1. Slide 8.4- 1 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

2. OBJECTIVES Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Systems of Linear Inequalities; Linear Programming Learn to graph linear systems of inequalities in two variables. Learn to graph a linear inequality in two variables. Learn to apply systems of linear inequalities to linear programming. SECTION 8.4 1 2 3

3. Slide 8.4- 3 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Definitions The statements x + y > 4, 2x + 3y < 7, y ≥ x, and x + y ≤ 9 are examples of linear inequalities in the variables x and y. A solution of an inequality in two variables x and y is an ordered pair (a, b) that results in a true statement when x is replaced by a, and y is replaced by b in the inequality. The set of all solutions of an inequality is called the solution set of the inequality. The graph of an inequality in two variables is the graph of the solution set of the inequality.

4. Slide 8.4- 4 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley PROCEDURE FOR GRAPHING A LINEAR INEQUALITY IN TWO VARIABLES Step 1.Replace the inequality symbol by an equals (=) sign. Step 2.Sketch the graph of the corresponding equation Step 1. Use a dashed line for the boundary if the given inequality sign is, and a solid line if the inequality symbol is ≤ or ≥.

5. Slide 8.4- 5 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley PROCEDURE FOR GRAPHING A LINEAR INEQUALITY IN TWO VARIABLES Step 3.The graph in Step 2 will divide the plane into two regions. Select a test point in the plane. Be sure that the test point does not lie on the graph of the equation in Step 1. Step 4.(i) If the coordinates of the test point satisfy the given inequality, then so do all the points of the region that contains

6. Slide 8.4- 6 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley PROCEDURE FOR GRAPHING A LINEAR INEQUALITY IN TWO VARIABLES the test point. Shade the region that contains the test point. (ii) If the coordinates of the test point do not satisfy the given inequality, shade the region that does not contain the test point. The shaded region (including the boundary if it is solid) is the graph of the given inequality.

7. Slide 8.4- 7 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley EXAMPLE 2 Graphing Inequalities Sketch the graph of each of the following inequalities. a. x ≥ 2 b. y < 3 c. x + y < 4 Solution a. Step 1 Change the ≥ to = : x = 2 Step 2 Graph x = 2 with a solid line. Step 3 Test (0, 0). 0 ≥ 2 is a false statement.

8. Slide 8.4- 8 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley EXAMPLE 2 Graphing Inequalities Solution continued Step 3 continued The region not containing (0, 0), together with the vertical line, is the solution set. Step 4 Shade the solution set.

9. Slide 8.4- 9 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley EXAMPLE 2 Graphing Inequalities Solution continued Step 2 Graph y = 3 with a dashed line. Step 3 Test (0, 0). 0 < 3 is a true statement. The region containing (0, 0) is the solution set. Step 4 Shade the solution set. b. Step 1 Change the < to = : y < 3

10. Slide 8.4- 10 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley EXAMPLE 2 Graphing Inequalities Solution continued Step 2 Graph x + y = 4 with a dashed line. Step 3 Test (0, 0). 0 < 4 is a true statement. The region containing (0, 0) is the solution set. Step 4 Shade the solution set. c. Step 1 Change the < to = : x + y = 4

11. Slide 8.4- 11 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley SYSTEMS OF LINEAR INEQUALITIES IN TWO VARIABLES An ordered pair (a, b) is a solution of a system of inequalities involving two variables x and y if and only if, when x is replaced by a and y is replaced by b in each inequality of the system, all resulting statements are true. The solution set of a system of inequalities is the intersection of the solution sets of all the inequalities in the system.

12. Slide 8.4- 12 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley EXAMPLE 3 Graphing a System of Two Inequalities Graph the solution set of the system of Solution Step 1 2x + 3y = 6 Step 2 Sketch as a dashed line by joining the points (0, 2) and (3, 0). inequalities: Step 3Test (0, 0). 2(0) + 3(0) > 6 is a false statement. Step 4 Shade the solution set.

13. Slide 8.4- 13 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley EXAMPLE 3 Graphing a System of Two Inequalities Solution continued Now graph the second inequality.

14. Slide 8.4- 14 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley EXAMPLE 3 Graphing a System of Two Inequalities Solution continued Step 2 Sketch as a solid line by joining the points (0, 0) and (1, 1). Step 3Test (1, 0). 2(0) – 3(1) > 6 is a false statement. Step 4 Shade the solution set. Step 1 y – x = 0

15. Slide 8.4- 15 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley EXAMPLE 3 Graphing a System of Two Inequalities Solution continued The graph of the solution set of inequalities (1) and (2) is the region where the shading overlaps.

