LINEAR PROGRAMMING: A GEOMETRIC APPROACH. Copyright Cengage Learning. All rights reserved.

Size: px

Start display at page:

Edwina Lynch
5 years ago
Views:

3 Graphing Linear Inequalities 3

4 Graphing Linear Inequalities A linear equation in two variables x and y ax + by + c = 0 a, b not both equal to zero has a solution set that may be exhibited graphically as points on a straight line in the xy-plane. We now show that there is also a simple graphical representation for linear inequalities in two variables: ax + by + c < 0 ax + by + c 0 ax + by + c > 0 ax + by + c 0 4

5 Graphing Linear Inequalities Before turning to a general procedure for graphing such inequalities, let s consider a specific example. Suppose we wish to graph 2x + 3y < 6 (1) We first graph the equation 2x + 3y = 6, which is obtained by replacing the given inequality < with an equality = (Figure 1). A straight line divides the xy-plane into two half-planes. Figure 1 5

6 Graphing Linear Inequalities Observe that this line divides the xy-plane into two half-planes: an upper half-plane and a lower half-plane. Let s show that the upper half-plane is the graph of the linear inequality 2x + 3y > 6 (2) whereas the lower half-plane is the graph of the linear inequality 2x + 3y < 6 (3) 6

7 Graphing Linear Inequalities To see this, let s write Inequalities (2) and (3) in the equivalent forms (4) and (5) The equation of the line itself is (6) 7

8 Graphing Linear Inequalities Now pick any point P(x, y) lying above the line L. Let Q be the point lying on L and directly below P (see Figure 1). A straight line divides the xy-plane into two half-planes. Figure 1 8

9 Graphing Linear Inequalities Since Q lies on L, its coordinates must satisfy Equation (6). In other words, Q has representation. Comparing the y-coordinates of P and Q and recalling that P lies above Q, so that its y-coordinate must be larger than that of Q, we have But this inequality is just Inequality (4) or, equivalently, Inequality (2). Similarly, we can show that every point lying below L must satisfy Inequality (5) and therefore Inequality (3). 9

10 Graphing Linear Inequalities This analysis shows that the lower half-plane provides a solution to our problem (Figure 2). The set of points lying below the dashed line satisfies the given inequality. Figure 2 10

11 Graphing Linear Inequalities (By convention, we draw the line as a dashed line to show that the points on L do not belong to the solution set.) Observe that the two half-planes in question are disjoint; that is, they do not have any points in common. Alternatively, there is a simpler method for determining the half-plane that provides the solution to the problem. To determine the required half-plane, let s pick any point lying in one of the half-planes. 11

12 Graphing Linear Inequalities For simplicity, pick the origin (0, 0), which lies in the lower half-plane. Substituting x = 0 and y = 0 (the coordinates of this point) into the given Inequality (1), we find 2(0) + 3(0) < 6 or 0 < 6, which is certainly true. This tells us that the required half-plane is the one containing the test point namely, the lower half-plane. 12

13 Graphing Linear Inequalities Next, let s see what happens if we choose the point (2, 3), which lies in the upper half-plane. Substituting x = 2 and y = 3 into the given inequality, we find 2(2) + 3(3) < 6 or 13 < 6, which is false. This tells us that the upper half-plane is not the required half-plane, as expected. Note, too, that no point P (x, y) lying on the line constitutes a solution to our problem, given the strict inequality <. 13

14 Graphing Linear Inequalities The following procedure is used for graphing a linear inequality in two variables. 14

15 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. The set of points lying on the line and in the upper half-plane satisfies the given inequality. Figure 3 15

16 Example 1 Solution cont d 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. 16

17 Graphing Systems of Linear Inequalities 17

18 Graphing Systems of Linear Inequalities 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. 18

19 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. 19

20 Example 4 Solution cont d These half-planes are shown in Figure 6. The set of points in the shaded area satisfies the system 4x + 3y 12 x y 0 Figure 6 20

21 Example 4 Solution cont d 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 21

22 Example 5 Sketch the solution set for the system x 0 y 0 x + y 6 0 2x + y

23 Example 5 Solution cont d 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. The set of points in the shaded region, including the x- and y-axes, satisfies the given inequalities. Figure 7 23

24 Example 5 Solution cont d 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 24

25 Graphing Systems of Linear Inequalities 25

Chapter 3 Linear Programming: A Geometric Approach

Chapter 3 Linear Programming: A Geometric Approach Section 3.1 Graphing Systems of Linear Inequalities in Two Variables y 4x + 3y = 12 4 3 4 x 3 y 12 x y 0 x y = 0 2 1 P(, ) 12 12 7 7 1 1 2 3 x We ve seen