Chapter 4. Linear Programming. Chapter Outline. Chapter Summary

Chapter 4 Linear Programming Chapter Outline Introduction Section 4.1 Mixture Problems: Combining Resources to Maximize Profit Section 4.2 Finding the Optimal Production Policy Section 4.3 Why the Corner Point Principle Works Section 4.4 Linear Programming: Life Is Complicated Section 4.5 A Transportation Problem: Delivering Perishables Section 4.6 Improving on the Current Solution Chapter Summary A fundamental management problem that a company faces is determining how much of each of its products to produce in order to maximize profit. This problem is called a mixture problem, and linear programming is a powerful technique used to solve such problems. To get started, we must know certain information about the mixture problem: what products are to be made what resources will be needed how much of each resource is available how much of each resource is used by each product how much profit is generated by each unit of each product The variables in the problem are the number of units of each product produced. They are assumed to be nonnegative (minimum constraints). A production policy is a specification of how many units of each product are to be made. To be achievable, a production policy has to satisfy the natural constraints of the problem, since no policy can use more of a resource than the company has available. Production policies satisfying all problem constraints are called feasible solutions to the problem, and the set of all feasible solutions is called the feasible region. The profit function is an expression built from the variables and per-unit profits that tell us how much profit is produced by a given feasible solution. To solve our mixture problem, we need to find a feasible solution that produces the largest possible profit. Linear programming applies in the case when the problem constraints are linear inequalities and the profit function is linear in the variables. In this case, the feasible region is polygonal in 2-space and polyhedral in 3-space. It will have an analogous structure in higher dimensions. The geometry of this situation is crucial to the solution. A fundamental result, the corner point principle, states that the optimal solution occurs at a corner of the feasible region. This result suggests a possible solution strategy. find all corners of the feasible region determine the profit at each corner point choose the production policy of any corner producing the maximum profit 55

56 Chapter 4 Computationally, this procedure suffers from the same problem as the brute force method applied to the TSP; for large problems there are just too many corners to check quickly. Fortunately, faster algorithms exist to solve linear programming problems. The simplex method developed by George Dantzig starts at some corner of the feasible region and then moves from corner to neighboring corner, always going to a corner where profit is the same or larger, until it locates a corner producing maximum profit. An alternative to the simplex method is Karmarker s algorithm, which bores into the interior of the feasible region to decide which corner to examine next. The simplex method is fast in practice but can be shown no more efficient than the brute force method on certain cleverly constructed problems. Karmarker s algorithm provides a computationally efficient method for solving linear programming problems. In the transportation problem, we are interested in meeting demands while minimizing cost. Information can be set up in a tableau. The tableau shows costs and rim conditions for the transporation problem. Using the Northwest Corner Rule (NCR), we can obtain a feasible solution. This solution, however, may not be optimal. By calculating and comparing the values of indicator cells, we can improve on the feasible solution. Using the stepping stone method, we look at indicator cells that have negative values and determine an optimal solution. Skill Objectives 1. Create a chart to represent the given information in a linear programming problem with two variables. 2. From its associated chart, write the constraints of a linear programming problem as linear inequalities. 3. List two implied constraints in every linear programming problem. 4. Formulate a profit equation for a linear-programming problem when given the per-unit profits. 5. Describe the graphical implications of the implied constraints x 0 and y 0. 6. Draw the graph of a line in a coordinate-axis system. 7. Graph a linear inequality in a coordinate-axis system. 8. Determine by a substitution process whether a point with given coordinates is contained in the graph of a linear inequality. 9. Indicate the feasible region for a linear programming problem by shading the graphical intersection of its constraints. 10. Locate the corner points of a feasible region from its graph. 11. Evaluate the profit function at each corner point of a feasible region. 12. Apply the corner point theorem to determine the maximum profit for a linear programming problem. 13. Interpret the corner point producing the maximum profit as the solution to the corresponding linear programming problem. 14. List two methods for solving linear programming problems with many variables. 15. Given a tableau, apply the Northwest Corner Rule to find a feasible solution. 16. Calculate the cost of the system found by the Northwest Corner Rule. 17. Calculate the value of indicator cells. 18. Use the stepping stone method to find an optimal solution.

Linear Programming 57 Teaching Tips 1. The material differs significantly from that in other texts. The basic ideas are introduced one at a time; first, mixture problems having just one resource, then resource constraints are added, followed by minimum quantities for products and, finally, two products and two resources. Along the way, the students are introduced to the profit function, making mixture charts, the feasible region, and the corner principle. Although this approach is nonstandard, the slow buildup of ideas enables students to comprehend the material more thoroughly. 2. Because many of the students taking this course have only an algebra background, a review of coordinate graphing is often helpful. This is especially true when graphing linear inequalities. The concept of selecting a test point often requires additional explanation. 3. A short review of the concept of set intersection may be helpful when indicating the feasible region on a linear programming graph. 4. The instructor will need to decide whether to require the student to find the intersection point of two boundary lines by solving a system of linear equations, or to have the coordinates of that point provided to the student. Requiring the solution will probably necessitate reviewing that procedure in class discussion. 5. When setting up the chart for a linear programming problem, you may want to interchange the rows and columns in the text example. If the variables name the columns and the resources the rows, then the inequalities can be written easily from the chart in a horizontal fashion. This may be easier for the student because of its visual simplicity. 6. When graphing several linear inequalities, students find it helpful to have colored pens or pencils to indicate the different half-planes. It is then easier to find the feasible region because the intersection is the region that contains all the colors. 7. Text Exercises 30 41 offer a variety of settings for linear programming problems. These can be used to advantage in working through examples with students. 8. You may choose to prepare in advance blank tableaux for students to use during lecture of applying the Northwest Corner Rule to find a feasible solution. Research Paper In this chapter, two methods for solving linear programming problems with many variables are briefly discussed. These are the simplex method and Karmarkar s algorithm. Prior to Karmarkar s algorithm (1984), Leonid Khaciyan presented a method guaranteeing to solve any linear programming problem in 1979. Students can further investigate these methods and their origins, discuss their similarities and differences, or prepare a general discussion of the contributions of these two mathematicians.

58 Chapter 4 Collaborative Learning Linear Programming You live on a farm and bake chocolate chip cookies and chocolate chip muffins in your home, and sell them to the public. One morning, you check your supplies and discover that you are low on flour and chocolate chips, but have plenty of the other ingredients. Your car is in the repair shop and there are 2 feet of snow on the ground, so there is no way to replenish your supplies until tomorrow. Each batch of cookies requires 2 pounds of flour and 1 pound of chocolate chips; each batch of muffins uses 3 pounds of flour and 2 pounds of chocolate chips. You have 60 pounds of flour and 36 pounds of chocolate chips on hand. Your profit on each batch of cookies is $8, and on muffins your profit is $13 per batch. How should you divide your baking between cookies and muffins in order to maximize your profit for the day? Hints: a. If no cookies are baked, how many batches of muffins can be baked? What will the profit be in this case? b. Answer the same questions if no muffins are baked. c. Which is more profitable: baking only cookies or only muffins? d. Is there a mix of cookies and muffins that makes even more money than either only muffins or only cookies?

Linear Programming 59 Solutions Skills Check: 1. a 2. c 3. b 4. b 5. b 6. c 7. c 8. c 9. b 10. c 11. b 12. c 13. c 14. c 15. c 16. c 17. c 18. a 19. a 20. c Exercises: 1. (a) 2x 3y 12 y-intercept: Substitute x 0. 20 3y 12 0 3y 12 12 3y 12 y 4 y-intercept is ( 0,4 ). 3 Graph: x-intercept: Substitute y 0. 2x 3( 0) 12 2x 0 12 12 2x 12 x 6 x-intercept is ( 6,0 ). 2 (b) 3x 5y 30 y-intercept: Substitute x 0. 30 5y 30 0 5y 30 30 5y 30 y 6 y-intercept is ( 0,6 ). 5 Graph: x-intercept: Substitute y 0. 3x 5( 0) 30 3x 0 30 30 3x 30 x 10 x-intercept is ( 10,0 ). 3 (c) 4x 3y 24 y-intercept: Substitute x 0. 40 3y 24 0 3y 24 24 3y 24 y 8 y-intercept is ( 0,8 ). 3 Graph: x-intercept: Substitute y 0. 4x 3( 0) 24 4x 0 24 24 4x 24 x 6 x-intercept is ( 6,0 ). 4 Continued on next page

60 Chapter 4 1. continued (d) 7x 4y 42 y-intercept: Substitute x 0. 7( 0) 4y 42 0 4y 42 42 4y 42 y 10.5 y-intercept is ( 0,10.5 ). 4 Graph: x-intercept: Substitute y 0. 7x 4( 0) 42 7x 0 42 42 7x 42 x 6 x-intercept is ( 6,0 ). 7 (e) x 3 This form represents a vertical line. y-intercept: None x-intercept: ( 3,0 ). Graph: (f) y 6 This form represents a horizontal line. y-intercept is ( 0,6 ). x-intercept: None Graph: 2. (a) 4x 3y 18 and x 0 Note: This situation is shown only for the first quadrant. The y-intercept of 4x 3y 18 can be found by substituting x 0. 40 3y 18 18 0 3y 18 3y 18 y 6 The y-intercept is ( 0,6 ). The x-intercept of 4x 3y 18 can be found by substituting y 0. 4x 3( 0) 18 18 4x 0 18 4x 18 x 4.5 The x-intercept is ( 4.5,0 ). x 0 represents a vertical line, namely the y-axis. Continued on next page 4 3

Linear Programming 61 2. (a) continued To find the point of intersection, substitute x 0 into 4x 3y 18. 40 3y 18 0 3y 18 18 3y 18 y 6 The point of intersection is therefore ( 0,6 ). 3 (b) 5x 3y 45 and y 5 Note: This situation is shown only for the first and fourth quadrants. The y-intercept of 5x 3y 45can be found by substituting x 0. 50 3y 45 0 3y 45 45 3y 45 y 15 The y-intercept is ( 0,15 ). The x-intercept of 5x 3y 45can be found by substituting y 0. 5x 3( 0) 45 5x 0 45 45 5x 45 x 9 The x-intercept is ( 9,0 ). y 5 represents a horizontal line, which lies below the x-axis. To find the point of intersection, substitute y 5 into 5x 3y 45. 5x 3( 5) 45 5x ( 15) 45 5x 15 45 60 5x 60 x 12 The point of intersection is therefore ( 12, 5 ). 3 5 5

