C H A P T E R 9 Inequalities and linear programming What is a linear inequalit? How do we solve linear inequalities? What is linear programming and how is it used? In Chapter 3, Linear graphs and models, ou learned how linear equations and their graphs are used to model practical situations, such as plant growth, service charges and flow problems. In this chapter ou will learn how linear inequalities and their graphs can be used to model a different set of practical situations, such as determining the mi of products in a supermarket to maimise profit, or designing a diet to provide maimum nutrition for minimum cost. This is known as linear programming. Linear programming requires ou to solve both linear equations and linear inequalities. You learned how to solve linear equations in Chapter 2, Linear relations and equations. You now need to learn how to solve linear inequalities. 9.1 Linear inequalities in one variable Linear inequalities in one variable and the number line An epression such as 9 3 21 is called a linear inequalit in one variable. It is an inequalit, not an equation, because it involves an inequalit sign ( ) rather than an equals sign ( = ). The sign means less than or equal to. The solution to the linear equation 3 = 9is = 3, and the solution to the linear equation 3 = 21 is 7. We can represent these solutions on a number line b putting a closed circle ( )onthe number line at = 3 and = 7asshown. = 3 = 7 3 2 1 0 1 2 3 4 5 6 7 8 9 When solving an inequalit and graphing its solution on a number line, we need to be careful about whether the end values of the solution are included in the range of possible values. 376 Cambridge Universit Press Uncorrected Sample Pages 978-0-521-74049-4 2008 Evans, Lipson, Jones, Aver, TI-Nspire & Casio ClassPad material prepared in collaboration with Jan Honnens & David Hibbard
End values included To solve the linear inequalit 9 3 21 Chapter 9 Inequalities and linear programming 377 we divide through b 3 and get 9 3 3 3 21 3 or 3 7 There is no single solution to this inequalit. An value of from 3 to 7 is a solution. For eample, = 3, = 3.5, = 4.95 and = 7 are all possible solutions. In fact, it is impossible to list ever possible solution, as there are an infinite number of solutions. However, we can represent all the possible solutions on a number line. This is done b marking the points = 3 and = 7 with a closed circle ( ) onthe number line. These points are then joined b drawing a solid line to indicate that all the values between = 3 and = 7 are also solutions, as shown below. 3 7 3 2 1 0 1 2 3 4 5 6 7 8 9 End values not included To solve the linear inequalit 9 < 3 < 21 we divide through b 3 to obtain the solution 3 < < 7 The sign < means less than. This means that = 3 and = 7 are not solutions, but all values between = 3 and = 7 are possible solutions. To represent this solution on a number line, mark in the points = 3 and = 7 with an open circle ( ). These two open circles are then joined b a solid line to indicate that all the values between 3 and 7 are solutions, but not = 3 and = 7. 3 < < 7 3 2 1 0 1 2 3 4 5 6 7 8 9 Note that 7 > > 3 represents the same values of as 3 < < 7. A galler of signs = as in a = b reads as a equals b > as in a > b reads as a is greater than b as in a b reads as a is greater than or equal to b < as in a < b reads as a is less than b as in a b reads as a is less than or equal to b Cambridge Universit Press Uncorrected Sample Pages 978-0-521-74049-4 2008 Evans, Lipson, Jones, Aver, TI-Nspire & Casio ClassPad material prepared in collaboration with Jan Honnens & David Hibbard
378 Essential Standard General Mathematics Eample 1 Solve the inequalit 10 < 5 40 Solving an inequalit and graphing the solution for and displa the solution on a number line. Solution 1 Write the inequalit. 10 < 5 40 2 Solve the inequalit for b dividing through b 5. or 10 5 < 5 5 40 5 3 Displa the solution on a number line. or 2 < 8 Draw a number line to include 2 and 8. 3 2 1 0 1 2 3 4 5 6 7 8 9 Mark the point = 2with an open circle. 3 2 1 0 1 2 3 4 5 6 7 8 9 Mark the point = 8 with a closed circle. Join the two points with a solid line. Write in the solution inequalit on the graph. Eample 2 Solve the inequalit 10 > 20 Solving an inequalit and graphing the solution for and displa the solution on a number line. 3 2 1 0 1 2 3 4 5 6 7 8 9 3 2 1 0 1 2 3 4 5 6 7 8 9 2 < 8 3 2 1 0 1 2 3 4 5 6 7 8 9 Solution 1 Write the inequalit. 10 > 20 10 2 Solve the inequalit for b dividing through b 10. or 10 > 20 10 3 Displa the solution on a number line. or > 2 Draw a number line to include 2. 3 2 1 0 1 2 3 4 5 6 7 8 9 Mark in the point = 2 with an open circle. 3 2 1 0 1 2 3 4 5 6 7 8 9 To indicate all values of greater than 2, draw a solid line from this point to the 3 2 1 0 1 2 3 4 5 6 7 8 9 right that potentiall goes on forever. > 2 Write in the solution inequalit on the 3 2 1 0 1 2 3 4 5 6 7 8 9 graph. Cambridge Universit Press Uncorrected Sample Pages 978-0-521-74049-4 2008 Evans, Lipson, Jones, Aver, TI-Nspire & Casio ClassPad material prepared in collaboration with Jan Honnens & David Hibbard