16. Slide 8.4- 16 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley LINEAR PROGRAMMING The process of finding a maximum or minimum value of a quantity is called optimization. A linear programming problem satisfies the following two conditions: 1.The quantity f to be maximized or minimized can be written as a linear expression in x and y. That is, f = ax + by, where a ≠ 0, b ≠ 0 are constants.

17. Slide 8.4- 17 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley LINEAR PROGRAMMING The inequalities that determine the region S are called constraints, the region S is called the set of feasible solutions, and f = ax + by is called the objective function. A point in S at which f attains its maximum (or minimum) value, together with the value of f at that point, is called an optimal solution. 2.The domain of the variables x and y is restricted to a region S that is determined by (is a solution set of) a system of linear inequalities.

18. Slide 8.4- 18 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley PROCEDURE FOR SOLVING A LINEAR PROGRAMMING PROBLEM Step 1.Write an expression for the quantity to be maximized or minimized. This expression is the objective function. Step 2.Write all constraints as linear inequalities. Step 3.Graph the solution set of the constraint inequalities. This set is the set of feasible solutions.

19. Slide 8.4- 19 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Step 4.Find all the vertices of the solution set in Step 3 by solving all pairs of corresponding equations for the constraint inequalities. Step 5.Find the values of the objective function at each of the vertices of Step 4. Step 6.The largest of the values (if any) in Step 5 is the maximum value of the objective function, and the smallest of the values (if any) in Step 5 is the minimum value of the objective function.

20. Slide 8.4- 20 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley EXAMPLE 5 Nutrition; Minimizing Calories Fat Albert wishes to go on a crash diet and needs your help in designing a lunch menu. The menu is to include two items: soup and salad. The vitamin units (milligrams) and calorie counts in each ounce of soup and salad are given in the table. ItemVitamin AVitamin CCalories Soup Salad 1111 3232 50 40

21. Slide 8.4- 21 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley EXAMPLE 5 Nutrition; Minimizing Calories Find the number of ounces of each item in the menu needed to provide the required vitamins with the fewest number of calories. The menu must provide at least: 10 units of Vitamin A 24 units of Vitamin C Solution a. State the problem mathematically. Step 1Write the objective function. x = ounces of soup y = ounces of salad

22. Slide 8.4- 22 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley EXAMPLE 5 Nutrition; Minimizing Calories Minimize the total number of calories. Solution continued Step 2Write the constraints. 1 ounce soup provides 1 unit vitamin A 1 ounce salad provides 1 unit vitamin A So the total vitamin A is x + y and this must be at least 10 units so x + y ≥ 10. 50 calories per ounce of soup: 50x 40 calories per ounce of salad: 40y Total calories = f = 50x + 40y

23. Slide 8.4- 23 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley EXAMPLE 5 Nutrition; Minimizing Calories Solution continued Step 2Write the constraints. (continued) 1 ounce soup provides 3 units vitamin C 1 ounce salad provides 2 units vitamin C So the total vitamin C is 3x + 2y and this must be at least 24 units so 3x + 2y ≥ 24. x and y cannot be negative so x ≥ 0 and y ≥ 0.

24. Slide 8.4- 24 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley EXAMPLE 5 Nutrition; Minimizing Calories Solution continued Find x and y such that the value of f = 50x + 40y is a minimum, with the restrictions b. Solve the linear programming problem. Summarize the information.

25. Slide 8.4- 25 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley EXAMPLE 5 Nutrition; Minimizing Calories Solution continued Step 3Graph the set of feasible solutions. The set is bounded by

26. Slide 8.4- 26 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley EXAMPLE 5 Nutrition; Minimizing Calories Solution continued Step 4Find the vertices. The vertices of the feasible solutions are A(10, 0), B(4, 6) and C(0, 12). A (10, 0) is obtained by solvingB (4, 6) is obtained by solvingC (0, 12) is obtained by solving

27. Slide 8.4- 27 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley EXAMPLE 5 Nutrition; Minimizing Calories Solution continued Step 5Find the value of f at the vertices. Vertex (x, y)Value of f = 50x + 40y (10, 0) (4, 6) (0, 12) 50(10) + 40(0) = 500 50(4) + 40(0) = 440 50(0) + 40(0) = 480 Step 6Find the maximum or minimum value of f. The smallest value of f is 440, which occurs when x = 4 and y = 6.

28. Slide 8.4- 28 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley EXAMPLE 5 Nutrition; Minimizing Calories Solution continued c. State the conclusion The lunch menu for Fat Albert should contain 4 ounces of soup and 6 oounces of salad. His intake of 440 calories will be as small as possible under the given constraints.