62 Chapter 4 3. Note: These situations are shown only for the first quadrant. (a) x y 10 and x 2y 14 The y-intercept of x y 10 can be found by substituting x 0. 0 y 10 y 10 The y-intercept is ( 0,10 ). The x-intercept of x y 10 can be found by substituting y 0. x 0 10 x 10 The x-intercept is ( 10,0 ). The y-intercept of x 2y 14 can be found by substituting x 0. 0 2y 14 2y 14 14 y 7 The y-intercept is ( 0,7 ). The x-intercept of x 2y 14 can be found by substituting y 0. The x-intercept is ( 14,0 ). 2 x 2 0 14 x 0 14 x 14 To find the point of intersection, we can multiply both sides of x y 10 by 1, and add the result to x 2y 14. x y 10 x 2y 14 y 4 Substitute y 4 into x y 10 to solve to x. x 4 10 x 6 The point of intersection is therefore ( 6,4 ). Continued on next page

Linear Programming 63 3. continued (b) y 2x 0 and x 4 x 4 represents a vertical line, which lies to the right of the y-axis. By substituting either x 0 or y 0, we see that y 2x 0 passes through ( 0,0 ), the origin. By substituting an arbitrary value (except 0) for one of the variables, we can find another point that lies on the graph of y 2x 0. Since we see the point of intersection with the graph of x 4, we could use this value. y 2 4 0 y 8 0 y 8 Thus, a second point on the graph of y 2x 0 intersection between the two lines. is 4,8. This is also the point of 4. Note: These situations are shown only for the first quadrant. (a) x 8 The graph of x 8 represents a vertical line with x-intercept ( 8,0 ). Since any value of x to the right of this line is greater than 8, we shade the portion to the right of this vertical line. (b) y 5 The graph of y 5 represents a horizontal line with y-intercept ( 0,5 ). Since any value of y above this line is greater than 5, we shade the portion above this horizontal line. Continued on next page

64 Chapter 4 4. continued (c) 5x 3y 15 The y-intercept of 5x 3y 15 can be found by substituting x 0. 5 0 3y 15 0 3y 15 3y 15 y 5 The y-intercept is ( 0,5 ). 15 The x-intercept of 5x 3y 15 can be found by substituting y 0. 5x 30 15 5x 0 15 5x 15 x 3 15 The x-intercept is ( 3,0 ). We draw a line connecting these points. Testing the point ( 0,0 ), we have the statement 5( 0) 3( 0) 15 or 0 15. This is a true statement, thus we shade the half-plane containing our test point, the down side of the line. 3 5 (d) 4x 5y 30 The y-intercept of 4x 5y 30 can be found by substituting x 0. 4 0 5y 30 0 5y 30 5y 30 y 6 The y-intercept is ( 0,6 ). 30 The x-intercept of 4x 5y 30 can be found by substituting y 0. 4x 5 0 30 4x 0 30 4x 30 x 7.5 30 The x-intercept is ( 7.5,0 ). We draw a line connecting these points. Testing the point ( 0,0 ), we have the statement 4( 0) 5( 0) 30 or 0 30. This is a true statement, thus we shade the half-plane containing our test point, the down side of the line. 4 5

Linear Programming 65 5. Note: These situations are shown only for the first quadrant. (a) x 4 The graph of x 4 represents a vertical line with x-intercept ( 4,0 ). Since any value of x to the right of this line is greater than 4, we shade the portion to the right of this vertical line. (b) y 9 The graph of y 9 represents a horizontal line with y-intercept ( 0,9 ). Since any value of y above this line is greater than 9, we shade the portion above this horizontal line. (c) 3x 2y 18 The y-intercept of 3x 2y 18 can be found by substituting x 0. 3 0 2y 18 0 2y 18 2y 18 y 9 The y-intercept is ( 0,9 ). 18 The x-intercept of 3x 2y 18 can be found by substituting y 0. 3x 2 0 18 3x 0 18 3x 18 x 6 18 The x-intercept is ( 6,0 ). We draw a line connecting these points. Testing the point ( 0,0 ), we have the statement 3( 0) 2( 0) 18 or 0 18. This is a true statement, thus we shade the half-plane containing our test point, the down side of the line. 3 2 Continued on next page

66 Chapter 4 5. continued (d) 7x 2y 42 The y-intercept of 7x 2y 42 can be found by substituting x 0. 7 0 2y 42 0 2y 42 2y 42 y 21 The y-intercept is ( 0,21 ). 42 The x-intercept of 7x 2y 42 can be found by substituting y 0. 7x 20 42 7x 0 42 7x 42 x 6 42 The x-intercept is ( 6,0 ). We draw a line connecting these points. Testing the point ( 0,0 ), we have the statement 7( 0) 2( 0) 42 or 0 42. This is a true statement, thus we shade the half-plane containing our test point, the down side of the line. 2 7 6. (a) 2x 4y 28 (b) 2x 0.5y 40 ( pruning) 0.5x 0.25y 2 ( shredding) 7. (a) 6x 4y 300 (b) 30x 72y 420 8. 12x 10y 640 ( beef ) 4x 3y 480 ( pork) 9. x 0; y 0; x 2y 12 The constraints of x 0 and y 0 indicate that we are restricted to the upper right quadrant created by the x-axis and y-axis. The y-intercept of x 2y 12 can be found by substituting x 0. 12 0 2y 12 2y 12 y 6 The y-intercept is ( 0,6 ). The x-intercept of x 2y 12 can be found by substituting y 0. x 2 0 12 x 0 12 x 12 The x-intercept is ( 12,0 ). We draw a line connecting these points. Testing the point ( 0,0 ), we have the statement 0 20 12or 0 12. This is a true statement, thus we shade the half-plane containing our test point, the down side of the line, which is contained in the upper right quadrant. 2

Linear Programming 67 10. x 0; y 0;2x y 10 The constraints of x 0 and y 0 indicate that we are restricted to the upper right quadrant created by the x-axis and y-axis. The y-intercept of 2x y 10 can be found by substituting x 0. The y-intercept is ( 0,10 ). 2 0 y 10 0 y 10 y 10 The x-intercept of 2x y 10 can be found by substituting y 0. 10 2x 0 10 2x 10 x 5 The x-intercept is ( 5,0 ). We draw a line connecting these points. Testing the point ( 0,0 ), we have the statement 2( 0) 0 10 or 0 10. This is a true statement, thus we shade the half-plane containing our test point, the down side of the line, which is contained in the upper right quadrant. 2 11. x 0; y 0;2x 5y 60 The constraints of x 0 and y 0 indicate that we are restricted to the upper right quadrant created by the x-axis and y-axis. The y-intercept of 2x 5y 60 can be found by substituting x 0. 2 0 5y 60 0 5y 60 y 12 The y-intercept is ( 0,12 ). 60 The x-intercept of 2x 5y 60 can be found by substituting y 0. 2x 50 60 2x 0 60 2x 60 x 30 60 The x-intercept is ( 30,0 ). We draw a line connecting these points. Testing the point ( 0,0 ), we have the statement 2( 0) 5( 0) 60 or 0 60. This is a true statement, thus we shade the half-plane containing our test point, the down side of the line, which is contained in the upper right quadrant. 5 2

68 Chapter 4 12. x 10; y 0; 3x 5y 120 The constraints of x 10 and y 0 indicate that we are restricted to the upper right quadrant, to the right of the vertical line x 10. The point of intersection between x 10 and 3x 5y 120 can be found by substituting x 10 into 3x 5y 120. 3 10 5y 120 30 5y 120 5y 90 y 18 90 Thus, the point of intersection is ( 10,18 ). The y-intercept of 3x 5y 120 can be found by substituting x 0. 3 0 5y 120 0 5y 120 y 24 The y-intercept is ( 0,24 ). 120 The x-intercept of 3x 5y 120 can be found by substituting y 0. 3x 5 0 120 3x 0 120 3x 120 x 40 120 The x-intercept is ( 40,0 ). We draw a line connecting these points. Testing the point ( 0,0 ), we have the statement 2( 0) 5( 0) 120 or 0 120. This is a true statement, thus we shade the half-plane containing our test point, the down side of the line, which is contained in the upper right quadrant to the right of the vertical line x 10. 5 3 5

Linear Programming 69 13. x 0; y 4; x y 20 The constraints of x 0 and y 4 indicate that we are restricted to the upper right quadrant, above the horizontal line y 4. The point of intersection between y 4 and x y 20 can be found by substituting y 4 into x y 20. Thus, the point of intersection is ( 16, 4 ). x 4 20 x 16 The y-intercept of x y 20 can be found by substituting x 0. 0 y 20 y 20 The y-intercept is ( 0,20 ). The x-intercept of x y 20 can be found by substituting y 0. x 0 20 x 20 The x-intercept is ( 20,0 ). We draw a line connecting these points. Testing the point ( 0,0 ), we have the statement 0 0 20 or 0 20. This is a true statement, thus we shade the half-plane containing our test point, the down side of the line, which is contained in the upper right quadrant above the horizontal line y 4.

70 Chapter 4 14. x 2; y 6;3x 2y 30 The constraints of x 2 and y 6 indicate that we are restricted to the upper right quadrant, to the right of the vertical line x 2 and above the horizontal line y 6. The point of intersection between x 2 and y 6 is ( 2,6 ). The point of intersection between x 2 and 3x 2y 30 can be found by substituting x 2 into 3x 2y 30. 3 2 2y 30 6 2y 30 2y 24 y 12 Thus, the point of intersection is ( 2,12 ). The point of intersection between 6 into 3x 2y 30. 24 y and 3x 2y 30 can be found by substituting y 6 18 x x x x 3 Thus, the point of intersection is ( 6,6 ). 3 26 30 3 12 30 3 18 6 The y-intercept of 3x 2y 30 can be found by substituting x 0. 3 0 2y 30 0 2y 30 2y 30 y 15 The y-intercept is ( 0,15 ). 30 The x-intercept of 3x 2y 30 can be found by substituting y 0. 3x 2 0 30 3x 0 30 3x 30 x 10 30 The x-intercept is ( 10,0 ). We draw a line connecting these points. Testing the point ( 0,0 ), we have the statement 3( 0) 2( 0) 30 or 0 30. This is a true statement, thus we shade the half-plane containing our test point, the down side of the line, which is contained in the upper right quadrant to the right of the vertical line x 2 and above the horizontal line y 6. 3 2 2