Chapter 9 Inequalities and linear programming 379 Linear inequalities in one variable and the coordinate plane We can also represent linear inequalities in one variable on the coordinate plane. If the equation = 3isplotted on a set of aes we will have a vertical straight line, located at = 3. While the value of changes along the line, for ever point on this line the value of is 3. Just as with graphing the solution of an inequalit on a number line, we need to be careful about whether the boundar lines (end values) of the solution are included in the range of possible values. Boundar line included If we tried to plot the inequalit 3, we would have to plot all the points in the plane that have an -value greater than or equal to 3. f course we cannot show each individual point. What we do is shade in the containing these points. The shaded starts at the vertical line = 3 and etends right forever. Some representative points that satisf the condition 3, and which are found in the shaded, have also been plotted. Boundar line not included The plot of the inequalit > 3issimilar to the plot of 3, but the line = 3isdrawn as a dashed line to indicate that it is not included in the. Some representative points that satisf the condition > 3have also been plotted. Note: For (5, 6), 5 > 3 (4, 1), 4 > 3 Therefore both points satisf the inequalit > 3. = 3 = 3 (3, 6) (3, 3) (3, 0) (3, 1) (3, 0) Required = 3 Required (5, 6) (4, 1) > 3 (5, 6) Eample 3 n the coordinate plane, plot the graphs of: a 4 b 1 < < 3 (4, 1) > 3 Plotting a linear inequalit in one variable on the coordinate plane Cambridge Universit Press Uncorrected Sample Pages 978-0-521-74049-4 2008 Evans, Lipson, Jones, Aver, TI-Nspire & Casio ClassPad material prepared in collaboration with Jan Honnens & David Hibbard
380 Essential Standard General Mathematics Solution a 4 1 Draw in a solid line = 4todefine the boundar of the shaded. 2 Shade the on and below the line = 4to represent all the points defined b 4. b 1 < < 3 1 Draw in a dashed line = 3todefine the upper boundar of the shaded. 2 Draw in a dashed line = 1todefine the lower boundar of the shaded. 3 Shade the between the lines = 3 and = 1to represent all the points defined b 1 < < 3. Eercise 9A (0, 4) (0, 3) (0, 1) < 4 = 4 = 3 1 < < 3 = 1 1 Which of the smbols <, = or > should be placed in the bo in each of the following? a 7 9 b 3 2 c 7 + 1 9 1 d 0.5 1 e 8 4 f 3 1 g 2 1 h 0 0.5 2 Represent each of the following inequalities on a number line. a 1 4 b 0 < < 4 c < 4 d 4 e 1 < 4 f 3 < 5 3 Write down an inequalit represented b each of the following graphs. a b 2 1 0 1 2 3 4 5 6 7 8 2 1 0 1 2 3 4 5 6 7 8 c d 2 1 0 1 2 3 4 5 6 7 8 2 1 0 1 2 3 4 5 6 7 8 e 2 1 0 1 2 3 4 5 6 7 8 4 Solve each of the following inequalities and represent its solution on a number line. a 3 15 b 20 < 100 c 2 > 4 d 9 36 e 12 6 < 24 f 10 < 5 25 5 A person becomes a teenager when the turn 13. The stop being a teenager when the turn 20. Let be the variable age (in ears). Cambridge Universit Press Uncorrected Sample Pages 978-0-521-74049-4 2008 Evans, Lipson, Jones, Aver, TI-Nspire & Casio ClassPad material prepared in collaboration with Jan Honnens & David Hibbard
Chapter 9 Inequalities and linear programming 381 a Write down an inequalit in terms of that defines a teenager. b Graph this inequalit on a number line. 6 Carr-on luggage in most passenger aircraft can weigh no more than 5 kg. Let be the variable weight (in kg). a Write down an inequalit in terms of that defines the acceptable weight for carr-on luggage. b Graph this inequalit on a number line. 7 Graph the following inequalities on the coordinate plane. a 1 b > 2 c 5 d > 1 e < 2 f 2 2 g 1 < < 2 h 3 < 5 i 3 < 0 9.2 Linear inequalities in two variables The inequalities > 2 2 < 2 2 are linear inequalities in two variables, and. To help us interpret these inequalities, we have drawn the graph of = 2. This line (red) separates the coordinate plane into two s. 4 2 10 8 6 4 2 0 2 2 (2, 8) = 2 (2, 4) 4 (4, 4) The line = 2 The line is defined b the equation = 2 and coloured red. Itincludes all the points that lie on the line. From the graph above we can see that the point (2, 4) lies on the line. The points (2, 8) and (4, 4) clearl do not lie on the line. Cambridge Universit Press Uncorrected Sample Pages 978-0-521-74049-4 2008 Evans, Lipson, Jones, Aver, TI-Nspire & Casio ClassPad material prepared in collaboration with Jan Honnens & David Hibbard
382 Essential Standard General Mathematics We can also show this b carring out the following tests, using the equation of the line. Test: (2, 4): = 4 2 = 2; so the point (2, 4) lies on the line = 2. (2, 8): = 8 2 = 6; 6 is greater than 2, so (8, 9) does not lie on = 2. (4, 4): = 4 4 = 0; 0 is less than 2, so (4, 4) does not lie on the line = 2. The s >2and 2 This is defined b the inequalit > 2 and coloured light blue. Itincludes all the points that lie above the line; the point (2, 8) is an eample. B including the line in this, we have a wa ofrepresenting the inequalit 2 This includes all the points on and above 10 8 (2, 8) the line. From the graph on the right we can see that the points (2, 4) and (2, 8) are eamples of points that lie in this. 6 4 2 (2, 4) (4, 4) the point (4, 4) clearl does not lie in the 4 2 0. 2 4 We can also show this b carring out the 2 following tests, using the equation of the line. Test: (2, 4): = 4 2 = 2; so the point (2, 4) lies in the 2. (2, 8): = 8 2 = 6; 6 is greater than 2, so the point (2, 8) lies in the 2 (4, 4): = 4 4 = 0; 0 is less than 2, so (4, 4) does not lie in the 2. = 2 The s <2and 2 This is defined b the inequalit < 2 and coloured purple. Itincludes all the points that lie below the line; the point (4, 4) is an eample. B including the line in this, we have a wa ofrepresenting the inequalit 2 This includes all the points on and below the line. From the graph on the right we can see that the points (2, 4) and (4, 4) are eamples of points that lie in this. 10 8 6 4 2 (2, 8) (2, 4) (4, 4) the point (2, 8) clearl does not lie in the 4 2 0. 2 2 4 = 2 Cambridge Universit Press Uncorrected Sample Pages 978-0-521-74049-4 2008 Evans, Lipson, Jones, Aver, TI-Nspire & Casio ClassPad material prepared in collaboration with Jan Honnens & David Hibbard
