4Vocabulary linear programming constraint feasible region objective function
5Green roofs are covered with plants instead of traditional materials like concrete or shingles to help lower heat and improve air quality.The plants landscape architects choose might depend on the price, the amount of water they require, and the amount of carbon dioxide they absorb.
6Linear programming is method of finding a maximum or minimum value of a function that satisfies a given set of conditions called constraints. A constraint is one of the inequalities in a linear programming problem. The solution to the set of constraints can be graphed as a feasible region.
7Example 1: Graphing a Feasible Region Yum’s Bakery bakes two breads, A and B. One batch of A uses 5 pounds of oats and 3 pounds of flour. One batch of B uses 2 pounds of oats and 3 pounds of flour. The company has 180 pounds of oats and 135 pounds of flour available. Write the constraints for the problem and graph the feasible region.
8Let x = the number of bread A, and y = the number of bread B. Example 1 ContinuedLet x = the number of bread A, andy = the number of bread B.Write the constraints:x ≥ 0The number of batches cannot be negative.y ≥ 0The combined amount of oats is less than or equal to 180 pounds.5x + 2y ≤ 180The combined amount of flour is less than or equal to 135 pounds.3x + 3y ≤ 135
9Graph the feasible region Graph the feasible region. The feasible region is a quadrilateral with vertices at (0, 0), (36, 0), (30, 15), and (0, 45).Check A point in the feasible region, such as (10, 10), satisfies all of the constraints.
10Graph the feasible region for the following constraints. Check It Out! Example 1Graph the feasible region for the following constraints.x ≥ 0The number cannot be negative.y ≥ 1.5The number is greater or equal to 220.127.116.11x + 5y ≤ 20The combined area is less than or equal to 20.3x + 2y ≤ 12The combined area is less than or equal to 12.
11Check It Out! Example 1 Continued Graph the feasible region. The feasible region is a quadrilateral with vertices at (0, 1.5), (0, 4), (2, 3), and (3, 1.5).Check A point in the feasible region, such as (2, 2), satisfies all of the constraints.
12In most linear programming problems, you want to do more than identify the feasible region. Often you want to find the best combination of values in order to minimize or maximize a certain function. This function is the objective function.The objective function may have a minimum, a maximum, neither, or both depending on the feasible region.
14More advanced mathematics can prove that the maximum or minimum value of the objective function will always occur at a vertex of the feasible region.
15Example 2: Solving Linear Programming Problems Yum’s Bakery wants to maximize its profits from bread sales. One batch of A yields a profit of $40. One batch of B yields a profit of $30. Use the profit information and the data from Example 1 to find how many batches of each bread the bakery should bake.
16Example 2 ContinuedStep 1 Let P = the profit from the bread.Write the objective function: P = 40x + 30yStep 2 Recall the constraints and the graph from Example 1.x ≥ 0y ≥ 05x + 2y ≤ 1803x + 3y ≤ 135
17Example 2 ContinuedStep 3 Evaluate the objective function at the vertices of the feasible region.(x, y)40x + 30yP($)(0, 0)40(0) + 30(0)(0, 45)40(0) + 30(45)1350(30, 15)40(30) + 30(15)1650(36, 0)40(36) + 30(0)1440The maximum value occurs at the vertex (30, 15).Yum’s Bakery should make 30 batches of bread A and 15 batches of bread B to maximize the amount of profit.
18Check your graph of the feasible region by using your calculator. Be sure to change the variables to x and y.Helpful Hint
19Check It Out! Example 2Maximize the objective function P = 25x + 30y under the following constraints.x ≥ 0y ≥ 1.52.5x + 5y ≤ 203x + 2y ≤ 12
20Check It Out! Example 2 Continued Step 1 Write the objective function: P= 25x + 30yStep 2 Use the constraints to graph.x ≥ 0y ≥ 1.52.5x + 5y ≤ 203x + 2y ≤ 12
21Check It Out! Example 2 Continued Step 3 Evaluate the objective function at the vertices of the feasible region.(x, y)25x + 30yP($)(0, 4)25(0) + 30(4)120(0, 1.5)25(0) + 30(1.5)45(2, 3)25(2) + 30(3)140(3, 1.5)25(3) + 30(1.5)The maximum value occurs at the vertex (2, 3).P = 140
22Example 3: Problem-Solving Application Sue manages a soccer club and must decide how many members to send to soccer camp. It costs $75 for each advanced player and $50 for each intermediate player. Sue can spend no more than $13,250. Sue must send at least 60 more advanced than intermediate players and a minimum of 80 advanced players. Find the number of each type of player Sue can send to camp to maximize the number of players at camp.