Linear Programming 71 15. For Exercise 9: x 0; y 0; x 2y 12 ( 2, 4 ): Since 2 0, the constraint x 0 is satisfied. Since 4 0, the constraint y 0 is satisfied. Since 2 2( 4) 2 8 10 12, the condition x 2y 12 is satisfied. Thus, ( 2, 4 ) is feasible. ( 10, 6 ): Since 10 0, the constraint x 0 is satisfied. Since 6 0, the constraint y 0 is satisfied. Since 10 2( 6) 10 12 22 12, > the condition x 2y 12 is not satisfied. Thus, ( 10,6 ) is not feasible. For Exercise 11: x 0; y 0;2x 5y 60 ( 2, 4 ): Since 2 0, the constraint x 0 is satisfied. Since 4 0, the constraint y 0 is satisfied. Since 2( 2) 5( 4) 4 20 24 60, the condition x 2y 12 is satisfied. Thus, ( 2, 4 ) is feasible. ( 10, 6 ): Since 10 0, the constraint x 0 is satisfied. Since 6 0, the constraint y 0 is satisfied. Since 2( 10) 5( 6) 20 30 50 60, the condition x 2y 12 is satisfied. Thus, ( 10,6 ) is feasible. For Exercise 13: x 0; y 4; x y 20 ( 2, 4 ): Since 2 0, the constraint x 0 is satisfied. Since 4 4, the constraint y 4 is satisfied. Since 2 4 6 20, the condition x y 20 is satisfied. Thus, ( 2, 4 ) is feasible. Note: It is on the boundary. ( 10, 6 ): Since 10 0, the constraint x 0 is satisfied. Since 6 4, the constraint y 4 is satisfied. Since 10 6 16 20, the condition x y 20 Thus, ( 10,6 ) is feasible. 16. For Exercise 10: x 0; y 0;2x y 10 ( 2, 4 ): Since 2 0, the constraint x 0 is satisfied. Since 4 0, the constraint y 0 is satisfied. Since 2( 2) 4 4 4 8 10, is satisfied. the condition 2x y 10 is satisfied. Thus, ( 2, 4 ) is feasible. ( 10, 6 ): Since 10 0, the constraint x 0 is satisfied. Since 6 0, the constraint y 0 is satisfied. Since 2( 10) 6 20 6 26 10, Thus, ( 10,6 ) is not feasible. Continued on next page > the condition 2x y 10 is not satisfied.

72 Chapter 4 16. continued For Exercise 12: x 10; y 0; 3x 5y 120 ( 2, 4 ): Since 2 < 10, the constraint x 10 is not satisfied. Thus, ( 2, 4 ) is not feasible. Note: There is no need to check the other constraints. ( 10, 6 ): Since 10 10, the constraint x 10 is satisfied. Since 6 0, the constraint y 0 is satisfied. Since 3( 10) 5( 6) 30 30 60 120, Thus, ( 10,6 ) is feasible. Note: It is on the boundary. For Exercise 14: x 2; y 6;3x 2y 30 the condition 3x 5y 120 is satisfied. ( 2, 4 ): Since 2 2, the constraint x 2 is satisfied. Since 4 < 6, the constraint y 6 is not satisfied. Thus, ( 2, 4 ) is not feasible. Note: There is no need to check the other constraint. ( 10, 6 ): Since 10 2, the constraint x 2 is satisfied. Since 6 6, the constraint y 6 is satisfied. Since 3( 10) 2( 6) 30 12 42 30, Thus, ( 10,6 ) is not feasible. 17. We wish to maximize $2.30 x $3.70 y. > the condition 3x 2y 30 is not satisfied. Corner Point Value of the Profit Formula: $2.30 x $3.70y ( 0,0) ( 0,30) ( 12,0) $2.30 0 $3.70 0 $0.00 $0.00 $0.00 $2.30 0 $3.70 30 $0.00 $111.00 $111.00* $2.30 12 $3.70 0 $27.60 $0.00 $27.60 Optimal production policy: Make 0 skateboards and 30 dolls for a profit of $111. 18. We wish to maximize $5.50 x $1.80 y. Corner Point Value of the Profit Formula: $5.50 x $1.80y ( 0,0) ( 0,30) ( 12,0) $5.50 0 $1.80 0 $0.00 $0.00 $0.00 $5.50 0 $1.80 30 $0.00 $54.00 $54.00 $5.50 12 $1.80 0 $66.00 $0.00 $66.00* Optimal production policy: Make 12 skateboards and 0 dolls for a profit of $66.

Linear Programming 73 19. Note: These situations are shown only for the first quadrant. (a) 5x 4y 22 and 5x 10y 40 The y-intercept of 5x 4y 22can be found by substituting x 0. 50 4y 22 0 4y 22 4y 22 22 y 5.5 The y-intercept is ( 0,5.5 ). The x-intercept of 5x 4y 22 can be found by substituting y 0. 5x 4( 0) 22 5x 0 22 5x 22 22 x 4.4 The x-intercept is ( 4.4,0 ). The y-intercept of 5x 10y 40 can be found by substituting x 0. 50 10y 40 0 10y 40 40 y 4 The y-intercept is ( 0,4 ). The x-intercept of 5x 10y 40 can be found by substituting y 0. 5x 10( 0) 40 5x 0 40 5x 40 40 x 8 The x-intercept is ( 8,0 ). To find the point of intersection, we can multiply both sides of 5x 4y 22 by 1, and adding the result to 5x 10y 40. 5x 4y 22 5x 10y 40 18 6 y 18 y 3 6 Substitute y 3 into 5x 10y 40 and solve for x. 5x 10 3 40 5x 30 40 5x 10 x 2 10 The point of intersection is therefore ( 2,3 ). 4 5 10 4 5 Continued on next page

74 Chapter 4 19. continued (b) x y 7 and 3x 4y 24 The y-intercept of x y 7 can be found by substituting x 0. 0 y 7 y 7 The y-intercept is ( 0,7 ). The x-intercept of x y 7 can be found by substituting y 0. x 0 7 x 7 The x-intercept is ( 7,0 ). The y-intercept of 3x 4y 24 can be found by substituting x 0. 30 4y 24 0 4y 24 24 y 6 The y-intercept is ( 0,6 ). The x-intercept of 3x 4y 24 can be found by substituting y 0. 3x 40 24 3x 0 24 3x 24 24 x 8 The x-intercept is ( 8,0 ). 4 3 To find the point of intersection, we can multiply both sides of x y 7 by 3, and adding the result to 3x 4y 24 3x 3y 21 3x 4y 24 y 3 Substitute y 3 into 3x 4y 24 and solve for x. 3x 43 24 3x 12 24 3x 12 x 4 12 The point of intersection is therefore ( 4,3 ). 3

Linear Programming 75 20. x 0; y 0;3x y 9; x 2y 8 The constraints of x 0 and y 0 indicate that we are restricted to the upper right quadrant created by the x-axis and y-axis. The y-intercept of 3x y 9 can be found by substituting x 0. The y-intercept is ( 0,9 ). 3 0 y 9 0 y 9 y 9 The x-intercept of 3x y 9 can be found by substituting y 0. 9 3x 0 9 3x 9 x 3 The x-intercept is ( 3,0 ). We draw a line connecting these points. Testing the point ( 0,0 ), we have the statement 3( 0) 0 9 or 0 9. This is a true statement. The y-intercept of x 2y 8 can be found by substituting x 0. 8 0 2y 8 2y 8 y 4 The y-intercept is ( 0,4 ). The x-intercept of x 2y 8 can be found by substituting y 0. x 2 0 8 x 0 8 x 8 The x-intercept is ( 8,0 ). We draw a line connecting these points. Testing the point ( 0,0 ), we have the statement 0 20 8or 0 8. This is a true statement. Thus, we shade the part of the plane in the upper right quadrant which is on the down side of both the lines 3x y 9 and x 2y 8. Three of the corner points, ( 0,0 ), ( 0,4 ), and ( 3,0 ) lie on the coordinate axes. The fourth corner point is the point of intersection between the lines 3x y 9 and x 2y 8. We can find this by multiplying both sides of 3x y 9 by 2, and adding the result to x 2y 8. 6x 2y 18 x 2y 8 10 5x 10 x 2 Substitute x 2 into x 2y 8 and solve for y. 6 2 2y 8 2y 6 y 3 The point of intersection is therefore ( 2,3 ). 3 2 2 5

76 Chapter 4 21. x 0; y 0;2x y 4;4x 4y 12 The constraints of x 0 and y 0 indicate that we are restricted to the upper right quadrant created by the x-axis and y-axis. The y-intercept of 2x y 4 can be found by substituting x 0. The y-intercept is ( 0,4 ). 2 0 y 4 0 y 4 y 4 The x-intercept of 2x y 4 can be found by substituting y 0. 4 2x 0 4 2x 4 x 2 The x-intercept is ( 2,0 ). We draw a line connecting these points. Testing the point ( 0,0 ), we have the statement 2( 0) 0 4 or 0 4. This is a true statement. The y-intercept of 4x 4y 12 can be found by substituting x 0. 4 0 4y 12 0 4y 12 y 3 The y-intercept is ( 0,3 ). 12 The x-intercept of 4x 4y 12 can be found by substituting y 0. 4x 40 12 4x 0 12 4x 12 x 3 12 The x-intercept is ( 3,0 ). We draw a line connecting these points. Testing the point ( 0,0 ), we have the statement 4( 0) 4( 0) 12 or 0 12. This is a true statement. Thus, we shade the part of the plane in the upper right quadrant which is on the down side of both the lines 2x y 4 and 4x 4y 12. Three of the corner points, ( 0,0 ), ( 0,3 ), and ( 2,0 ) lie on the coordinate axes. The fourth corner point is the point of intersection between the lines 2x y 4 and 4x 4y 12. We can find this by multiplying both sides of 2x y 4 by 4, and adding the result to 4x 4y 12. 8x 4y 16 4x 4y 12 4 4x 4 x 1 Substitute x 1 into 2x y 4 and solve for y. () The point of intersection is therefore ( 1,2 ). 2 4 2 1 y 4 2 y 4 y 2 4 4