Chapter 9 Inequalities and linear programming 383 We can also show this b carring out the following tests using the equation of the line. Test: (2, 4): = 4 2 = 2; so the point (2, 4) lies in the 2. (2, 8): = 8 2 = 6; 6 is greater than 2, so (8, 9) does not lie in the 2. (4, 4): = 4 4 = 0; 0 is less than 2, so (4, 4) lies in the 2. We now have a graphical wa of representing inequalities. Linear inequalities can be represented b s in the coordinate plane If the inequalit sign is: or, the line defining the is included, indicated b using a solid line to indicate the boundar <or>,the line defining the is not included, indicated b using a dashed line to indicate the boundar. Graphing a linear inequalit in two variables Eample 4 Graphing a linear inequalit in two variables Sketch the graph of the 3 + 2 18. Solution 1 Find the intercepts for the boundar line 3 + 2 = 18. Find the -intercept. Substitute = 0 into the equation and solve for. Find the -intercept. Substitute = 0 into the equation and solve for. 2 n a labelled set of aes, draw a straight line through the two intercepts. Use a solid line to indicate that the line is included in the. Label the line. 3 + 2 = 18 When = 0, 2 = 18 = 9 -intercept is (0, 9). When = 0, 3 = 18 = 6 -intercept is (6, 0). 3 Use a test point to determine whether the required lies above or below the line. Note: The origin (0, 0) is usuall a good point to test. (0, 9) 3 + 2 = 18 (6, 0) Test (0, 0): 3 + 2 = 3(0) + 2(0) = 0 0 < 18, so (0, 0) lies in the 3 + 2 18. Cambridge Universit Press Uncorrected Sample Pages 978-0-521-74049-4 2008 Evans, Lipson, Jones, Aver, TI-Nspire & Casio ClassPad material prepared in collaboration with Jan Honnens & David Hibbard
384 Essential Standard General Mathematics 4 As (0, 0) is below the line, the required lies on and below the line. Shade in the on and below the line. Label the. Eample 5 (0, 9) 3 + 2 18 Graphing a linear inequalit in two variables Sketch the graph of the 4 5 > 20. Solution 1 Find the intercepts for the boundar line 4 5 = 20. Find the -intercept. Substitute = 0 into the equation and solve for. Find the -intercept. Substitute = 0 into the equation and solve for. 2 n a labelled set of aes, draw a straight line through the two intercepts. Use a dashed line to indicate that the boundar line is not included in the. Label the line. 3 Use a test point to determine whether the required lies above or below the line. 4 As (0, 0) is above the line, the required lies below the line. Shade in the on and below the line. Label the. 3 + 2 = 18 (6, 0) 4 5 = 20 When = 0, 5 = 20 = 4 -intercept is (0, 4). When = 0, 4 = 20 = 5 -intercept is (5, 0). (0, 4) 4 5 = 20 (5, 0) Test (0, 0): 4 5 = 4(0) 5(0) = 0 0 < 20, so (0, 0) does not lie in the 4 5 > 20. (0, 4) 4 5 = 20 (5, 0) 4 5 > 20 Cambridge Universit Press Uncorrected Sample Pages 978-0-521-74049-4 2008 Evans, Lipson, Jones, Aver, TI-Nspire & Casio ClassPad material prepared in collaboration with Jan Honnens & David Hibbard
Chapter 9 Inequalities and linear programming 385 Summar: Plotted linear inequalities Graph the inequalit as if it contained an equals (=) sign. Draw a solid line if the inequalit is or. Draw a dashed line if the inequalit is < or >. Pick a point not on the line to use as a test point. The origin is a good test point, provided the boundar line does not pass through the origin. Substitute the test point into the inequalit. If the point makes the inequalit true, shade the containing the test point. If not, shade the not containing the test point. Eercise 9B 1 Test to see whether the point (0, 0) lies in the following s. a + 0 b + < 4 c 2 + > 2 d 3 2 3 e 2 > 5 f 3 < 6 2 Test to see whether the point (1, 2) lies in the following s. a + 0 b + < 0 c 2 + > 2 d 3 2 3 e 2 + 3 > 5 f 5 2 8 3 Graph the following inequalities. a 5 b 2 4 c < 3 d + 10 e 3 + 9 f 5 + 3 15 g 3 5 < 15 h 2 5 > 5 i > 3 9.3 s In Chapter 2, Linear relations and equations, ou learned how to solve pairs of simultaneous linear equations graphicall. Foreample, to solve the pair of linear equations 3 + 2 = 18 (0, 9) = 3 3 + 2 = 18 graphicall, we simpl plot their graphs and find the point of intersection. (0, 3) (4, 3) = 3 The solution is the point on the coordinate plane that (6, 0) is common to both graphs. This is the point (4, 3), the point where the two lines intersect. From this, we concludes that = 4 and = 3. When we tr to solve the pair of simultaneous linear inequalities 3 + 2 18 3 Cambridge Universit Press Uncorrected Sample Pages 978-0-521-74049-4 2008 Evans, Lipson, Jones, Aver, TI-Nspire & Casio ClassPad material prepared in collaboration with Jan Honnens & David Hibbard
386 Essential Standard General Mathematics graphicall, there is not a single solution, but man solutions. The solutions are all the points that lie in the in the coordinate plane that is common to both inequalities. The common to both inequalities is called the feasible.it is called the feasible because all the points in this are possible solutions of the pair of simultaneous linear inequalities. The feasible (solution ) for a set of inequalities is determined b finding the common to all of the inequalities involved. This process is illustrated below for the inequalities (0, 9) 3 + 2 18 and 3. 3 + 2 18 (6, 0) The shaded pink is defined b the inequalit 3 + 2 18. (0, 3) > 3 The shaded blue is defined b the inequalit 3. (0, 9) (0, 3) (4, 3) 3 + 2 18 3 (6, 0) The shaded purple is the feasible. It is the common to the inequalities 3 + 2 18 and 3. The method we have used to graphicall determine the feasible is called shading in. Sometimes this method of finding the common to a set of inequalities can quickl become mess and impractical when we have too man inequalities. Fortunatel, for the sort of applications ou will meet in this chapter, the required will lie in the first quadrant and involve onl a small number of inequalities so that the shading in method is appropriate. Eample 6 Graphing a feasible Graph the feasible for the following four simultaneous inequalities: 0, 0, + 8, 3 + 5 30 Solution Because 0 and 0, the feasible is restricted to the first quadrant. 1 Graph the inequalit + 8inthe first quadrant. Plot the boundar line + = 8, (0, 8) marking and labelling the -intercept (0, 8) and the -intercept (8, 0). + = 8 Shade in the bounded b the - and -aes and the line. Here it has been shaded blue. (8, 0) Cambridge Universit Press Uncorrected Sample Pages 978-0-521-74049-4 2008 Evans, Lipson, Jones, Aver, TI-Nspire & Casio ClassPad material prepared in collaboration with Jan Honnens & David Hibbard
