University:
The University Of Western Ontario
Course:
MATH 3020A | Introduction to Abstract Algebra
Academic year:

2024
Views:

334

Pages:

2
Author:

Kyuwon J.

Statement A small factory produces two types of products: A and B. The production of each product requires two resources: labor hours and raw materials. Each unit of product A requires 2 labor hours and 3 units of raw materials, while each unit of product B requires 4 labor hours and 2 units of raw materials. The factory has a total of 120 labor hours and 100 units of raw materials available per week. The profit per unit is $30 for product A and $50 for product B. Formulate and solve a linear programming problem to maximize the factory’s weekly profit. Step 1: Define the Decision Variables Let: x = Number of units of product A to produce per week. y = Number of units of product B to produce per week. Step 2: Formulate the Objective Function The objective is to maximize the total profit: Maximize Z = 30x + 50y Step 3: Formulate the Constraints 1. Labor hours constraint: 2x + 4y ≤ 120 2. Raw materials constraint: 3x + 2y ≤ 100 3. Non-negativity constraints: x ≥ 0, y ≥ 0 Step 4: Solve the Problem To solve, graph the constraints and find the feasible region. Identify the vertices of the feasible region and evaluate the objective function at each vertex. Graphing the Constraints • From the labor hours constraint 2x + 4y ≤ 120, the boundary line is: y= 120 − 2x = 30 − 0.5x 4 • From the raw materials constraint 3x + 2y ≤ 100, the boundary line is: y= 100 − 3x = 50 − 1.5x 2 Vertices of the Feasible Region The intersection points of the constraints are: 1. Intersection of 2x + 4y = 120 and 3x + 2y = 100: Solve: 2x + 4y = 120 and 3x + 2y = 100 Multiply the second equation by 2: 6x + 4y = 200 Subtract the first equation: ⇒ 4x = 80 1 x = 20 Substitute x = 20 into 2x + 4y = 120: 2(20) + 4y = 120 ⇒ 40 + 4y = 120 ⇒ y = 20 Thus, one vertex is (20, 20). 2. Intersection of 2x + 4y = 120 and x = 0: Substitute x = 0 into 2x + 4y = 120: 4y = 120 ⇒ y = 30 Thus, another vertex is (0, 30). 3. Intersection of 3x + 2y = 100 and y = 0: Substitute y = 0 into 3x + 2y = 100: 3x = 100 ⇒ x= 100 ≈ 33.33 3 Thus, another vertex is (33.33, 0). Evaluate the Objective Function at Each Vertex • At (20, 20): Z = 30(20) + 50(20) = 600 + 1000 = 1600 • At (0, 30): Z = 30(0) + 50(30) = 0 + 1500 = 1500 • At (33.33, 0): Z = 30(33.33) + 50(0) ≈ 1000 Step 5: Conclusion The maximum profit is achieved at (20, 20), where: Z = 1600 Thus, the factory should produce 20 units of product A and 20 units of product B to maximize profit. 2