answer:To maximize the profit, we need to find the optimal combination of products A and B that the company should produce, given the constraints on machine hours. We can use linear programming to solve this problem. Let x be the number of units of product A and y be the number of units of product B. Objective function (profit): P = 5x + 4y Constraints: 1. 2x + y ≤ 20 (machine X hours) 2. x + 3y ≤ 10 (machine Y hours) 3. x ≥ 0 4. y ≥ 0 Now, we'll find the feasible region by graphing the constraints: 1. y ≤ -2x + 20 2. y ≤ -(1/3)x + 10/3 The feasible region is the area where all constraints are satisfied. The vertices of the feasible region are the points where the constraints intersect. We'll find these vertices by solving the systems of equations: Intersection of constraint 1 and 2: 2x + y = 20 x + 3y = 10 Solving this system, we get x = 2 and y = 8/3. Intersection of constraint 1 and x-axis (y = 0): 2x + 0 = 20 x = 10 Intersection of constraint 2 and y-axis (x = 0): 0 + 3y = 10 y = 10/3 Now, we'll evaluate the profit function P at each of these vertices: 1. P(2, 8/3) = 5(2) + 4(8/3) = 10 + 32/3 = 62/3 ≈ 20.67 2. P(10, 0) = 5(10) + 4(0) = 50 3. P(0, 10/3) = 5(0) + 4(10/3) = 40/3 ≈ 13.33 The maximum profit occurs at the vertex (10, 0), with a profit of 50. Therefore, the company should produce 10 units of product A and 0 units of product B to maximize the profit.

answer:Let x be the number of units of Product A and y be the number of units of Product B. The objective function to maximize the profit is: P(x, y) = 8x + 10y Subject to the following constraints: 1. Resource X constraint: x + y ≤ 500 2. Resource Y constraint: 2x + y ≤ 700 3. Product A constraint: 0 ≤ x ≤ 100 4. Product B constraint: 0 ≤ y ≤ 150 First, we need to find the feasible region by plotting these constraints on a graph. 1. x + y = 500 y = 500 - x 2. 2x + y = 700 y = 700 - 2x Now, let's find the intersection points of these lines: Intersection of (1) and (2): 500 - x = 700 - 2x x = 200 So, the intersection point is (200, 300). Now, let's check the intersection points with the product constraints: Intersection of (1) and x = 100: y = 500 - 100 = 400 Intersection of (2) and x = 100: y = 700 - 2(100) = 500 Intersection of (1) and y = 150: x = 500 - 150 = 350 Intersection of (2) and y = 150: x = (700 - 150) / 2 = 275 Now, we need to find the feasible region. Since x ≤ 100 and y ≤ 150, the feasible region is a polygon with vertices (0, 0), (100, 0), (100, 150), (0, 150), and (100, 400). Now, we need to find the maximum profit within this feasible region. We can do this by evaluating the profit function at each vertex: P(0, 0) = 8(0) + 10(0) = 0 P(100, 0) = 8(100) + 10(0) = 800 P(100, 150) = 8(100) + 10(150) = 2300 P(0, 150) = 8(0) + 10(150) = 1500 P(100, 400) is not in the feasible region since it violates the constraint y ≤ 150. The maximum profit is achieved at the vertex (100, 150), with a profit of 2300. The company should produce 100 units of Product A and 150 units of Product B to achieve this profit.

answer:Let x be the number of units of P1 and y be the number of units of P2. The constraints for the raw materials are: 2x + y ≤ 5 (raw material A) x + 3y ≤ 12 (raw material B) x ≥ 0, y ≥ 0 (non-negativity) The objective function to maximize the profit is: Profit = 3x + 5y We will solve this linear programming problem using the graphical method. First, we will find the feasible region by plotting the constraints on a graph. 1. 2x + y ≤ 5 y ≤ -2x + 5 2. x + 3y ≤ 12 y ≤ (-1/3)x + 4 The feasible region is the intersection of these inequalities, which is a quadrilateral with vertices A(0,0), B(0,4), C(3,2), and D(2,1). Now, we will evaluate the objective function at each vertex: A(0,0): Profit = 3(0) + 5(0) = 0 B(0,4): Profit = 3(0) + 5(4) = 20 C(3,2): Profit = 3(3) + 5(2) = 19 D(2,1): Profit = 3(2) + 5(1) = 11 The maximum profit occurs at vertex B(0,4), with a profit of 20. Therefore, the company should produce 0 units of P1 and 4 units of P2 to maximize its profit.

answer:Let x be the number of chairs and y be the number of tables. We have the following constraints: 1. x ≥ 100 (at least 100 chairs) 2. y ≥ 60 (at least 60 tables) 3. x + y ≤ 200 (maximum of 200 products) The profit function is P(x, y) = 30x + 50y, which we want to maximize. First, we need to find the feasible region based on the constraints. Since x ≥ 100 and y ≥ 60, we are in the first quadrant of the coordinate plane. The line x + y = 200 is the boundary for the maximum number of products. The feasible region is a triangle with vertices at (100, 60), (100, 140), and (200, 0). Now we need to check the profit at each vertex: 1. P(100, 60) = 30(100) + 50(60) = 3000 + 3000 = 6000 2. P(100, 140) = 30(100) + 50(140) = 3000 + 7000 = 10000 3. P(200, 0) = 30(200) + 50(0) = 6000 + 0 = 6000 The maximum profit is 10,000, which occurs when the factory produces 100 chairs and 140 tables.