2 Graph the inequalit 3 + 5 30 in the first quadrant. Plot the boundar line 3 + 5 = 30, marking and labelling the -intercept (0, 6) and the -intercept (10, 0). Shade in the bounded b the - and -aes and the line. Here it has been shaded pink, but it becomes purple where it overlaps the blue. 3 The overlap (purple) is the feasible. Label the overlap the. To complete the feasible, find the coordinates of the point where the two boundar lines intersect, b solving the simultaneous equations + = 8 3 + 5 = 30 The lines intersect at the point (5, 3). Mark this point on the graph. Eample 7 Graphing a feasible Chapter 9 Inequalities and linear programming 387 (0, 8) (0, 6) (0, 8) (0, 6) + = 8 3 + 5 = 30 (8, 0) + = 8 Graph the feasible for the following four simultaneous inequalities: 0, 0, + 2 10, 6 + 4 36 Solution Because 0 and 0, the feasible is restricted to the first quadrant. 1 Graph the inequalit + 2 10 in the first quadrant. Plot the boundar line + 2 = 10, marking and labelling the -intercept (0, 5) and the -intercept (10, 0). (0, 5) Shade in the bounded b the - and -aes and the line. Here it + 2 = 10 has been shaded blue. (10, 0) (5, 3) 3 + 5 = 30 (8, 0) (10, 0) (10, 0) Cambridge Universit Press Uncorrected Sample Pages 978-0-521-74049-4 2008 Evans, Lipson, Jones, Aver, TI-Nspire & Casio ClassPad material prepared in collaboration with Jan Honnens & David Hibbard
388 Essential Standard General Mathematics 2 Graph the inequalit 6 + 4 36 in the first quadrant. Plot the boundar line 6 + 4 = 36, marking and labelling the -intercept (0, 9) and the -intercept (6, 0). Shade in the bounded b the - and -aes and the line. Here it has been shaded pink, but it becomes purple where it overlaps the blue. 3 The overlap (purple) is the feasible. Label the overlap the. To complete the feasible, find the coordinates of the point where the two boundar lines intersect, b solving the simultaneous equations + 2 = 10 6 + 4 = 36 The lines intersect at the point (4, 3). Mark this point on the graph. (0, 9) (0, 5) + 2 = 10 (0, 9) How to graph a feasible using a TI-Nspire CAS 6 + 4 = 36 (6, 0) (0, 5) + 2 = 10 (4, 3) 6 + 4 = 36 (6, 0) (10, 0) (10, 0) Graph the feasible for the following four simultaneous inequalities: 0, 0, + 2 10, 6 + 4 36 Because 0 and 0, the feasible is restricted to the first quadrant. We take this into account when setting the viewing window on the calculator. Steps 1 To graph the inequalities + 2 10 and 6 + 4 36 using a graphics calculator, first we need to rearrange both inequalities so that is the subject. Hence, (10 ) + 2 10 becomes 2 (36 6) 6 + 4 36 becomes 4 Cambridge Universit Press Uncorrected Sample Pages 978-0-521-74049-4 2008 Evans, Lipson, Jones, Aver, TI-Nspire & Casio ClassPad material prepared in collaboration with Jan Honnens & David Hibbard
Chapter 9 Inequalities and linear programming 389 2 pen a new document (b pressing / + N) and select 2: Graphs & Geometr. a Use the backspace ke ( )to delete the f 1() = and tpe in >= (10 ) 2. Press enter. This plots the inequalit + 2 10. b Repeat the above but this time tpe in >= (36 6) 4. Press enter. This plots the inequalit 6 + 4 36 c Press / + to hide the entr line. d The inequalities 0 and 0 indicate that the feasible is restricted to the first quadrant. This is best achieved b resetting the viewing window. 3 To reset the viewing window, press b/4:window/1:window Settings. Using to move between the entr boes, enter the following values: XMin: 0 XMa: 12 XScale: Auto YMin: 0 YMa: 10 YScale: Auto 4 Pressing enter confines the plot to the first quadrant. The graphs will appear as shown. The feasible is the more heavil shaded. Note: It ma be necessar to grab and move the graph labels if the overlap with other labels. Cambridge Universit Press Uncorrected Sample Pages 978-0-521-74049-4 2008 Evans, Lipson, Jones, Aver, TI-Nspire & Casio ClassPad material prepared in collaboration with Jan Honnens & David Hibbard
390 Essential Standard General Mathematics 5 To complete the feasible, we need to know the coordinates of the corner points. a Press b/6:points & Lines/3:Intersection Point/s. b Move the cursor to one of the graphs and press.now move to the other graph and press. The point of intersection (4, 3) will be displaed. Press to eit the Intersection Point tool. The other two points, (0, 9) and (10, 0), can be determined from the equations of the boundar lines. How to graph a feasible using the ClassPad Graph the feasible for the following four simultaneous inequalities: 0, 0, + 2 10, 6 + 4 36 Because 0 and 0, the feasible is restricted to the first quadrant. We take this into account when setting the viewing window on the calculator. Steps 1 From the application menu, locate and open the Graph and Table ( ) built-in application. To graph the inequalities + 2 10 and 6 + 4 36, first we need to rearrange both inequalities so that is the subject. Hence, (10 ) + 2 10 becomes 2 (36 6) 6 + 4 36 becomes 4 Cambridge Universit Press Uncorrected Sample Pages 978-0-521-74049-4 2008 Evans, Lipson, Jones, Aver, TI-Nspire & Casio ClassPad material prepared in collaboration with Jan Honnens & David Hibbard
Chapter 9 Inequalities and linear programming 391 2 To set the calculator to draw the correct inequalit, tap the down arrow ( ) adjacent to in the toolbar and select. Adjacent to 1: tpe (10 )/2. Press E. Adjacent to 2: tpe (36 6)/4. Press E. Adjacent to 3: tpe 0. Press E. 3 To enter the 0 inequalit, tap the down arrow ( ) adjacent to in the toolbar and select. Adjacent to 4: tpe 0. Press E. Tap on the View Window icon (6)inthe toolbar to set the graph viewing window. Cambridge Universit Press Uncorrected Sample Pages 978-0-521-74049-4 2008 Evans, Lipson, Jones, Aver, TI-Nspire & Casio ClassPad material prepared in collaboration with Jan Honnens & David Hibbard
