Name: Thus, y-intercept is (0,40) (d) y-intercept: Set x = 0: Cover the x term with your finger: 2x + 6y = 240 Solve that equation: 6y = 24 y = 4

Transcription

1 Name: GRAPHING LINEAR INEQUALITIES IN TWO VARIABLES SHOW ALL WORK AND JUSTIFY ALL ANSWERS. 1. We will graph linear inequalities first. Let us first consider (a) First, we will graph the boundar equation: = 240 (b) This line is in standard form, so the intercepts are eas to find. (c) -intercept: Set = 0: Cover the term with our finger: 2 +6 = 240 Solve that equation: 2 = 24 = 12 Thus, -intercept is (120,0) (d) -intercept: Set = 0: Cover the term with our finger: = 240 Solve that equation: 6 = 24 = 4 Thus, -intercept is (0,40) (e) We need two points to determine a line, so we can plot these and draw the line through these points. Note: You can also change this slope-intercept ( = m + b) form, but that is more work. This is much easier. (f) We will use the coordinates of our intercepts to determine the scale of our graph. The bigger the graph, the better. Our values were 40 and 120. Let s count b 20 s and go to 140. Notice we have no negative values. (g) After ou have labeled the aes, plot the points (120,0) and (0,40). (h) Now we shift attention back to the orginal inequalit: (i) Since the inequalit is less than or equal to we will use a solid line to connect the intercepts we found above. A solid line is used antime the inequalit is or (j) We now need to determine which region of the plane to shade. To do that, we will use a test point that does not la on the line. Often the origin, (0,0), is a nice test point. (k) Since our inequalit does not lie on the line we will use the origin. Substitute = 0 and = 0 into the inequalit and determine if the resulting statement is true or false. 2( ) + 6( )? 240 (h) Did ou get a true statement? Yes, 0 is alwas less than 240! Since our statement is true, we shade the region that contains the test point. In this problem, we will shade the region below and to the left of the line. That s it! Admire our work and let s move on! c M. Yosko & R. Winkenwerder This page ma not be scanned, copied, or redistributed without permission. Page 1

2 2. Now let s tr another: 18 8 > 144 (a) First, we will graph the boundar equation: 18 8 = 144 (b) This line is in standard form, so the intercepts are eas to find. (c) -intercept: Set = 0: Cover the term with our finger: 18 8 = 144 Solve that equation: Thus, -intercept is (,0) (d) -intercept: Set = 0: Cover the term with our finger: 18 8 = 144 Solve that equation: Thus, -intercept is (0, ) (e) Now we have the two intercepts (8,0) and (0, 18). This time, let s count b 2 s and go to 18 (f) After ou have labeled the aes, plot the points above. (g) Now look at the original inequalit: 18 8 > 144 (h) Since the inequalit is strictl greater than we will use a dashed line to connect the intercepts we found above. The inequalities < and > alwas use a dashed line. Connect the intercepts with a dashed line. (i) Since the line does not pass through the origin we can use it again. Note: You can use an point not on the line, but the origin is often the nicest because it minimizes the computation in evaluating the inequalit. (j) Substitute = 0 and = 0 into the inequalit and determine if the resulting statement is true or false. 18( ) 8( )? > 144 (g) This time, ou should have got a false statement. This means we shade the region that does NOT contain our test point. Here we will shade the region below and to the right of the line. Now let s take the problems to the net level... c M. Yosko & R. Winkenwerder This page ma not be scanned, copied, or redistributed without permission. Page 2

3 { Graph 3 > 3 (a) Let s find the information for each inequalit separatel. (b) (c) 3 > 3 i. Find the -intercept: Let = 0 (,0) i. Find the -intercept: Let = 0 (,0) ii. Find the -intercept: Let = 0 (0, ) ii. Find the -intercept: Let = 0 (0, ) iii. Look at the inequalit and make a decision on the tpe of line to use iv. Choose a test point: v. Evaluate the test point in the inequalit. 5( ) + 3( )? 15 iii. Look at the inequalit and make a decision on the tpe of line to use. 3 > 3 iv. Choose a test point: v. Evaluate the test point in the inequalit. 3( ) ( )? > 3 vi. Is the statement true or false? Where should ou shade? vi. Is the statement true or false? Where should ou shade? (d) Since this is our first eample graphing two inequalities on one plane, let s stick to a scale of 1 for each the horizontal and vertical ais. (e) There is going to be a lot of shading, and it can be eas to shade the wrong region b mistake. So we are going to use arrows to indicate which regions should be shaded before we shade the final feasible region. (f) Transfer the information from the inequalit to the coordinate plane below. Rather than shading the entire region, use small below the line and along the entire line to indicate the region we need to shade for this inequalit (g) Now transfer the information from the inequalit 3 > 3 to the coordinate plane below. This time use small below the line and along the line to indicate the region we need to graph. (h) Finall, we can shade! We need to shade the region that satisfies BOTH inequalities. We are looking for the overlap of the two areas. c M. Yosko & R. Winkenwerder This page ma not be scanned, copied, or redistributed without permission. Page 3

4 Graph (a) Find the information for each inequalit. (b) (c) intercept: (, 0) -intercept: (0, ) -intercept: (, 0) -intercept: (0, ) Test Point: Evaluate the test point: Test Point: Evaluate the test point: Is the statement true or false? Where should ou shade? Is the statement true or false? Where should ou shade? (d) The third row of inequalities, 0 and 0, actuall make it easier to find the feasible region. i. The inequalit 0 means can take on positive values or zero. We know is positive in quadrants I or IV, and is zero for an point on the -ais. ii. Likewise, the inequalit 0 means can take on positive values or zero. We know is positive in quadrants I or II, and is zero for an point on the -ais. iii. If we shade those areas on the plane, it means we are looking for an points in Quadrant I, OR on the positive -ais, OR on the positive -ais, OR the origin itself. iv. So the third row of inequalities is telling us to restrict our feasible region to the upper, right-hand portion of the plane. (e) Transfer the information from each of the inequalities. Again, use small arrows indicating where ou will look of the feasible region (f) Use a scaling of 10 units for this graph. Then each midpoint of the grid lines is a multiple of 5. c M. Yosko & R. Winkenwerder This page ma not be scanned, copied, or redistributed without permission. Page 4

5 Graph Recommended scaling: 5 units Graph Recommended scaling: 10 units c M. Yosko & R. Winkenwerder This page ma not be scanned, copied, or redistributed without permission. Page 5