23Understand the Problem Example 3 Continued1Understand the ProblemThe answer will be in two parts—the number of advanced players and the number of intermediate players that will be sent to camp.
24Understand the Problem 1Understand the ProblemList the important information:Advanced players cost $75. Intermediate players cost $50.Sue can spend no more than $13,250.Sue must send at least 60 more advanced players than intermediate players.There needs to be a minimum of 80 advanced players.Sue wants to send the maximum number of players possible.
252Make a PlanLet x = the number of advanced players and y = the number of intermediate players. Write the constraints and objective function based on the important information.x ≥ 80The number of advanced players is at least 80.The number of intermediate players cannot be negative.y ≥ 0x – y ≥ 60There are at least 60 more advanced players than intermediate players.The total cost must be no more than $13,250.75x + 50y ≤ 13,250Let P = the number of players sent to camp.The objective function is P = x + y.
26Solve3Graph the feasible region, and identify the vertices. Evaluate the objective function at each vertex.P(80, 0) = (80) + (0) = 80P(80, 20) = (80) + (20) = 100P(176, 0) = (176) + (0) = 176P(130,70) = (130) + (70) = 200
27Check the values (130, 70) in the constraints. Look Back4Check the values (130, 70) in the constraints.x ≥ 80y ≥ 0130 ≥ 8070 ≥ 0x – y ≥ 6075x + 50y ≤ 13,250(130) – (70) ≥ 6075(130) + 50(70) ≤ 13,25060 ≥ 6013,250 ≤ 13,250
28Check It Out! Example 3A book store manager is purchasing new bookcases. The store needs 320 feet of shelf space. Bookcase A provides 32 ft of shelf space and costs $200. Bookcase B provides 16 ft of shelf space and costs $125. Because of space restrictions, the store has room for at most 8 of bookcase A and 12 of bookcase B. How many of each type of bookcase should the manager purchase to minimize the cost?
29Understand the Problem 1Understand the ProblemThe answer will be in two parts—the number of bookcases that provide 32 ft of shelf space and the number of bookcases that provide 16 ft of shelf space.List the important information:Bookcase A cost $200. Bookcase B cost $125.The store needs at least 320 feet of shelf space.Manager has room for at most 8 of bookcase A and 12 of bookcase B.Minimize the cost of the types of bookcases.
302Make a PlanLet x represent the number of Bookcase A and y represent the number of Bookcase B. Write the constraints and objective function based on the important information.x ≥ 0The number of Bookcase A cannot be negative.y ≥ 0The number of Bookcase B cannot be negative.x ≤ 8There are 8 or less of Bookcase A.y ≤ 12There are 12 or less of Bookcase B.32x + 16y ≤ 320The total shelf space is at least 320 feet.Let P = The number of Bookcase A and Bookcase B. The objective function is P = 200x + 125y.
31Solve3Graph the feasible region, and identify the vertices. Evaluate the objective function at each vertex.P(4, 12) = (800) + (1500) = 2300P(8, 12) = (1600) + (1500) = 3100P(8, 4) = (1600) + (500) = 2100
33Lesson Quiz1. Ace Guitars produces acoustic and electric guitars. Each acoustic guitar yields a profit of $30, and requires 2 work hours in factory A and 4 work hours in factory B. Each electric guitar yields a profit of $50 and requires 4 work hours in factory A and 3 work hours in factory B. Each factory operates for at most 10 hours each day. Graph the feasible region. Then, find the number of each type of guitar that should be produced each day to maximize the company’s profits.