392 Essential Standard General Mathematics 4 To complete the, the corner points need to be found. From the Analsis menu item, select G-solve, then Intersect. Use the up and directions from the blue oval directional button on the front of the calculator to select the equations for 1 and 2. When an equation has been selected, press E to confirm its choice. The equation of each line is displaed in a window at the bottom of the graphing screen. 5 After the second equation has been selected and confirmed, the intersection point will be displaed on the screen, indicated b a cursor in the shape of a small cross. In this case, (4, 3). The other two boundar points are (0, 9) and (10, 0). Eercise ********* 9C Graph the feasible for each of the following sets of linear inequalities. 1 0, 0, + 10 2 0, 0, 2 + 3 12 3 0, 0, 3 + 5 15 4 0, 0, + 6, 2 + 3 15 5 0, 0, 3 + 6, + 2 7 Cambridge Universit Press Uncorrected Sample Pages 978-0-521-74049-4 2008 Evans, Lipson, Jones, Aver, TI-Nspire & Casio ClassPad material prepared in collaboration with Jan Honnens & David Hibbard
6 0, 0, 5 + 2 20, 5 + 6 30 7 0, 0, 4 + 12, 3 + 6 30 8 0, 0, 2 0, + 30 9.4 Linear programming bjective functions and constraints Chapter 9 Inequalities and linear programming 393 An objective function is a quantit that ou are tring to make as large as possible (for eample, profits) or as small as possible (for eample, the amount of material needed to make a dress). f course, there are alwas factors, such as the resources available or the requirements of the dress pattern, that limit how much profit ou can make or how little material ou can use to make a dress. These are called constraints. The linear programming problem The process of maimising or minimising a linear quantit, subject to a set of constraints, is at the heart of linear programming. The linear programming problem From the mathematical point of view, linear programming can be viewed as finding the point, or points, in a feasible that gives the maimum or minimum value of some linear epression. Finding the maimum value of an objective function The aim is to find the maimum value of the objective function P = 2 + 3, subject to the constraints: 0, 0 + 8 3 + 5 30 as shown b the feasible opposite. (0, 8) B(0, 6) A(0, 0) + = 8 C(5, 3) 3 + 5 = 30 (10, 0) D(8, 0) At first, this seems like an insurmountable problem, as there is an infinite number of points in the to choose from. Fortunatel, we can make use of the corner point principle to help us solve the problem. The corner point principle In linear programming problems, the maimum or minimum value of a linear objective function will occur at one of the corners of the feasible. Note: If two corners give the same maimum or minimum value, then all points along a line joining the two points will also give the same maimum or minimum value. Cambridge Universit Press Uncorrected Sample Pages 978-0-521-74049-4 2008 Evans, Lipson, Jones, Aver, TI-Nspire & Casio ClassPad material prepared in collaboration with Jan Honnens & David Hibbard
394 Essential Standard General Mathematics This means that we onl need to evaluate the objective function at each of the corner points, labelled A, B, C and D, and find which gives the maimum value. It helps to set up a table as follows. bjective function Points P =2 +3 A (0, 0) P = 2 0 + 3 0 = 0 B (0, 6) P = 2 0 + 3 6 = 18 C (5, 3) P = 2 5 + 3 3 = 19 D (8, 0) P = 2 8 + 3 0 = 16 Thus, the maimum value of the objective function, P = 19, occurs when = 5 and = 3. Eample 8 Find the minimum value of the objective function C = 5 + 2, subject to the constraints: 0, 0 + 2 10 6 + 4 36 as displaed in the feasible opposite. Solution 1 Set up a table for the objective function. 2 Evaluate the objective function at each of the corners A, B and C. 3 Identif the corner point giving the minimum value and write our answer. Eercise 9D Finding the minimum value of an objective function A (0, 9) (0, 5) + 2 = 10 B(4, 3) 6 + 4 = 36 (0, 6) C (10, 0) bjective function Points C = 5 + 2 A(0, 9) C = 5 0 + 2 9 = 18 B(4, 3) C = 5 4 + 2 3 = 26 C(10, 0) C = 5 10 + 2 0 = 50 The minimum value is C = 18, which occurs when = 0and = 9. For each of the following objective functions and feasible s, find the maimum or minimum value (as required) and the point at which it occurs. 1 P = 4 + 2 (maimum) 2 P = 3 + 4 (maimum) B(0, 5) C(3, 3) A(0, 0) D(6, 0) B(0, 10) A(0, 0) C(2, 12) D(6, 8) E(12, 0) Cambridge Universit Press Uncorrected Sample Pages 978-0-521-74049-4 2008 Evans, Lipson, Jones, Aver, TI-Nspire & Casio ClassPad material prepared in collaboration with Jan Honnens & David Hibbard
Chapter 9 Inequalities and linear programming 395 3 C = 3 + 5 (minimum) 4 C = + (minimum) A(0, 10) B(3, 2.5) C(6, 0) 5 P = + 2 (maimum) A(0, 0) B(10, 20) C(30, 0) A(0, 12) B(2, 4) C(10, 0) 6 C = 2 + 2 (minimum) A(0, 5) 9.5 Linear programming applications B(5, 0) Younow have all the technical skills necessar to set up and solve a basic linear programming problem. Eample 9 Setting up and solving a maimising problem A manufacturer makes two sorts of orange-flavoured chocolates: House Brand and range Delights. 1kgofHouse Brand contains 0.3 kg of chocolate and 0.7 kg of orange fill. 1kgofrange Delights contains 0.5 kg of chocolate and 0.5 kg of orange fill. 300 kg of chocolate and 350 kg of orange fill are available to the manufacturer each da. The profit is $7.50 per kilogram on House Brand and $10 per kilogram on range Delights. How much of each tpe of orange-flavoured chocolate should be made each da to maimise profit? Cambridge Universit Press Uncorrected Sample Pages 978-0-521-74049-4 2008 Evans, Lipson, Jones, Aver, TI-Nspire & Casio ClassPad material prepared in collaboration with Jan Honnens & David Hibbard
