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question:A farmer has 60 acres of land to plant crops. The farmer wants to maximize the profit by planting two crops: corn and wheat. The profits per acre for corn and wheat are 300 and 200, respectively. The farmer has 2000 pounds of fertilizer and 1000 hours of labor available. It takes 5 pounds of fertilizer and 2 hours of labor to produce an acre of corn, and 3 pounds of fertilizer and 1 hour of labor to produce an acre of wheat. How many acres of each crop should the farmer plant to maximize the profit?
answer:Let x be the number of acres of corn and y be the number of acres of wheat. We want to maximize the profit function P(x, y) = 300x + 200y, subject to the following constraints: 1. x + y ≤ 60 (total acres constraint) 2. 5x + 3y ≤ 2000 (fertilizer constraint) 3. 2x + y ≤ 1000 (labor constraint) 4. x ≥ 0, y ≥ 0 (non-negativity constraint) We can use the method of linear programming to solve this problem. First, we'll find the feasible region by graphing the constraints: 1. y ≤ 60 - x 2. y ≤ (2000 - 5x) / 3 3. y ≤ 1000 - 2x Now, we'll find the vertices of the feasible region by solving the systems of equations formed by the intersections of the constraint lines: Vertex 1: Intersection of constraints 1 and 2: x + y = 60 5x + 3y = 2000 Solving this system, we get x = 40, y = 20. Vertex 2: Intersection of constraints 1 and 3: x + y = 60 2x + y = 1000 Solving this system, we get x = 20, y = 40. Vertex 3: Intersection of constraints 2 and 3: 5x + 3y = 2000 2x + y = 1000 Solving this system, we get x = 50, y = 10. Now, we'll evaluate the profit function P(x, y) at each vertex: P(40, 20) = 300(40) + 200(20) = 12000 + 4000 = 16,000 P(20, 40) = 300(20) + 200(40) = 6000 + 8000 = 14,000 P(50, 10) = 300(50) + 200(10) = 15000 + 2000 = 17,000 The maximum profit of 17,000 is achieved when the farmer plants 50 acres of corn and 10 acres of wheat.
question:A furniture company produces chairs and tables. The company has 180 hours of carpentry worktime, 110 hours of sanding worktime, and 220 hours of painting worktime. It takes 6 hours to produce a chair and 8 hours to produce a table. The profit per chair is 30 and the profit per table is 20. Determine the number of chairs and tables the company should produce in order to maximize their profit.
answer:Let x be the number of chairs and y be the number of tables produced. We are given the following constraints: 1. Carpentry worktime: 6x + 8y ≤ 180 2. Sanding worktime: x + y ≤ 110 3. Painting worktime: 2x + 4y ≤ 220 4. x ≥ 0, y ≥ 0 (non-negativity constraints) We want to maximize the profit function P(x, y) = 30x + 20y. First, we will solve the inequality constraints to find the feasible region: 1. 6x + 8y ≤ 180 Divide by 2: 3x + 4y ≤ 90 y ≤ (90 - 3x) / 4 2. x + y ≤ 110 y ≤ 110 - x 3. 2x + 4y ≤ 220 Divide by 2: x + 2y ≤ 110 y ≤ (110 - x) / 2 Now, we will find the vertices of the feasible region by solving the system of equations formed by the intersection of the constraint lines: A. Intersection of constraint 1 and 2: 3x + 4y = 90 x + y = 110 Solve for x: x = 110 - y Substitute x in the first equation: 3(110 - y) + 4y = 90 330 - 3y + 4y = 90 y = 240 / 1 = 240 x = 110 - 240 = -130 (not valid, as x ≥ 0) B. Intersection of constraint 1 and 3: 3x + 4y = 90 x + 2y = 110 Solve for x: x = 110 - 2y Substitute x in the first equation: 3(110 - 2y) + 4y = 90 330 - 6y + 4y = 90 -2y = -240 y = 120 x = 110 - 2(120) = -130 (not valid, as x ≥ 0) C. Intersection of constraint 2 and 3: x + y = 110 x + 2y = 110 Subtract the first equation from the second: y = 0 x = 110 So, the feasible region has only one vertex (110, 0). Now, we will evaluate the profit function P(x, y) at this vertex: P(110, 0) = 30(110) + 20(0) = 3300 Therefore, the company should produce 110 chairs and 0 tables to maximize their profit, which will be 3300.
question:A local farm has 600 acres of land available to grow two crops: corn and wheat. A bushel of corn requires 2 acres of land and 1 unit of fertilizers. A bushel of wheat requires 1 acre of land and 2 units of fertilizers. The farm has 1,000 units of fertilizers available. If the profit per bushel of corn is 4 and the profit per bushel of wheat is 6, how many bushels of each crop should the farm produce to maximize profit?