Linear Programming 77 22. x 0; y 2; 5x y 14; x 2y 10 The constraints of x 0 and y 2 indicate that we are restricted to the upper right quadrant, above the horizontal line y 2. The y-intercept of 5x y 14 can be found by substituting x 0. The y-intercept is ( 0,14 ). 5 0 y 14 0 y 14 y 14 The x-intercept of 5x y 14 can be found by substituting y 0. 14 5x 0 14 5x 14 x 2.8 The x-intercept is ( 2.8,0 ). We draw a line connecting these points. Testing the point ( 0,0 ), we have the statement 5( 0) 0 14 or 0 14. This is a true statement. The y-intercept of x 2y 10 can be found by substituting x 0. 10 0 2y 10 2y 10 y 5 The y-intercept is ( 0,5 ). The x-intercept of x 2y 10 can be found by substituting y 0. x 2 0 10 x 0 10 x 10 The x-intercept is ( 10,0 ). We draw a line connecting these points. Testing the point ( 0,0 ), we have the statement 0 20 10or 0 10. This is a true statement. Thus, we shade the part of the plane in the upper right quadrant which is on the down side of both the lines 5x y 14 and x 2y 10, which is also above the horizontal line y 2. Two of the corner points, ( 0,2 ) and ( 0,5 ), lie on the coordinate axes. The third corner point is the point of intersection between the lines 5x y 14 and y 2. We can find this by 12 substituting y 2 into 5x y 14 to solve for y. We have 5x 2 14 5x 12 x 2.4. The point of intersection is therefore ( 2.4,2 ). The fourth corner point is the point of intersection between the lines 5x y 14 and x 2y 10. We can find this by multiplying both sides of 5x y 14 by 2, and adding the result to x 2y 10. 10x 2y 28 x 2y 10 18 9x 18 x 2 9 8 Substitute x 2 into x 2y 10 and solve for y. We have 2 2y 10 2y 8 y 4. The point of intersection is therefore ( 2,4 ). 5 2 5 2

78 Chapter 4 23. x 4; y 0;5x 4y 60; x y 13 The constraints of x 4 and y 0 indicate that we are restricted to the upper right quadrant, to the right of the vertical line x 4. The y-intercept of 5x 4y 60 can be found by substituting x 0. 5 0 4y 60 0 4y 60 y 15 The y-intercept is ( 0,15 ). 60 The x-intercept of 5x 4y 60 can be found by substituting y 0. 5x 40 60 5x 0 60 5x 60 x 12 60 The x-intercept is ( 12,0 ). We draw a line connecting these points. Testing the point ( 0,0 ), we have the statement 5( 0) 4( 0) 60 or 0 60. This is a true statement. The y-intercept of x y 13 can be found by substituting x 0. 0 y 13 y 13 The y-intercept is ( 0,13 ). The x-intercept of x y 13 can be found by substituting y 0. x 0 13 x 13 The x-intercept is ( 13, 0 ). We draw a line connecting these points. Testing the point ( 0,0 ), we have the statement 0 0 13 or 0 13. This is a true statement. Thus, we shade the part of the plane in the upper right quadrant which is on the down side of both the lines 5x 4y 60 and x y 13, which is also to the right of the vertical line x 4. Two of the corner points, ( 4,0 ) and ( 12,0 ), lie on the coordinate axes. The third corner point is the point of intersection between the lines x y 13 and x 4. We can find this by substituting x 4 into x y 13 to solve to y. We have, 4 y 13 y 9. The point of intersection is therefore ( 4,9 ). The fourth corner point is the point of intersection between the lines 5x 4y 60 and x y 13. We can find this by multiplying both sides of x y 13 by 4, and adding the result to 5x 4y 60. 4x 4y 52 5x 4y 60 x 8 Substitute x 8 into x y 13 and solve for y. We have 8 y 13 y 5. The point of intersection is therefore ( 8,5 ). 4 5

Linear Programming 79 24. For Exercise 21: x 0; y 0;2x y 4;4x 4y 12 ( 4, 2 ): Since 4 0, the constraint x 0 is satisfied. Since 2 0, the constraint y 0 is satisfied. Since 2( 4) 2 8 2 10 4, > the condition 2x y 4 is not satisfied. There is no need to check the fourth constraint. Thus, ( 4, 2 ) is not feasible. ( 1, 3 ): Since 1 0, the constraint x 0 is satisfied. Since 3 0, the constraint y 0 is satisfied. Since 2() 1 3 2 3 5> 4, the condition 2x y 4 is not satisfied. There is no need to check the fourth constraint. Thus, ( 1,3 ) is not feasible. For Exercise 23: x 4; y 0;5x 4y 60; x y 13 ( 4, 2 ): Since 4 4, the constraint x 4 is satisfied. Since 2 0, the constraint y 0 is satisfied. Since 5( 4) 4( 2) 20 8 28 60, the condition 5x 4y 60 is satisfied. Since 4 2 6 13, the condition x y 13 is satisfied. Thus, ( 4, 2 ) is feasible. ( 1, 3 ): Since 1 < 4, the constraint x 4 is not satisfied. There is no need to check the other three constraints. Thus, ( 1,3 ) is not feasible. 25. Maximize P 3x 2y subject to x 3; y 2; x y 10;2x 3y 24 We need to first graph the feasible region while finding the corner points. The constraints of x 3 and y 2 indicate that we are restricted to the upper right quadrant, to the right of the vertical line x 3 and above the horizontal line y 2. The point of intersection between x 3 and y 2 is ( 3,2 ). The y-intercept of x y 10 can be found by substituting x 0. 0 y 10 y 10 The y-intercept is ( 0,10 ). The x-intercept of x y 10 can be found by substituting y 0. x 0 10 x 10 The x-intercept is ( 10,0 ). The y-intercept of 2x 3y 24 can be found by substituting x 0. 2 0 3y 24 0 3y 24 3y 24 y 8 The y-intercept is ( 0,8 ). 24 The x-intercept of 2x 3y 24 can be found by substituting y 0. 2x 30 24 2x 0 24 2x 24 x 12 The x-intercept is ( 12,0 ). Continued on next page 24 2 3

80 Chapter 4 25. continued Testing the point 0,0 in both x y 10 and 2x 3y 24, we have the following. 0 0 0 10 and 2( 0) 3( 0) 0 24 Since these are both true statements, we shade the down side of the line of both lines which is contained in the upper right quadrant to the right of the vertical line x 3 and above the horizontal line y 2. The point of intersection between x 3 and 2x 3y 24 can be found by substituting x 3 18 into 2x 3y 24. We have 2( 3) 3y 24 6 3y 24 3y 18 y 6. Thus, the point of intersection is ( 3,6 ). The point of intersection between 2 y and x y 10 can be found by substituting y 2 into x y 10. We have, x 2 10 x 8. 8,2. The final corner point is the point of intersection between x y 10 and 2x 3y 24. We can find this by multiplying both sides of x y 10 by 2, and adding the result to 2x 3y 24. 2x 2y 20 2x 3y 24 y 4 Substitute y 4 into x y 10 and solve for x. We have, x 4 10 x 6. The point of intersection is therefore ( 6,4 ). Thus, the point of intersection is 3 We wish to maximize P 3x 2 y. Corner Point Value of the Profit Formula: 3x 2y ( 3,6) ( 3,2) ( 8,2) 3( 3) 3( 3) 38 2( 6) 2( 2) 22 9 9 24 12 4 4 21 13 28* 6,4 36 24 18 8 26 The maximum value occurs at the corner point ( 8,2 ), where P is equal to 28.

Linear Programming 81 26. Maximize P 3x 2y subject to x 2; y 3;3x y 18;6x 4y 48 We need to first graph the feasible region while finding the corner points. The constraints of x 2 and y 3 indicate that we are restricted to the upper right quadrant, to the right of the vertical line x 2 and above the horizontal line y 3. The point of intersection between x 2 and y 3 is ( 2,3 ). The y-intercept of 3x y 18 can be found by substituting x 0. The y-intercept is ( 0,18 ). 3 0 y 18 0 y 18 y 18 The x-intercept of 3x y 18 can be found by substituting y 0. 18 3x 0 18 3x 18 x 6 The x-intercept is ( 6,0 ). The y-intercept of 6x 4y 48 can be found by substituting x 0. 6 0 4y 48 0 4y 48 4y 48 y 12 The y-intercept is ( 0,12 ). 48 The x-intercept of 6x 4y 48 can be found by substituting y 0. 6x 4 0 48 6x 0 48 6x 48 x 8 48 The x-intercept is ( 8,0 ). Testing the point ( 0,0 ) in both 3x y 18 and 6x 4y 48, 3( 0) 0 0 18 and 6( 0) 4( 0) 0 48 3 4 6 we have the following. Since these are both true statements, we shade the down side of the line of both lines which is contained in the upper right quadrant to the right of the vertical line x 2 and above the horizontal line y 3. The point of intersection between x 2 and 6x 4y 48 can be found by substituting x 2 36 into 6x 4y 48. We have, 6( 2) 4y 48 12 4y 48 4y 36 y 9. Thus, the point of intersection is ( 2,9 ). The point of intersection between 3 y and 3x y 18 can be found by substituting y 3 into 15 3x y 18. We have 3x 3 18 3x 15 x 5. Thus, the point of intersection is ( 5,3 ). The final corner point is the point of intersection between 3x y 18 and 6x 4y 48. We can find this by multiplying both sides of 3x y 18 by 4, and adding the result to 6x 4y 48. 12x 4y 72 6x 4y 48 24 6x 24 x 4 Substitute x 4 into 3x y 18 The point of intersection is therefore ( 4,6 ). Continued on next page 3 6 and solve for y. We have, 3 4 y 18 12 y 18 y 6. 4

82 Chapter 4 26. continued We wish to maximize P 3x 2 y. Corner Point Value of the Profit Formula: 3x 2y ( 2,3) ( 5,3) ( 4,6) 3( 2) 3( 5) 34 2( 3) 2( 3) 26 6 15 12 6 6 12 0 9* 0 2,9 32 29 6 18 12 The maximum value occurs at the corner point ( 5,3 ), where P is equal to 9. 27. Maximize P 5x 2y subject to x 2; y 4; x y 10 We need to first graph the feasible region while finding the corner points. The constraints of x 2 and y 4 indicate that we are restricted to the upper right quadrant, to the right of the vertical line x 2 and above the horizontal line y 4. The point of intersection between x 2 and y 4 is ( 2,4 ). The y-intercept of x y 10 can be found by substituting x 0. 0 y 10 y 10 The y-intercept is ( 0,10 ). The x-intercept of x y 10 can be found by substituting y 0. x 0 10 x 10 The x-intercept is ( 10,0 ). Testing the point 0,0 in x y 10, we have the following. 0 0 0 10 Since this is a true statement, we shade the down side of this line which is contained in the upper right quadrant to the right of the vertical line x 2 and above the horizontal line y 4. The point of intersection between x 2 and x y 10 can be found by substituting x 2 into x y 10. We have 2 y 10 y 8. 2,8. The point of intersection between y 4 and x y 10 can be found by substituting y 4 into x y 10. We have x 4 10 x 6. Continued on next page Thus, the point of intersection is Thus, the point of intersection is 6,4.