396 Essential Standard General Mathematics Solution 1 Define and. Let be the amount (in kg) of House Brand made each da. Let be the amount (in kg) of range Delights made each da. 2 Write down the constraints. and cannot be negative. 300 kg of chocolate is available. 350 kg of orange fill is available. 3 Graph the feasible defined b the constraints. Mark in each of the corner points and label with their coordinates. Use a calculator to determine the point of intersection. 4 Write down the objective function (in dollars). Call it P, for profit. 5 Determine the maimum profit b evaluating the objective function at each corner of the feasible. Constraints: 0, 0 0.3 + 0.5 300 (chocolate) 0.7 + 0.5 350 (orange fill) B(0, 600) A(0, 0) bjective function: P = 7.5 + 10 (0, 700) C (125, 525) 0.3 + 0.5 = 300 D(500, 0) (1000, 0) 0.7 + 0.5 = 350 bjective function Point P = 7.5 + 10 A(0, 0) P = 7.5 0 + 10 0 = $0 B(0, 600) P = 7.5 0 + 10 600 = $6000 C(125, 525) P = 7.5 125 + 10 525 = $6187.50 D(500, 0) P = 7.5 500 + 10 0 = $3750 6 Write our answer to the question. Themaimum profit is $6187.50, which is obtained b making 125 kg of House Brand and 525 kg of range Delights. Cambridge Universit Press Uncorrected Sample Pages 978-0-521-74049-4 2008 Evans, Lipson, Jones, Aver, TI-Nspire & Casio ClassPad material prepared in collaboration with Jan Honnens & David Hibbard
Chapter 9 Inequalities and linear programming 397 Eample 10 Setting up and solving a minimising problem SpeedGro and Powerfeed are two popular brands of home garden fertiliser. The both contain the nutrients X, Y and Z, needed for health plant growth. 1kgofSpeedGro contains 30 units of X, 50units of Y and 10 units of Z. 1kgofPowerfeed contains 20 units of X, 20units of Y and 20 units of Z. Agardener calculates that he needs a fertiliser containing at least 160 units of nutrient X, 200 units of nutrient Y and 80 units of nutrient Z. Speedgro costs $8 per kg and Powerfeed costs $6 per kg. How much of each tpe of fertiliser should he bu to meet his needs at the minimum cost? Solution 1 Define and. Let be the amount (in kg) of SpeedGro needed. Let be the amount (in kg) of Powerfeed needed. 2 Write down the constraints. Constraints: and cannot be negative. 0, 0 At least 160 units of X are needed. At least 200 units of Y are needed. At least 80 units of Z are needed. 30 + 20 160 (nutrient X) 50 + 20 200 (nutrient Y) 10 + 20 80 (nutrient Z) 3 Graph the feasible defined b the constraints. Mark in each of the corner A(0, 10) points and label with their coordinates. (0, 8) Use a calculator to determine the points of intersection. B(2, 5) (0, 4) C (4, 2) D(8, 0) (4, 0) (5.3, 0) 50 + 20 = 200 10 + 20 = 80 30 + 20 = 160 4 Write down the objective function (in dollars). Call it C, for cost. bjective function: C = 8 + 6 5 Determine the minimum cost b bjective function evaluating the objective function at Point C = 8 + 6 each corner of the feasible. A(0, 10) C = 8 0 + 6 10 = $60 B(2, 5) C = 8 2 + 6 5 = $46 C(4, 2) C = 8 4 + 6 2 = $44 D(8, 0) C = 8 8 + 6 0 = $64 6 Write our answer to the question. Theminimum cost is $44, which is achieved b buing 4 kg of SpeedGro and 2 kg of Powerfeed. Cambridge Universit Press Uncorrected Sample Pages 978-0-521-74049-4 2008 Evans, Lipson, Jones, Aver, TI-Nspire & Casio ClassPad material prepared in collaboration with Jan Honnens & David Hibbard
398 Essential Standard General Mathematics Eercise 9E 1 Afactor makes two products: Wigits and Gigits. Two different machines are used. To make a Wigit takes 1 hour on Machine 1 and 2 hours on Machine 2. To make a Gigit takes 1 hour on Machine 1 and 4 hours on Machine 2. Up to 8 hours of Machine 1 time and up to 24 hours of Machine 2 time are available each da. The factor makes a profit of $200 for each Wigit and $360 for each Gigit it produces. a Let be the number of Wigits made each da. Let be the number of Gigits made each da. The constraints for this problem are: 0, + 8 + 4 (Machine 1 time) (Machine 2 time) Determine the missing information. b The feasible is shown on the right. Some information is missing. Determine the missing information. (0, 8) + = 8 B C A(0, 0) D(8, 0) = 24 ( ) c The objective function is give b P = 200 +, where P stands for profit (in dollars). Determine the missing information. d How man Wigits and Gigits should be made each da to maimise profit, and what is this profit? 2 An outdoor clothing manufacturer makes two sorts of jackets: Polarbear and Polarfo. To make a Polarbear jacket takes2mofmaterial. The time taken to make a Polarbear jacket is 2.4 hours. To make a Polarfo jacket takes2mofmaterial. The time taken to make a Polarfo jacket is 3.2 hours. The manufacturer has 520 m of material available and 672 hours of worker time to make the jackets. The manufacturer makes a profit of $36 for each Polarbear jacket and $42 for each Polarfo jacket it produces. Cambridge Universit Press Uncorrected Sample Pages 978-0-521-74049-4 2008 Evans, Lipson, Jones, Aver, TI-Nspire & Casio ClassPad material prepared in collaboration with Jan Honnens & David Hibbard
a Let be the number of Polarbear jackets made. Let be the number of Polarfo jackets made. The constraints for this problem are:, + 2 (material availabilit) + 672 (worker time availabilit) Determine the missing information. b The feasible is shown on the right. Some information is missing. Determine the missing information. Chapter 9 Inequalities and linear programming 399 B(0, 210) = 520 C ( ) = 672 E ( ) A(0, 0) D ( ) c The objective function is given b P = +, where P stands for profit (in dollars) Determine the missing information. d What is the maimum profit that can be made, and how man Polarbear jackets and Polarfo jackets should be made each da to achieve this profit? 3 Following a natural disaster, the arm plans to use helicopters to transport medical teams and their equipment into a remote area. The have two tpes of helicopter: Redhawks and Blackjets. Redhawks carr 45 people and 3 tonnes of equipment. Blackjets carr 30 people and 4 tonnes of equipment. At least 450 people and 36 tonnes of equipment need to be transported. Redhawks cost $3600 per hour to run and Blackjets cost $3200 per hour to run. a Let be the number of Redhawks. Let be the number of Blackjets. The constraints for this problem are: 0, 0 (people) (equipment) Determine the missing information. Cambridge Universit Press Uncorrected Sample Pages 978-0-521-74049-4 2008 Evans, Lipson, Jones, Aver, TI-Nspire & Casio ClassPad material prepared in collaboration with Jan Honnens & David Hibbard