answer:Let x be the number of bushels of corn and y be the number of bushels of wheat. We have the following constraints: Land constraint: 2x + y ≤ 600 (1) Fertilizer constraint: x + 2y ≤ 1000 (2) The objective function to maximize is the profit function, P(x, y) = 4x + 6y. First, we need to find the feasible region by solving the constraint inequalities. From constraint (1), we get y ≤ -2x + 600. From constraint (2), we get y ≤ -0.5x + 500. Now, we need to find the vertices of the feasible region. These vertices are the points where the constraint lines intersect. We have three lines: y = -2x + 600, y = -0.5x + 500, and y = 0 (x-axis). Intersection of y = -2x + 600 and y = -0.5x + 500: -2x + 600 = -0.5x + 500 -1.5x = -100 x = 200/3 ≈ 66.67 y = -2(200/3) + 600 = 400/3 ≈ 133.33 Intersection of y = -2x + 600 and y = 0: 0 = -2x + 600 x = 300 Intersection of y = -0.5x + 500 and y = 0: 0 = -0.5x + 500 x = 1000 Now, we have three vertices: (66.67, 133.33), (300, 0), and (1000, 0). However, the point (1000, 0) is not in the feasible region because it doesn't satisfy the land constraint. So, we have two vertices to consider: (66.67, 133.33) and (300, 0). Now, we evaluate the profit function at these vertices: P(66.67, 133.33) = 4(66.67) + 6(133.33) ≈ 1066.67 P(300, 0) = 4(300) + 6(0) = 1200 The maximum profit is 1,200, which occurs when the farm produces 300 bushels of corn and 0 bushels of wheat.
question:A furniture manufacturer produces two types of chairs, A and B. Each chair A requires 3 kg of wood and 2 hours of labor, while each chair B requires 4 kg of wood and 3 hours of labor. The manufacturer has 300 kg of wood and 210 hours of labor available for production. Each chair A generates a profit of 15, while each chair B generates a profit of 20. How many chairs of each type should the manufacturer produce in order to maximize their profit?
answer:Let x be the number of chairs of type A and y be the number of chairs of type B. The constraints for the problem are: 1. Wood constraint: 3x + 4y ≤ 300 2. Labor constraint: 2x + 3y ≤ 210 3. Non-negativity constraint: x ≥ 0, y ≥ 0 The objective function to maximize the profit is: P(x, y) = 15x + 20y First, we'll find the feasible region by solving the constraint inequalities. 1. Wood constraint: 3x + 4y ≤ 300 y ≤ (300 - 3x) / 4 2. Labor constraint: 2x + 3y ≤ 210 y ≤ (210 - 2x) / 3 Now, we'll find the corner points of the feasible region: A. Intersection of the wood constraint and the x-axis (y = 0): 3x + 4(0) ≤ 300 x ≤ 100 So, point A is (100, 0). B. Intersection of the labor constraint and the x-axis (y = 0): 2x + 3(0) ≤ 210 x ≤ 105 Since x ≤ 100 from the wood constraint, point B is not in the feasible region. C. Intersection of the wood constraint and the y-axis (x = 0): 3(0) + 4y ≤ 300 y ≤ 75 So, point C is (0, 75). D. Intersection of the labor constraint and the y-axis (x = 0): 2(0) + 3y ≤ 210 y ≤ 70 Since y ≤ 75 from the wood constraint, point D is not in the feasible region. E. Intersection of the wood constraint and the labor constraint: 3x + 4y = 300 2x + 3y = 210 Solving this system of equations: y = (300 - 3x) / 4 2x + 3((300 - 3x) / 4) = 210 Multiplying both sides by 4: 8x + 9(300 - 3x) = 840 Expanding and simplifying: 8x + 2700 - 27x = 840 -19x = -1860 x = 98 Substituting x back into the equation for y: y = (300 - 3(98)) / 4 y = (300 - 294) / 4 y = 6 / 4 y = 1.5 So, point E is (98, 1.5). Now, we'll evaluate the profit function P(x, y) at each of the corner points: P(A) = 15(100) + 20(0) = 1500 P(C) = 15(0) + 20(75) = 1500 P(E) = 15(98) + 20(1.5) = 1470 + 30 = 1500 The maximum profit is 1500, which can be achieved by producing 100 chairs of type A and 0 chairs of type B, or by producing 0 chairs of type A and 75 chairs of type B.