Linear Programming 83 27. continued We wish to maximize P 5x 2 y. Corner Point Value of the Profit Formula: 5x 2y ( 2,4) ( 2,8) ( 6,4) 5 2 2 4 10 8 18 5 2 2 8 10 16 26 5 6 2 4 30 8 38* The maximum value occurs at the corner point ( 6,4 ), where P is equal to 38. 28. (a) x 0; y 0;7x 4y 13 The constraints of x 0 and y 0 indicate that we are restricted to the upper right quadrant created by the x-axis and y-axis. The y-intercept of 7x 4y 13 can be found by substituting x 0. 13 7( 0) 4y 13 0 4y 13 4y 13 y 4 13 The y-intercept is ( 0, ). For graphing purposes, treat 13 as 1 3. 4 4 The x-intercept of 7x 4y 13 can be found by substituting y 0. 13 7x 4( 0) 13 7x 0 13 7x 13 x 7 13 The x-intercept is (,0 7 ). For graphing purposes, treat 13 as 6 1. 7 7 We draw a line connecting these points. Testing the point ( 0,0 ), we have the statement 7( 0) 4( 0) 13 or 0 13. This is a true statement, thus we shade the half-plane containing our test point, the down side of the line, which is contained in the upper right quadrant. 4 Continued on next page

84 Chapter 4 28. continued (b) According to the optimal production policy, we are looking for the maximum of P 21x 11 y, given our constraints. Corner Point Value of the Profit Formula: 21x 11y ( 0,0) 13 ( 0, 4 ) 13 (,0) 7 13 21 0 110 0 0 0 13 143 3 21 0 11 0 35 4 4 4 21 11 0 39 0 39* 13 The maximum value occurs at the corner point (,0. 7 ) 7 13 29. (a) The optimal corner point for Exercise 28 being (,0 7 ), the coordinates of Q (b) ( 2,0 ) is not a feasible point. 13 (c) The profit at ( 2,0 ) is greater than the profit at 7,0. 2,0. (d) Since 0 0 and 3 0 the constraints x 0 and y 0 are satisfied, respectively. Also, since 7( 0) 4( 3) 0 12 12 13, R ( 0,3) is feasible. The profit at R is the constraint 7x 4y 13 is satisfied. Thus, 21 0 113 33. This is less than the profit at Q. (e) Solving maximization problems involving linear constraints but where the variables are required to take on integer values cannot be solved by first solving the associated linear programming problem and rounding the answer to the nearest integers. This example shows the rounded value used to obtain an integer solution may not be feasible. Even if the rounded value is feasible it may not be optimal. "Integer programming" is unfortunately a much harder problem to solve than linear programming. For Exercises 30 41, part (e) (using a simplex algorithm program) will not be addressed in the solutions. 30. (a) Let x be the number of shirts and y be the number of vests. Cloth (600 yds) Minimums Profit Shirts, x items 3 100 $5 Vests, y items 2 30 $2 (b) Profit formula: P $5 x $2y x 100 and y 30 minimums ;3x 2y 600 cloth (c) Feasible region: Constraints: Continued on next page

Linear Programming 85 30. continued Corner points: One corner point is the point of intersection between x 100 and y 30, namely ( 100,30 ). Another is the point of intersection between x 100 and 3x 2y 600. Substituting x 100 into 3x 2y 600, we have the following. 3( 100) 2y 600 300 2y 600 2y 300 y 150 Thus, ( 100,150 ) is the second corner point. The third corner point is the point of intersection between y 30 and 3x 2y 600. Substituting y 30 into 3x 2y 600, we have the following. 3x 2 30 600 3x 60 600 3x 540 x 180 Thus, ( 180,30 ) is the third corner point. (d) We wish to maximize $5 x $2 y. Corner Point Value of the Profit Formula: $5 x $2y ( 100,30) ( 100,150) ( 180,30) $5 100 $2 30 $500 $60 $560 $5 100 $2 150 $500 $300 $800 $5 180 $2 30 $900 $60 $960* Optimal production policy: Make 180 shirts and 30 vests. With zero minimums, the feasible region looks like the following. We wish to maximize $5 x $2 y. Corner Point Value of the Profit Formula: $5 x $2y ( 0,0) ( 0,300) ( 200,0) $5 0 $20 $0 $0 $0 $5 0 $2 300 $0 $600 $600 $5 200 $2 0 $1000 $0 $1000* The production policy changes to make 200 shirts and no vests.

86 Chapter 4 31. (a) Let x be the number of oil changes and y be the number tune-ups. Time (8,000 min) Minimums Profit Oil changes, x 20 0 $15 Tune-ups, y 100 0 $65 (b) Profit formula: P $15 x $65y Constraints: x 0 and y 0 ( minimums );20x 100y 8,000 ( time) (c) Feasible region: Corner points: The corner points are intercepts on the axes. These are ( 0,0 ), ( 400,0 ). (d) We wish to maximize $15 x $65 y. Corner Point Value of the Profit Formula: $15 x $65y ( 0,0) ( 0,80) ( 400,0) $15 0 $65 0 $0 $0 $0 $15 0 $65 80 $0 $5200 $5200 $15 400 $65 0 $6000 $0 $6000* Optimal production policy: Schedule 400 oil changes and no tune-ups. With non-zero minimums, the constraints are as follows. x 50 and y 20 ( minimums ); 20x 100y 8, 000 ( time) The feasible region looks like the following. 0,80, and Corner points: One corner point is the point of intersection between x 50 and y 20, namely 50, 20. Another is the point of intersection between x 50 and 20x 100y 8000. Substituting x 50 into 20x 100y 8000, we have the following. 20( 50) 100y 8000 1000 100y 8000 100y 7000 y 70 Thus, ( 50,70) is the second corner point. The third corner point is the point of intersection between y 20 and 20x 100y 8000. Substituting y 20 into 20x 100y 8000, we have the following. 20x 100 20 8000 20x 2000 8000 20x 6000 x 300 Thus, ( 300,20 ) is the third corner point. Continued on next page

Linear Programming 87 31. continued We wish to maximize $15 x $65 y. Corner Point Value of the Profit Formula: $15 x $65y ( 50, 20) ( 50,70) ( 300,20) $15 50 $65 20 $750 $1300 $2050 $15 50 $65 70 $750 $4550 $5300 $15 300 $65 20 $4500 $1300 $5800* Optimal production policy: Schedule 300 oil changes and 20 tune-ups. 32. (a) Let x be the number of minutes processing mail orders and y be the number of minutes processing voice-mail orders. Time (90 min) Minimums Profit Mail order, x minutes 10 0 $30 Voice-mail order, y minutes 15 0 $40 (b) Profit formula: P $30 x $40y Constraints: x 0 and y 0 ( minimums );10x 15y 90 ( time) (c) Feasible region: Corner points: The corner points are intercepts on the axes. These are ( 0,0 ), ( 9,0 ). (d) We wish to maximize $30 x $40 y. Corner Point Value of the Profit Formula: $30 x $40y ( 0,0) ( 0,6) ( 9,0) $30 0 $40 0 $0 $0 $0 $30 0 $40 6 $0 $240 $240 $30 9 $40 0 $270 $0 $270* Optimal production policy: Process 9 mail orders and no voice-mail orders. Continued on next page 0,6,and

88 Chapter 4 32. continued With non-zero minimums, the constraints are x y x y The feasible region looks like the following. 3 and 2 minimums ;10 15 90 time. Corner points: One corner point is the point of intersection between x 3 and y 2, namely 3, 2. Another is the point of intersection between x 3 and 10x 15y 90. Substituting x 3 into 10x 15y 90 we have the following. 10( 3) 15y 90 30 15y 90 15y 60 y 4 Thus, ( 3, 4 ) is the second corner point. The third corner point is the point of intersection between y 2 and 10x 15y 90. Substituting y 2 into 10x 15y 90 we have the following. 10x 15 2 90 10x 30 90 10x 60 x 6 Thus, ( 6,2 ) is the third corner point. We wish to maximize $30 x $40 y. Corner Point Value of the Profit Formula: $30 x $40y ( 3,2) ( 3,4) ( 6,2) $30 3 $40 2 $90 $80 $170 $30 3 $40 4 $90 $160 $250 $30 6 $40 2 $180 $80 $260* Optimal production policy: Process 6 mail orders and 2 voice-mail orders. 33. (a) Let x be the number of routine visits and y be the number of comprehensive visits. Doctor Time (1800 min) Minimums Profit Routine, x visits 5 0 $30 Comprehensive, y visits 25 0 $50 (b) Profit formula: P $30 x $50y Constraints: x 0 and y 0 ( minimums );5x 25y 1800 ( time) (c) Feasible region: Continued on next page

Linear Programming 89 33. (c) continued Corner points: The corner points are intercepts on the axes. These are ( 0,0 ), ( 360,0 ). (d) We wish to maximize $30 x $50 y. Corner Point Value of the Profit Formula: $30 x $50y ( 0,0) ( 0,72) ( 360,0) $30 0 $50 0 $0 $0 $0 $30 0 $50 72 $0 $3600 $3600 $30 360 $50 0 $10,800 $0 $10,800* Optimal production policy: Schedule 360 routine visits and no comprehensive visits. With non-zero minimums, the constraints are as follows. x 20 and y 30 ( minimums );5x 25y 1800 ( time) The feasible region looks like the following. 0,72,and Corner points: One corner point is the point of intersection between x 20 and y 30, namely 20,30. Another is the point of intersection between x 20 and 5x 25y 1800. Substituting x 20 into 5x 25y 1800, we have the following. 5( 20) 25y 1800 100 25y 1800 25y 1700 y 68 Thus, ( 20,68 ) is the second corner point. The third corner point is the point of intersection between y 30 and 5x 25y 1800. Substituting y 30 into 5x 25y 1800, we have the following. 5x 25 30 1800 5x 750 1800 5x 1050 x 210 Thus, ( 210,30 ) is the third corner point. We wish to maximize $30 x $50 y. Corner Point Value of the Profit Formula: $30 x $50y ( 20,30) ( 20,68) ( 210, 30) $30 20 $50 30 $600 $1500 $2100 $30 20 $50 68 $600 $3400 $4000 $30 210 $50 30 $6300 $1500 $7800* Optimal production policy: Schedule 210 routine visits and 30 comprehensive visits. 34. (a) Let x be the number of loaves of multigrain bread and y be the number of loaves of herb bread. Breads (600 loaves) Minimums Profit Multigrain, x loaves 1 100 $8 Herb, y loaves 1 200 $10 (b) Profit formula: P $8 x $10y x 100 and y 200 minimums ; x y 600 breads Constraints: Continued on next page

90 Chapter 4 34. continued (c) Feasible region: Corner points: One corner point is the point of intersection between x 100 and y 200, namely ( 100,200 ). Another is the point of intersection between x 100 and x y 600. Substituting x 100 into x y 600, we have the following. 100 y 600 y 500 Thus, ( 100,500 ) is the second corner point. The third corner point is the point of intersection between y 200 and x y 600. Substituting y 200 into x y 600, we have the following. Thus, ( 400,200 ) is the third corner point. (d) We wish to maximize $8 x $10 y. x 200 600 x 400 Corner Point Value of the Profit Formula: $8 x $10y ( 100, 200) ( 100, 500) ( 400,200) $8 100 $10 200 $800 $2000 $2800 $8 100 $10 500 $800 $5000 $5800* $8 400 $10 200 $3200 $2000 $5200 Optimal production policy: Make 100 multigrain and 500 herb loaves. With zero minimums, the feasible region looks like the following. We wish to maximize $8 x $10 y. Corner Point Value of the Profit Formula: $8 x $10y ( 0,0) ( 0,600) ( 600,0) $8 0 $10 0 $0 $0 $0 $8 0 $10 600 $0 $6000 $6000* $8 600 $10 0 $4800 $0 $4800 Optimal production policy: Make no multigrain and 600 herb loaves.