400 Essential Standard General Mathematics b The feasible is shown on the right. Some information is missing. Determine the missing information. ( ) C(12, 0) ( ) = 36 = 450 c The objective function is given b C = +, where C stands for cost (in dollars). Determine the missing information. d How man Redhawks and Blackjets should be used to minimise the cost per hour, and what is this cost? A(0, 15) B( ) 4 Asawmill produces both construction grade and furniture grade timber. To produce 1 cubic metre of construction grade timber takes 2 hours of sawing and 3 hours of planing. To produce 1 cubic metre of furniture grade timber takes 2 hours of sawing and 6 hours of planing. Up to 8 hours of sawing time and 18 hours of planing time are available each da. The sawmill makes a profit of $500 per cubic metre of construction grade timber and $600 per cubic metre of furniture grade timber it produces. a Write down the constraints and profit function for this problem. b Draw a diagram. c Find how much construction grade and furniture grade timber the sawmill should make each da to maimise its profit. What is this profit? 5 Two breakfast cereal mies, Healthstart and Wakeup, are available in bulk. Each kilogram of Healthstart contains 12 mg of vitamin B1 and 40 mg of vitamin B2. Each kilogram of Wakeup contains 20 mg of vitamin B1 and 25 mg of vitamin B2. Youwant a mi of the two that contains at least 15 mg of vitamin B1 and 30 mg of vitamin B2. Healthstart costs $5 a kilogram and Wakeup costs $4.50 per kilogram. a Write down the constraints and cost function for this problem. b Draw a diagram. c Find the miture of these two cereals that will meet our needs at minimum cost. What is this cost? Cambridge Universit Press Uncorrected Sample Pages 978-0-521-74049-4 2008 Evans, Lipson, Jones, Aver, TI-Nspire & Casio ClassPad material prepared in collaboration with Jan Honnens & David Hibbard
Chapter 9 Inequalities and linear programming 401 Ke ideas and chapter summar Linear inequalit Displaing linear inequalities in one variable on a number line Displaing linear inequalities in one variable on the coordinate plane Displaing linear inequalities in two variables on the coordinate plane Linear programming A linear inequalit involves one or two of the signs >,, <or, but not an equals sign ( = ). A linear inequalit in one variable can be represented on a number line b a solid coloured line ending at one or two circles. The line represents all the possible solutions of the inequalit. An open circle ( ) indicates that the end value is not included in the inequalit (for < or >). A closed circle ( ) indicates that the end value is included in the inequalit (for or ). Linear inequalities in one variable can be represented on a coordinate plane b a shaded bounded b one or two lines parallel to the -or-aes. The represents all the possible solutions of the inequalit. A dashed line indicates that the line is not included in the inequalit (for < or >). A solid line indicates that the line is included in the inequalit (for or ). A linear inequalit in two variables can be represented on a coordinate plane b a shaded bounded b a line at an angle to the - and -aes. The represents all the possible solutions of the inequalit. The boundar line is dashed if it is not included in the inequalit (for < or >), but solid if it is included (for or ). A reference point, often the origin (0, 0), can be used to help decide whether the required lies above or below the line. When solving simultaneous inequalities, the in the coordinate plane that is common to all the inequalities is called the feasible.itrepresents all the possible solutions to the simultaneous inequalities. The feasible can be found graphicall (for a small number of inequalities) b shading in the required s for all the inequalities and determining where the all overlap. Agraphics calculator can be used to graph a feasible. Linear programming involves maimising or minimising a linear quantit subject to the constraints represented b a set of linear inequalities. The constraints (e.g. requirements, resources) define the feasible in which the quantit is to be maimised or minimised. Review Cambridge Universit Press Uncorrected Sample Pages 978-0-521-74049-4 2008 Evans, Lipson, Jones, Aver, TI-Nspire & Casio ClassPad material prepared in collaboration with Jan Honnens & David Hibbard
402 Essential Standard General Mathematics Review bjective function Corner point principle Skills check The constraints 0 and 0together restrict the feasible to the positive (first) quadrant. The objective function is a linear epression representing the quantit to be maimised (e.g. profit) or minimised (e.g. cost) in a linear programming problem. The corner point principle states that, in linear programming problems, the maimum or minimum value of a linear objective function will occur at one of the corners of the feasible, or on a line on the boundar of the feasible joining two of the corners. Having completed this topic ou should be able to: represent a linear inequalit in one variable on a number line represent a linear inequalit in one or two variables on the coordinate plane know the meaning of the terms feasible, constraint and objective function as the relate to linear programming determine the maimum or minimum value of an objective function for a given feasible set up and solve basic linear programming problems. Multiple-choice questions 1 The inequalit displaed on the number line on the right is: A 1 7 B 1 < < 7 C 1 < 7 D 1 < 7 E 1 > > 7 2 The inequalit displaed on the number line on the right is: A < 5 B 5 C > 5 D 5 E 0 > > 5 3 The inequalit displaed on the coordinate plane on the right is: A < 8 B 8 C < 8 D 8 E 0 > > 8 1 0 1 2 3 4 5 6 7 8 1 0 1 2 3 4 5 6 7 8 (0, 8) = 8 Cambridge Universit Press Uncorrected Sample Pages 978-0-521-74049-4 2008 Evans, Lipson, Jones, Aver, TI-Nspire & Casio ClassPad material prepared in collaboration with Jan Honnens & David Hibbard
