the objective function Z = 300x + 200y at each corner point:
| Corner Point | x | y | Z = 300x + 200y |
|---|---|---|---|
| A (2, 4) | 2 | 4 | 300(2) + 200(4) = 600 + 800 = 1400 |
| B (1.8, 4.6) | 1.8 | 4.6 | 300(1.8) + 200(4.6) = 540 + 920 = 1460 |
| C (5, 1.5) | 5 | 1.5 | 300(5) + 200(1.5) = 1500 + 300 = 1800 |
The minimum cost is ₹1400, which occurs at point A (2, 4).
Optimal Solution
The department should purchase 2 boxes of type A and 4 boxes of type B to minimize the total cost, which will be ₹1400. This solution ensures that the department has sufficient quantities of chemicals X, Y, and Z for the practical examination.
Conclusion
In conclusion, utilizing Linear Programming and the graphical method, we determined the optimal procurement strategy for the Chemistry department. Purchasing 2 boxes of type A and 4 boxes of type B results in the minimal cost of ₹1400 while satisfying the required quantities of each chemical. This approach demonstrates the power of mathematical modeling in optimizing resource allocation within an educational institution. Further refinement could involve considering potential discounts for bulk purchases or exploring alternative suppliers.
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.