Linear Programming 91 35. (a) Let x be the number of hours spent on math courses and y be the number of hours spent on other courses. Time (48 hr) Minimums Value Points Math,x courses 12 0 2 Other,y courses 8 0 1 (b) Value Point formula: V 2x y x 0 and y 0 minimums ;12x 8y 48 time (c) Feasible region: Constraints: Corner points: The corner points are intercepts on the axes. These are ( 0,0 ), ( 4,0 ). (d) We wish to maximize 2 x y. Corner Point Value of the Value Point Formula: 2x y ( 0,0) ( 0,6) ( 4,0) 0,6,and 2 0 0 0 0 0 2 0 6 0 6 6 2 4 0 8 0 8* Optimal production policy: Take four math courses and no other courses. x 2 and y 2 minimums ;12x 8y 48 time. With non-zero minimums, the constraints are The feasible region looks like the following. Corner points: One corner point is the point of intersection between x 2 and y 2, namely 2,2. Another is the point of intersection between x 2 and 12x 8y 48. Substituting x 2 into 12x 8y 48, we have the following. Thus, ( 2,3 ) is the second corner point. Continued on next page 12 2 8y 48 24 8y 48 8y 24 y 3

92 Chapter 4 35. continued The third corner point is the point of intersection between y 2 and 12x 8y 48. Substituting y 2 into 12x 8y 48, we have the following. 8 Thus, 3 32 8 2 12x 8 2 48 12x 16 48 12x 32 x 2, 2 is the third corner point. We wish to maximize 2 x y. Corner Point Value of the Profit Formula: 2x y ( 2,2) ( 2,3) 8 (, 2) 3 2 2 2 4 2 6 2 2 3 4 3 7 2 2 5 2 7 * 8 1 1 3 3 3 12 3 3 2 Optimal production policy: Take 2 math courses and 2 other courses. However, one cannot 3 take a fractional part of a course. So given the constraint on study time, the student should take two math courses and 2 other courses. 36. (a) Let x be the number of Hot sites maintained and y be the number of Cool sites maintained. Layout (12 hr) Content (16 hr) Minimums Profit Hot, x sites 1.5 1 0 $50 Cool, y sites 1 2 0 $250 (b) Profit formula: P $50 x $250y x 0 and y 0 minimums ;1.5x y 12 layout ; x 2y 16 content (c) Feasible region: Constraints: Corner points: Three of the corner points are intercepts on the axes. These are ( 0,0 ), ( 0,8 ), and ( 8,0 ). The final corner point is the point of intersection between 1.5x y 12 and x 2y 16. We can find this by multiplying both sides of 1.5x y 12 by 2, and adding the result to x 2y 16. 3x 2y 24 x 2y 16 8 2x 8 x 4 2 Substitute x 4 into x 2y 16 and solve for y. We have 4 2y 16 2y 12 y 6. The point of intersection is therefore 4,6. Continued on next page

Linear Programming 93 36. continued (d) We wish to maximize $50 x $250 y. Corner Point Value of the Profit Formula: $50 x $250y ( 0,0) ( 0,8) ( 8,0) ( 4,6) $50 0 $250 0 $0 $0 $0 $50 0 $250 8 $0 $2000 $2000* $50 8 $250 0 $400 $0 $400 $50 4 $250 6 $200 $1500 $1700 Optimal production policy: Maintain no hot and 8 cool sites. With non-zero minimums, the constraints are as follows. x 2 and y 3 minimums ;1.5x y 12 layout ; x 2y 16 content The feasible region looks like the following. Corner points: One corner point is the point of intersection between x 2 and y 3, namely 2,3. Another is the point of intersection between x 2 and x 2y 16. Substituting x 2 into x 2y 16, we have the following. 2 2y 16 2y 14 y 7 Thus, ( 2,7 ) is the second corner point. The third corner point is the point of intersection between y 3 and 1.5x y 12. Substituting y 3 into 1.5x y 12, we have the following. 1.5x 3 12 1.5x 9 x 6 Thus, ( 6,3 ) is the third corner point. The fourth corner point is the point of intersection between 1.5x y 12 and x 2y 16. In part (a) this was found to be ( 4,6 ). We wish to maximize $50 x $250 y. Corner Point Value of the Profit Formula: $50 x $250y ( 2,3) ( 2,7) ( 6,3) ( 4,6) $50 2 $250 3 $100 $750 $850 $50 2 $250 7 $100 $1750 $1850* $50 6 $250 3 $300 $750 $1050 $50 4 $250 6 $200 $1500 $1700 Optimal production policy: Maintain 2 hot and 7 cool sites.

94 Chapter 4 37. (a) Let x be the number of Grade A batches and y be the number of Grade B batches. Scrap cloth (100 lb) Scrap Paper (120 lb) Minimums Profit Grade A, x batches 25 10 0 $500 Grade B, y batches 10 20 0 $250 (b) Profit formula: P $500 x $250y x 0 and y 0 minimums ;25x 10y 100 cloth ;10x 20y 120 paper (c) Feasible region: Constraints: Corner points: Three of the corner points are intercepts on the axes. These are ( 0,0 ), ( 0,6 ),and ( 4,0 ). The final corner point is the point of intersection between 25x 10y 100 and 10x 20y 120. We can find this by multiplying both sides of 25x 10y 100 by 2, and adding the result to 10x 20y 120. 50x 20y 200 10x 20y 120 80 40x 80 x 2 40 Substitute x 2 into 25x 10y 100 and solve for y. We have the following. The point of intersection is therefore ( 2,5 ). 25 2 10y 100 50 10y 100 10y 50 y 5 (d) We wish to maximize $500 x $250 y. Corner Point Value of the Profit Formula: $500 x $250y ( 0,0) ( 0,6) ( 4,0) ( 2,5) $500 0 $250 0 $0 $0 $0 $500 0 $250 6 $0 $1500 $1500 $500 4 $250 0 $2000 $0 $2000 $500 2 $250 5 $1000 $1250 $2250* Optimal production policy: Make 2 grade A and 5 grade B batches. With non-zero minimums x 1 and y 1, there will be no change because the optimal production policy already obeys these non-zero minimums.

Linear Programming 95 38. (a) Let x be the number of modest houses and y be the number of deluxe houses. Space (100 acres) Money in thousands ($2600) Minimums Profit in thousands Modest, x houses 1 $20 0 $25 Deluxe, y houses 1 $40 0 $60 (b) Profit formula: P $25 x $60y Constraints: x 0 and y 0 ( minimums) x y 100 ( space) 20x 40y 2600 available funds (c) Feasible region: Corner points: Three of the corner points are intercepts on the axes. These are ( 0,0 ), ( 0,65 ),and ( 130, 0 ). The final corner point is the point of intersection between x y 100 and 20x 40y 2600. We can find this by multiplying both sides of x y 100 by 20, and add the result to 20x 40y 2600. 20x 20y 2000 20x 40y 2600 20y 600 y 30 Substitute y 30 into x y 100 and solve for y. We have x 30 100 x 70. The point of intersection is therefore ( 70,30 ). (d) We wish to maximize $25 x $60 y. Corner Point Value of the Profit Formula: $25 x $60y (in thousands) ( 0,0) ( 0,65) ( 100,0) ( 70,30) $25 0 $60 0 $0 $0 $0 $25 0 $60 65 $0 $3900 $3900* $25 100 $60 0 $2500 $0 $2500 $25 70 $60 30 $1750 $1800 $3550 Optimal production policy: Build no modest and 65 deluxe houses. Continued on next page

96 Chapter 4 38. continued With non-zero minimums, the constraints are as follows. x 20 and y 20 minimums ; x y 100 space ;20x 40y 2600 available funds The feasible region looks like the following. Corner points: One corner point is the point of intersection between x 20 and y 20, namely 20, 20. Another is the point of intersection between y 20 and x y 100. Substituting y 20 into x y 100, we have x 20 100 x 80. 80, 20 is the second corner point. The third corner point is the point of intersection between x 20 and 20x 40y 2600. Substituting x 20 into 20x 40y 2600, we have the following. Thus, 20( 20) 40y 2600 400 40y 2600 40y 2200 y 55 Thus, ( 20,55 ) is the third corner point. The fourth corner point is the point of intersection between x y 100 and 20x 40y 2600. In part (a) this was found to be ( 70,30 ). We wish to maximize $25 x $60 y. Corner Point Value of the Profit Formula: $25 x $60y(in thousands) ( 20, 20) ( 20,55) ( 80, 20) ( 70,30) $25 20 $60 20 $500 $1200 $1700 $25 20 $60 55 $500 $3300 $3800* $25 80 $60 20 $2000 $1200 $3200 $25 70 $60 30 $1750 $1800 $3550 Optimal production policy: Make 20 modest and 55 deluxe houses.