Chapter 9 Inequalities and linear programming 403 4 The inequalit displaed on the coordinate plane on the right is: A 3 < < 10 B 3 < 10 C 3 10 D 3 < < 10 E 3 < 10 5 The equation of the line displaed on the right is: A 4 + 5 = 4 B 4 5 = 4 C 5 + 4 = 20 D 4 + 5 = 20 E 4 5 = 20 6 The equation of the line displaed on the right is: A 4 5 = 0 B 4 + 5 = 0 C 5 4 = 20 D 5 + 4 = 20 E 5 = 20 7 The displaed on the right (including the line) represents the inequalit: A 5 + 2 < 20 B 5 + 2 20 C 5 + 2 > 20 D 5 + 2 20 E 2 + 5 > 20 = 3 (0, 10) = 10 (3, 0) (10, 0) (0, 4) (0, 0) 5 + 2 = 20 8 The displaed on the right (not including the line) represents the inequalit: A 3 6 B 3 < 6 C 3 6 D 3 > 6 E 3 > 6 (4, 0) (0, 6) ( 2, 0) (5, 0) (5, 4) 3 = 6 Review Cambridge Universit Press Uncorrected Sample Pages 978-0-521-74049-4 2008 Evans, Lipson, Jones, Aver, TI-Nspire & Casio ClassPad material prepared in collaboration with Jan Honnens & David Hibbard
404 Essential Standard General Mathematics Review 9 The two lines shown on the right intersect at the point: A (1, 1.3) B (2, 1.5) C (1.2, 1.2) D (2.4, 1.2) E (3, 2.4) (0, 3) (0, 2) 10 The feasible displaed on the right (including the line) is defined b the inequalities: (0, 5) A 0, 0, < 5 B 0, 0, 5 C 0, 0, + < 5 D 0, 0, + 5 E 0, 0, + 5 11 The feasible displaed on the right (including the lines) is defined b the inequalities: A 0, 0, + 2 10, 4 + 12 B 0, 0, + 2 10, 4 + 12 C 0, 0, + 2 > 10, 4 + > 12 D 0, 0, + 2 < 10, 4 + < 12 E 0, 0, + 2 10, 4 + 12 (0, 12) 3 + 4 = 12 (0, 5) (2, 4) (3, 0) 4 + = 12 + 3 = 6 (4, 0) (6, 0) (5, 0) 12 For the feasible displaed in Question 11, the minimum value of the objective function, C = 2 +, is: A 5 B 6 C 8 D 12 E 20 13 For the feasible displaed on the right, the maimum value of the objective function, P = 4 + 3, is: A 0 B 40 C 42 D 48 E 60 (0, 10) (0, 12) (6, 6) (0, 0) (12, 0) (10, 0) + 2 = 10 The following information relates to Questions 14 to 16 An outdoor clothing manufacturer makes two stles of all-weather coats: long and short. To make a short coat,2mofmaterial are required. The time taken to make a short coat is 2.5 hours. To make a long coat,3mofmaterial are required. The time taken to make a long coat is 3.5 hours. (15, 0) Cambridge Universit Press Uncorrected Sample Pages 978-0-521-74049-4 2008 Evans, Lipson, Jones, Aver, TI-Nspire & Casio ClassPad material prepared in collaboration with Jan Honnens & David Hibbard
Chapter 9 Inequalities and linear programming 405 The manufacturer has 450 m of material available and 700 hours of worker time to make the coats. The manufacturer makes a profit of $40 for each short coat and $48 for each long coat. Let be the number of short coats made. Let be the number of long coats made. 14 The constraints that relate to the amount of material available are: A 0, 0, 2 + 3 450 B 0, 0, 2 + 3 450 C 0, 0, 2.5 + 3.5 700 D 0, 0, 2.5 + 3.5 700 E 0, 0, 40 + 48 700 15 The constraints that relate to the amount of time available are: A 0, 0, 2 + 3 450 B 0, 0, 2 + 3 450 C 0, 0, 2.5 + 3.5 700 D 0, 0, 2.5 + 3.5 700 E 0, 0, 40 + 48 700 16 The objective function P is: A P = 2.5 + 3.5 B P = 2 + 3 C P = 3 + 3.5 D P = 40 + 48 E P = 450 + 700 Short-answer questions 1 Plot the inequalit 2 < 4onanumber line. 2 Plot the inequalit 1 < 5onthe coordinate plane. 3 Plot the inequalit 5 + 4 < 40 on the coordinate plane. 4 Plot the defined b the inequalities: 0, 0, 3 + 5 60 5 Plot the defined b the inequalities: 0, 0, 2 + 3 30, + 4 20 Etended-response questions 1 Agarden products compan makes two sorts of fertiliser: Standard Grade and Premium Grade. There are two main ingredients: nitrate and phosphate. To make a tonne of Standard Grade fertiliser takes 0.8 tonnes of nitrate and 0.2 tonnes of phosphate. To make a tonne of Premium Grade fertiliser takes 0.7 tonnes of nitrate and 0.3 tonnes of phosphate. The compan has 56 tonnes of nitrate and 21 tonnes of phosphate. The compan makes a profit of $600 per tonne on Standard Grade fertiliser and $750 per tonne on Premium Grade fertiliser. Review Cambridge Universit Press Uncorrected Sample Pages 978-0-521-74049-4 2008 Evans, Lipson, Jones, Aver, TI-Nspire & Casio ClassPad material prepared in collaboration with Jan Honnens & David Hibbard
406 Essential Standard General Mathematics Review a Write the constraints and profit function for this problem. b Draw a diagram. c Find how much of each tpe of fertiliser the compan should make to maimise its profit. What will this profit be? 2 Two foods fed to animals contain both vitamin A and vitamin B. 1kgofFood A contains 3 units of vitamin A and 4 units of vitamin B. 1kgofFood B contains 5 units of vitamin A and 3 units of vitamin B. The dail vitamin requirement of each animal is at least 15 units of vitamin A and at least 12 units of vitamin B. Food A costs $0.30 per kg and Food B costs $0.24 per kg. a Write the constraints and cost function for this problem. b Draw a diagram. c Find how much of each tpe of food should be fed to the animals each da to minimise cost. What is this cost? Cambridge Universit Press Uncorrected Sample Pages 978-0-521-74049-4 2008 Evans, Lipson, Jones, Aver, TI-Nspire & Casio ClassPad material prepared in collaboration with Jan Honnens & David Hibbard