1. Copyright © Cengage Learning. All rights reserved. 3 LINEAR PROGRAMMING: A GEOMETRIC APPROACH Warm Up: Graph the inequality 2x + 3y < 6

2. Copyright © Cengage Learning. All rights reserved. 3.1 Graphing Systems of Linear Inequalities in Two Variables

3. 3 Graphing Linear Inequalities

4. 4 Solution. (Figure 2). Figure 2 The set of points lying below the dashed line satisfies the given inequality.

5. 5 Example 1 Determine the solution set for the inequality 2x + 3y  6. Solution: Replacing the inequality  with an equality =, we obtain the equation 2x + 3y = 6, whose graph is the straight line shown in Figure 3. Figure 3 The set of points lying on the line and in the upper half-plane satisfies the given inequality.

6. 6 Example 1 – Solution Instead of a dashed line as before, we use a solid line to show that all points on the line are also solutions to the inequality. Picking the origin as our test point, we find 2(0) + 3(0)  6, or 0  6, which is false. So we conclude that the solution set is made up of the half-plane that does not contain the origin, including (in this case) the line given by 2x + 3y = 6. cont’d

7. 7 Graphing Systems of Linear Inequalities

8. 8 By the solution set of a system of linear inequalities in the two variables x and y, we mean the set of all points (x, y) satisfying each inequality of the system. The graphical solution of such a system may be obtained by graphing the solution set for each inequality independently and then determining the region in common with each solution set.

9. 9 Example 4 Determine the solution set for the system 4x + 3y  12 x – y  0 Solution: Proceeding as in the previous examples, you should have no difficulty locating the half-planes determined by each of the linear inequalities that make up the system.

10. 10 Example 4 – Solution These half-planes are shown in Figure 6. cont’d Figure 6 The set of points in the shaded area satisfies the system 4x + 3y  12 x – y  0

11. 11 Example 4 – Solution The intersection of the two half-planes is the shaded region. A point in this region is an element of the solution set for the given system. The point P, the intersection of the two straight lines determined by the equations, is found by solving the simultaneous equations 4x + 3y = 12 x – y = 0 cont’d

12. 12 Example 5 Sketch the solution set for the system x  0 y  0 x + y – 6  0 2x + y – 8  0

13. 13 Example 5 – Solution The first inequality in the system defines the right half-plane—all points to the right of the y-axis plus all points lying on the y-axis itself. The second inequality in the system defines the upper half-plane, including the x-axis. The half-planes defined by the third and fourth inequalities are indicated by arrows in Figure 7. Figure 7 The set of points in the shaded region, including the x- and y-axes, satisfies the given inequalities. cont’d

14. 14 Example 5 – Solution Thus, the required region—the intersection of the four half-planes defined by the four inequalities in the given system of linear inequalities—is the shaded region. The point P is found by solving the simultaneous equations x + y – 6 = 0 and 2x + y – 8 = 0 cont’d

15. 15 Graphing Systems of Linear Inequalities

16. 16 Practice p. 165 Self-Check Exercises #1 & 2