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Advanced Mathematical Concepts Chapter 2 Lesson 2-7 Example 1 MANUFACTURING Suppose a sporting goods factory can turn out 5000 balls each week. To meet the needs of its regular customers, the factory must produce 1500 basketballs and 2000 baseballs. If the profit for each basketball is $15 and the profit for each baseball is $5, how many of each type of ball should the factory produce to maximize profit? Define variables. Let x = the number of basketballs produced. Let y = the number of baseballs produced. Write inequalities. x 1500 y 2000 x + y 5000 Graph the system. The vertices are (1500, 2000), (3000, 2000), and (1500, 3500). Write an expression. Since profit is $15 per ball for basketballs and $5 per ball for baseballs, the profit function is P(x, y) = 15x + 5y. Substitute values. P(1500, 2000) = 15(1500) + 5(2000) = 32,500 P(3000, 2000) = 15(3000) + 5(2000) = 55,000 P(1500, 3500) = 15(1500) + 5(3500) = 40,000 Answer the problem. The maximum profit occurs when 3000 basketballs and 2000 baseballs are produced. Advanced Mathematical Concepts Chapter 2 Example 2 MANUFACTURING The Happy Times Toy Shop makes wooden cars and wooden tops. Each car requires 5 hours of woodworking and 2 hours of painting. Each top requires 3 hours of woodworking and 1 hour of painting. The profit is $10 on each car and $6 on each top. There are 60 hours available each week for woodworking and 35 hours available for painting. How many of each item should be produced in order to maximize profit? Define variables. Write inequalities. Let c = the number of cars produced. Let t = the number of tops produced. c0 t0 5c + 3t 60 No more than 60 hours woodworking. 2c + t 35 No more than 35 hours painting. Graph the system. The vertices are (0, 0), (12, 0), and (0, 20). Write an expression. Since profit on each car is $10 and the profit on each top is $6, the profit function is P(c, t) = 10c + 6t. Substitute values. P(0, 0) = 10(0) + 6(0) or 0 P(12, 0) = 10(12) + 6(0) or 120 P(0, 20) = 10(0) + 6(20) or 120 Answer the problem. The problem has alternate optimal solutions. The shop will make the same profit if they produce 12 cars and 0 tops as it will from producing 0 cars and 6 tops.