Linear Programming 97 39. (a) Let x be the number of cartons of regular soda and y be the number of cartons of diet soda. Cartons (5000) Money ($5400) Minimums Profit Regular, x cartons 1 $1.00 600 $0.10 Diet, y cartons 1 $1.20 1000 $0.11 (b) Profit formula: P $0.10 x $0.11y Constraints: x 600 and y 1000 ( minimums) x y 5000 cartons ;1.00x 1.20y 5400 money (c) Feasible region: Corner points: One corner point is the point of intersection between x 600 and y 1000, namely ( 600,1000 ). Another is the point of intersection between y 1000 and x y 5000. Substituting y 1000 into x y 5000, we have x 1000 5000 x 4000. Thus, ( 4000,1000 ) is the second corner point. The third corner point is the point of intersection between x 600 and 1.00x 1.20y 5400. Substituting x 600 into 1.00x 1.20y 5400, we have the following. 1.00( 600) 1.20y 5400 600 1.20y 5400 1.20y 4800 y 4000 Thus, ( 600,4000 ) is the third corner point. The fourth corner point is the point of intersection between x y 5000 and 1.00x 1.20y 5400. We can find this by multiplying both sides of x y 5000 by 1, and adding the result to 1.00x 1.20y 5400. x y 5000 1.00x 1.20y 5400 0.20y 400 y 2000 Substitute y 2000 into x y 5000 and solve for y. We have x 2000 5000 or x 3000. The point of intersection is therefore ( 3000,2000 ). (d) We wish to maximize $0.10 x $0.11 y. Corner Point Value of the Profit Formula: $0.10 x $0.11y ( 600,1000) ( 4000,1000) ( 600, 4000) ( 3000,2000) $0.10 600 $0.11 1000 $60 $110 $170 $0.10 4000 $0.11 1000 $400 $110 $510 $0.10 600 $0.11 4000 $60 $440 $500 $0.10 3000 $0.11 2000 $300 $220 $520* Optimal production policy: Make 3000 cartons of regular and 2000 cartons of diet. With zero minimums there is no change in the optimal production policy because the corresponding corner point does not touch either line from a minimum constraint.

98 Chapter 4 40. (a) Let x be the number of pheasants and y be the number of partridges. Bird count (100) Cost ($2400) Minimums Profit Pheasants, x 1 $20 0 $14 Partridges, y 1 $30 0 $16 (b) Profit formula: P $14 x $16y x 0 and y 0 minimums ; x y 100 bird count ;20x 30y 2400 money (c) Feasible region: Constraints: Corner points: Three of the corner points are intercepts on the axes. These are ( 0,0 ), ( 0,80 ), and ( 100, 0 ). The final corner point is the point of intersection between x y 100 and 20x 30y 2400. We can find this by multiplying both sides of x y 100 by 20, and adding the result to 20x 30y 2400. 20x 20y 2000 20x 30y 2400 10y 400 y 40 Substitute y 40 into x y 100 and solve for y. We have x 40 100 x 60. The point of intersection is therefore ( 60, 40 ). (d) We wish to maximize $14 x $16 y. Corner Point Value of the Profit Formula: $14 x $16y ( 0,0) ( 0,80) ( 100,0) ( 60, 40) $14 0 $16 0 $0 $0 $0 $14 0 $16 80 $0 $1280 $1280 $14 100 $16 0 $1400 $0 $1400 $14 60 $16 40 $840 $640 $1480* Optimal production policy: Raise 60 pheasants and 40 partridges. With nonzero minimum of x 20 and y 10, there is no change because the optimal production policy obeys these minimums.

Linear Programming 99 41. (a) Let x be the number of desk lamps and y be the number of floor lamps. Labor (1200 hr) Money ($4200) Minimums Profit Desk, x lamps 0.8 $4 0 $2.65 Floor, y lamps 1.0 $3 0 $4.67 (b) Profit formula: P $2.65 x $4.67y x 0 and y 0 minimums 0.8x 1.0y 1200 ( labor) 4x 3y 4200 money Constraints: (c) Feasible region: Corner points: Three of the corner points are intercepts on the axes. These are ( 0,0 ), ( 0,1200 ), and ( 1050,0 ). The final corner point is the point of intersection between 0.8x 1.0y 1200 and 4x 3y 4200. We can find this by multiplying both sides of 0.8x 1.0y 1200 by 3, and adding the result to 4x 3y 4200. 2.4x 3y 3600 4x 3y 4200 1.6x 600 x 375 Substitute x 375 into 0.8x 1.0y 1200 and solve for y. We have the following. The point of intersection is therefore ( 375,900 ). (d) We wish to maximize $2.65 x $4.67 y. 0.8 375 y 1200 300 y 1200 y 900 Corner Point Value of the Profit Formula: $2.65 x $4.67y ( 0,0) ( 0,1200) ( 1050, 0) ( 375, 900) $2.65 0 $4.67 0 $0.00 $0.00 $0.00 $2.65 0 $4.67 1200 $0.00 $5604.00 $5604.00* $2.65 1050 $4.67 0 $2782.50 $0.00 $2782.50 $2.65 375 $4.67 900 $993.75 $4203.00 $5196.75 Optimal production policy: Make no desk lamps and 1200 floor lamps. Continued on next page

100 Chapter 4 41. continued With non-zero minimums, the constraints are as follows. x 150 and y 200 minimums ;0.8x 1.0y 1200 labor ;4x 3y 4200 money The feasible region looks like the following. Corner points: One corner point is the point of intersection between x 150 and y 200, namely ( 150, 200 ). Another is the point of intersection between y 200 and 4x 3 y 4200. Substituting y 200 into 4x 3 y 4200, we have the following. 4x 3( 200) 4200 4x 600 4200 4x 3600 x 900 Thus, ( 900, 200 ) is the second corner point. The third corner point is the point of intersection between x 150 and 0.8x 1.0y 1200. Substituting x 150 into 0.8x 1.0y 1200, we have the following. 0.8 150 y 1200 120 y 1200 y 1080 Thus, ( 150,1080 ) is the third corner point. The fourth corner point is the point of intersection x y and 4x 3y 4200. In part (a) this was found to be ( 375,900 ). between 0.8 1.0 1200 We wish to maximize $2.65 x $4.67 y. Corner Point Value of the Profit Formula: $2.65 x $4.67y (in thousands) ( 150,200) ( 150,1080) ( 375,900) ( 900, 200) $2.65 150 $4.67 200 $397.50 $934.00 $1331.50 $2.65 150 $4.67 1080 $397.50 $5043.60 $5441.10* $2.65 375 $4.67 900 $993.75 $4203.00 $5196.75 $2.65 900 $4.67 200 $2385.00 $934.00 $3319.00 Optimal production policy: Make 150 desk lamps and 1080 floor lamps.

Linear Programming 101 For Exercises 42 45, part (c) (using a simplex algorithm program) will not be addressed in the solutions. 42. (a) Let x be the number of toy A, y be the number of toy B, and z be the number of toy C. Shaper (50) Smoother (40) Painter (60) Minimums Profit Toy A, x 1 2 1 0 $4 Toy B, y 2 1 3 0 $5 Toy C, z 3 2 1 0 $9 (b) Profit formula: P $4 x $5 y $9z Constraints: x 0, y 0, and z 0 ( minimums) x 2y 3z 50 ( shaper) 2x y 2z 40( smoother) x 3y z 60( painter) (c) Optimal product policy: Make 5 toy A, no toy B, 15 toy C for a profit of $155. 43. (a) Let x be the number of chairs, y be the number of tables, and z be the number of beds. Chis (80 hr) Sue (200 hr) Juan (200) Minimums Profit Chairs, x 1 3 2 0 $100 Tables, y 3 5 4 0 $250 Beds, z 5 4 8 0 $350 (b) Profit formula: P $100 x $250 y $350z Constraints: x 0, y 0, and z 0 ( minimums) x 3y 5z 80 Chris ;3x 5y 4z 200 Sue ;2x 4y 8z 200 Juan (c) Optimal product policy: Make 50 chairs, 10 tables, and no beds each month for a profit of 7500 in one month. 44. (a) Let x be the number of boxes of Special Mix, y be the number of boxes of Regular Mix, and z be the number of boxes of Purist Mix. Chocolate (1000 lb) Nuts (200 lb) Fruit (100 lb) Minimums Profit Special, x boxes 3 1 1 0 $10 Regular, y boxes 4 0.5 0 0 $6 Purist, z boxes 5 0 0 0 $4 (b) Profit formula: P $10 x $6 y $4z Constraints: x 0, y 0, and z 0 ( minimums) 3x 4y 5z 1000 ( chocolate) x 0.5y 0z 200 ( nuts) x 0y 0z 100( fruit) (c) Optimal product policy: Make 100 boxes of Special, 175 boxes of Regular, and 0 boxes of Purist for a profit of $2050.

102 Chapter 4 45. (a) Let w be the number of pounds of Excellent coffee, x be the number of pounds of Southern coffee, y be the number of pounds of World coffee, and z be the number of pounds of Special coffee. African (17,600 oz) Brazilian (21,120 oz) Columbian (12,320 oz) Minimums Profit Excellent, w pounds 0 0 16 0 $1.80 Southern, x pounds 0 12 4 0 $1.40 World, y pounds 6 8 2 0 $1.20 Special, z pounds 10 6 0 0 $1.00 (b) Profit formula: P $1.80 w $1.40 x $1.20 y $1.00z Constraints: w 0, x 0, y 0, and z 0 ( minimums) 0w 0x 6y 10z 17,600 ( African) 0w 12x 8y 6z 12,120 ( Brazilian) 16w 4x 2y 6z 12,320( Columbian ) (c) Optimal product policy: Make 470 pounds of Excellent, none of Southern, 2400 pounds of World, and 320 pounds for a profit of $4046. 46. Minimize C 4x 7 y subject to x 3; y 2; x y 10;2x 3y 24 The corner points are ( 3,6 ),( 3,2 ),( 8,2 ), and ( 6,4 ). We wish to minimize C 4x 7 y. Corner Point Value of the cost: 4x 7 y ( 3,6) ( 3,2) ( 8,2) 4( 3) 4( 3) 48 7( 6) 7( 2) 72 12 12 32 42 14 14 54 26* 46 6,4 46 74 24 28 52 The minimum value occurs at the corner point ( 3,2 ), where C is equal to 26.

