Section 7.1 Graphing Linear Inequalities in Two Variables

Transcription

1 Section 7.1 Graphing Linear Inequalities in Two Variables Eamples of linear inequalities in two variables include + <, 3+ > 6, and 1 A solution of a linear inequalit is an ordered pair that satisfies the inequalit. For eample (,) is a solution of 3 6 (Check b substituting for and for.) A linear inequalit has an infinite number of solutions, one for ever choice of a value for. The best wa to show these solutions is to sketch the graph of the inequalit, which consists of all points in the plane whose coordinates satisf the inequalit. EXAMPLE: Graph the inequalit 3 6. Solution: First, solve the inequalit for : = 1. 3 This inequalit has the same solutions as the original one. To solve it, note that the points on the line = 1. 3 certainl satisf Plot some points, and graph this line, as in the Figure below (left). The points on the line satisf equals The points satisfing is greater than 1. 3 are the points above the line (because the have larger second coordinates than the points on the line; see the Figure below (middle)). Similarl, the points satisfing < 1. 3 lie below the line (because the have smaller second coordinates), as shown in the Figure below (middle). The line = 1. 3 is the boundar line. Thus, the solutions of 1. 3 are all points on or above the line = The line and the shaded region of the Figure below (right) make up the graph of the inequalit Points satisf > = Points satisf 3 < EXAMPLE: Graph + <. 1

2 EXAMPLE: Graph + <. Solution: First obtain an equivalent inequalit b solving for : < + < + < + =.+1 The boundar line is =. + 1, but it is not part of the solution, since points on the line do not satisf <.+1. To indicate this, the line is drawn dashed in the Figure below. The points below the boundar line are the solutions of <.+1, because the have smaller second coordinates than the points on the line =.+1. The shaded region in the Figure below (ecluding the dashed line) is the graph of the inequalit <.+1. <. + 1 The Eamples above show that the solutions of a linear inequalit form a half-plane consisting of all points on one side of the boundar line (and possibl the line itself). When an inequalit is solved for, the inequalit smbol immediatel tells whether the points above (>), on (=), or below (<) the boundar line satisf the inequalit, as summarized here. When graphing b hand, draw the boundar line = m+b solid when it is included in the solution ( or inequalities) and dashed when it is not part of the solution (> or < inequalities). EXAMPLE: Graph 1. Solution: Solve the inequalit for : =.+ Thegraphconsistsofallpointsonorbelowtheboundarline.+, as shown in the Figure on the right.

3 EXAMPLE: Graph each of the given inequalities. (a) Solution: The boundar line is the horizontal line =. The graph consists of all points on or above this line (see the Figure below (left)). (b) 1 Solution: This inequalit does not fit the pattern discussed earlier, but it can be solved b a similar technique. Here, the boundar line is the vertical line = 1, and it is included in the solution. The points satisfing < 1 are all points to the left of this line (because the have -coordinates smaller than 1). So the graph consists of the points that are on or to the left of the vertical line = 1, as shown in the Figure below (right). 1 1 An alternative technique for solving inequalities that does not require solving for is illustrated in the net eample. EXAMPLE: Graph 6. Solution: The boundar line is = 6, which can be graphed b plotting its - and -intercepts. The graph contains the half-plane above or below this line. To determine which, choose a test point an point not on the boundar line, sa, (,). Letting = and = in the inequalit produces () () 6, a false statement. Therefore, (,) is not in the solution. So the solution is the half-plane that does not include (,), as shown in the Figure below. If a different test point is used, sa, (3,), then substituting = 3 and = in the inequalit produces () (3) 6, a true statement. Therefore, the solution of the inequalit is the half-plane containing (3, ), as shown in the Figure below. (3, ) 3 (, ) 3 3

4 Sstems of Inequalities Real-world problems often involve man inequalities. For eample, a manufacturing problem might produce inequalities resulting from production requirements, as well as inequalities about cost requirements. A set of at least two inequalities is called a sstem of inequalities. The graph of a sstem of inequalities is made up of all those points which satisf all the inequalities of the sstem. EXAMPLE: Graph the sstem 3+ 1 Solution: First, solve each inequalit for : Then the original sstem is equivalent to this one: 3+1 The solutions of the first inequalit are the points on or below the line = 3+1 (see the Figure below (left)). The solutions of the second inequalit are the points on or above the line = / (see the Figure below (middle)). So the solutions of the sstem are the points that satisf both of these conditions, as shown in the Figure below (right) = The shaded region in the Figure above (right) is sometimes called the region of feasible solutions, or just the feasible region, since it consists of all the points that satisf (are feasible for) ever inequalit of the sstem. EXAMPLE: Graph the feasible region for the sstem 1 + 8,

5 EXAMPLE: Graph the feasible region for the sstem 1 + 8, Solution: Begin b solving the first two inequalities for : Then the original sstem is equivalent to this one: , The inequalities and restrict the graph to the first quadrant. So the feasible region consists of all points in the first quadrant that are on or above the line =. and on or below the line =.+ (see the Figure on the right). =. =. + EXAMPLE: Graph the feasible region for the sstem 1 3, Solution: Solve the first inequalit for : Then the original sstem is equivalent to this one: + 1 =.+.+ 3, One can show that the feasible region consists of all points that lie on or be low the line.+ and on or to the right of the vertical line = 3 and on or above the horizontal line = as shown in the Figure on the right = 3 =. + = 3 6

6 EXAMPLE: Midtown Manufacturing Compan makes plastic plates and cups, both of which require time on two machines. Producing a unit of plates requires 1 hour on machine A and on machine B, while producing a unit of cups requires 3 hours on machine A and 1 on machine B. Each machine is operated for at most 1 hours per da. Write a sstem of inequalities epressing these conditions, and graph the feasible region. Solution: Let represent the number of units of plates to be made and represent the number of units of cups. Then make a chart that summarizes the given information. Time on Machine Number of Units A B Plates 1 Cups 3 1 Maimum Time 1 1 Available We must have and because the compan cannot produce a negative number of cups or plates. On machine A, producing units of plates requires a total of 1 = hours, while producing units of cups requires 3 = 3 hours. Since machine A is available no more than 1 hours a da, Similarl, the requirement that machine B be used no more than 1 hours a da gives So we must solve the sstem The feasible region is shown in the Figure below , 1 /! " # + 1 $ $ 1 6