Linear Programming 103 47. Minimize C 3x 11y subject to x 2; y 3;3x y 18;6x 4y 48 The corner points are ( 2,3 ),( 5,3 ),( 4,6 ), and ( 2,9 ). We wish to minimize C 3x 11 y. Corner Point Value of the Profit Formula: 3x 11y ( 2,3) ( 5,3) ( 4,6) 3( 2) 3( 5) 34 11( 3) 11( 3) 116 6 15 12 33 33 66 39* 48 78 2,9 32 119 6 99 105 The minimum value occurs at the corner point ( 2,3 ), where C is equal to 39. 48. The only feasible point with integer coordinates for the collection of inequalities x 0, y 0 and x y 0.5 0,0. is 49. (a) Let x be the number of business calls and y be the charity calls. Time (240 min) Minimums Profit Business, x calls 4 0 $0.50 Charity, y calls 6 0 $0.40 (b) Profit formula: P $0.50 x $0.40y x 0 and y 0 minimums ; 4x 6y 240 time (c) Feasible region: Constraints: Corner points: The corner points are intercepts on the axes. These are ( 0,0 ), ( 60,0 ). Continued on next page 0,40,and

104 Chapter 4 49. continued (d) We wish to maximize $0.50 x $0.40 y. Corner Point Value of the Profit Formula: $0.50 x $0.40y ( 0,0) ( 0,40) $0.50( 0 ) $0.50( 0 ) $0.40( 0 ) $0.40( 40 ) $0.00 $0.00 $0.00 $16.00 $0.00 $16.00 60,0 $0.50 60 $0.40 0 $30.00 $0.00 $30.00* Optimal production policy: Make 60 business and no charity calls. With non-zero minimums, the constraints are as follows. x 12 and y 10 minimums ; 4x 6y 240 time The feasible region looks like the following. Corner points: One corner point is the point of intersection between x 12 and y 10, namely 12,10. Another is the point of intersection between x 12 and 4x 6y 240. Substituting x 12 into 4x 6y 240, we have the following. 4( 12) 6y 240 48 6y 240 6y 192 y 32 Thus, ( 12,32) is the second corner point. The third corner point is the point of intersection between y 10 and 4x 6y 240. Substituting y 10 into 4x 6y 240 we have the following. 4x 6 10 240 4x 60 240 4x 180 x 45 Thus, ( 45,10 ) is the third corner point. We wish to maximize $0.50 x $0.40 y. Corner Point Value of the Profit Formula: $0.50 x $0.40y ( 12,10) ( 12,32) $0.50( 12 ) $0.50( 12 ) $0.40( 10 ) $0.40( 32 ) $6.00 $6.00 $4.00 $12.80 $10.00 $18.80 45,10 $0.50 45 $0.40 10 $22.50 $4.00 $26.50* Optimal production policy: Make 45 business and 10 charity calls.

Linear Programming 105 50. (a) Let x be the number of gallons of premium and y be the number of gallons of regular. High Octane Low Octane Minimums Profit (500 gal.) (600 gal.) Premium, x gallons 0.5 0.5 0 $0.40 Regular, y gallons 0.25 0.75 0 $0.30 (b) Profit formula: P $0.40 x $0.30y Constraints: x 0 and y 0 ( minimums) 0.5x 0.25y 500 high octane 0.5x 0.75y 600 low octane (c) Feasible region: Corner points: Three of the corner points are intercepts on the axes. These are ( 0,0 ), ( 0,800 ), and ( 1000,0 ). The final corner point is the point of intersection between 0.5x 0.25y 500 and 0.5x 0.75y 600. We can find this by multiplying both sides of 0.5x 0.25y 500 by 1, and adding the result to 0.5x 0.75y 600. 0.5x 0.25y 500 0.5x 0.75y 600 0.50y 100 y 200 Substitute y 200 into 0.5x 0.25y 500 and solve for x. We have the following. The point of intersection is therefore ( 900, 200 ). 0.5x 0.25 200 500 0.5x 50 500 0.5x 450 x 900 (d) We wish to maximize $0.40 x $0.30 y. Corner Point Value of the Profit Formula: $0.40 x $0.30y ( 0,0) ( 0,800) ( 1000,0) $0.40( 0 ) $0.40( 0 ) $0.40( 1000 ) $0.30( 0 ) $0.30( 800 ) $0.30( 0 ) $0 $0 $400 $0 $240 $0 $0 $240 $400 900,200 $0.40 900 $0.30 200 $360 $60 $420* Optimal production policy: Make 900 gallons of premium and 200 gallons of regular. With non-zero minimums x 100 and y 100, there will be no change because the optimal production policy already obeys these non-zero minimums.

106 Chapter 4 51. (a) Let x be the number of bikes and y be the number of wagons. Machine (12 hr) Paint(16 hr.) Minimums Profit Bikes, x 2 4 0 $12 Wagons, y 3 2 0 $10 (b) Profit formula: P $12 x $10y x 0 and y 0 minimums ; 2x 3y 12 machine ; 4x 2y 16 paint (c) Feasible region: Constraints: Corner points: Three of the corner points are intercepts on the axes. These are ( 0,0 ), ( 0,4 ), and ( 4,0 ). The final corner point is the point of intersection between 2x 3y 12 and 4x 2y 16. We can find this by multiplying both sides of 2x 3y 12 by 2, and adding the result to 4x 2y 16. 4x 6y 24 4x 2y 16 4y 8 y 2 Substitute y 2 into 4x 2y 16 and solve for x. We have the following. The point of intersection is therefore ( 3,2 ). 4x 22 16 4x 4 16 4x 12 x 3 (d) We wish to maximize $12 x $10 y. Corner Point Value of the Profit Formula: $12 x $10y ( 0,0) ( 0,4) ( 4,0) $12( 0 ) $12( 0 ) $12( 4 ) $10( 0 ) $10( 4 ) $10( 0 ) $0 $0 $48 $0 $40 $0 $0 $40 $48 3,2 $12 3 $10 2 $36 $20 $56* Optimal production policy: Make 3 bikes and 2 wagons. With non-zero minimums x 2 and y 2, there will be no change because the optimal production policy already obeys these non-zero minimums.

Linear Programming 107 52. (a) (b) Continued on next page

108 Chapter 4 52. continued (c) (d) Answers will vary. For (a) we might have two companies that supply a total of 11 units of lemon ices, and two stores that require 11 units of lemon ices. For (b) we have two coal mines which can supply a total of 9 units of coal, and three coal burning power plants which require 9 units of coal. For (c) we have three orchards which can supply 10 units of apples and two supermarket chains which require 10 units of apples. (e) For (a) the cost is 2( 3) 6( 5) 3( 2) 6 30 6 42. For (b) the cost is 1( 11) 1( 6) 2( 9) 5( 2) 11 6 18 10 45. For (c) the cost is 1( 17) 1( 1) 4( 9) 4( 6) 17 1 36 24 78.

Linear Programming 109 53. (a) (b) The cost for this solution is 1() 1 1() 3 2( 4) 1( 5) 1 3 8 5 17. (c) The indicator value for cell ( I,2 ) is 7 1 3 4 5. The indicator value for cell ( I,3 ) is 2 1 3 5 1.

110 Chapter 4 54. (a) (b) The indicator value for cell ( II,1 ) is 3 13 1 4 13. The indicator value for cell ( I,3 ) is 2 1 4 1 4. Continued on next page

Linear Programming 111 54. continued (c) The cost for this solution is 1( 13) 1() 1 2( 4) 1() 1 13 1 8 1 23. We get a cheaper solution, as shown below, by shipping using cell ( II,1 ). The new solution has cost 2() 1 1( 3) 1( 4) 1() 1 2 3 4 1 10. This saves 13 units of cost over the previous solution, as would be expected because the indicator value is 13, and we shipped one more unit via cell ( II,1 ). 55. (a) The graph is a tree because it is connected and has no circuit. (b) If we add the edge joining Vertex I to Vertex 2 we get the circuit 2, I, 1, II, 2. If we add the edge from Vertex I to Vertex 3 we get the circuit 3, I, 1, II, 3. (c) For the circuit 2, I, 1, II, 2 it corresponds to the following circuit of cells. ( I,2 ),( I,1 ),( II,1 ),( II,2 ),( I,2 ) For the circuit 3, I, 1, II, 3 it corresponds to the following circuit of cells. ( I,3 ),( I,1 ),( II,1 ),( II,3 ),( I,3 )

112 Chapter 4 56. (a) In row I the minimum cost is 1 cell I,2 and for row II the minimum cost is 2 cell II,1. Thus, the row with the minimum value is row 1. Since this minimum cost occurs in column 2, we try shipping as much as we can via this cell. This means we can cell I,2. So we cross out column 2, and reduce the rim condition for row I ship 2 units in to a 1. We now have a 2 2 tableau. We find the minimum cost for row I is 3 (there is a tie for two different cells) and the minimum cost for row II is 2, in cell ( II,1 ). So the row with minimum value is row II; so in this row we ship as much as we can via cell (II,1). We can ship 4 via this cell, so now we cross out column 1. The rim value for this row is reduced from 5 to 1. We now have a tableau with one column (column 3) and two rows; row I has a minimum of 3 and row II a minimum of 4; so we ship first via the minimum row cost (row I whose cost is 3) in cell ( I,3 ) (1 unit) and cross out row I, and finally, in cell ( II,3 ), 1 unit (cost 4). II,1 with a 4; and So the cells with circled numbers are: ( I,2 ) with a 2; ( I,3 ) with a 1; ( II,3 ) with a 1. (b) The cost of the solution above is 2( 1) 1( 3) 4( 2) 1( 4) 2 3 8 4 17. Continued on next page

Linear Programming 113 56. continued (c) The cost of this solution is 3 13 1 2 2 5 2 4 39 2 10 8 59, which is 42 units more expensive than the solution we found in (b). It is often worth using a more complex method to find a better initial solution to a problem than using a simple method, which finds a poor initial solution, as the Northwest Corner Rule does for this problem.

114 Chapter 4 57. (a) (i) (ii) Since the rim values are the same, the end result will be the same (relative to the new table) Continued on next page

Linear Programming 115 57. (a) continued (iii) (b) For (i) the indicator value for each non-circled cell is calculated as follows. cell ( II,1 ): 4 6 3 1 0 cell ( II, 2 ): 5 6 3 2 0 The tableau shown is optimal. However, there are also other optimal tableaux, as can be seen from the fact that the indicator values for each of the cells that have no circled entries are 0. Continued on next page

116 Chapter 4 57. (b) continued For (ii) the indicator value for each non-circled cell is calculated as follows. cell ( II,1 ): 3 1 4 6 0 cell ( II, 2 ): 2 1 4 5 0 The tableau shown is optimal, although there are also other optimal tableaux. For (iii) the indicator value for each non-circled cell is calculated as follows. cell ( II,1 ): 2 4 1 6 3 cell ( II, 2 ): 3 4 1 5 5 The tableau shown is not optimal. The current cost is as follows. Continued on next page 2( 6) 1( 5) 2( 1) 4( 4) 12 5 2 16 35

Linear Programming 117 57. (b) continued Since cell ( II,1 ) has a more negative indicator value, we can reduce the cost more by using that cell. Increasing by 2, we obtain the following tableau. The cost is now 1( 5) 4( 1) 2( 4) 2( 2) 5 4 8 4 21. We can reduce the cost more by using cell ( II,2 ). Increasing by 1, we obtain the following tableau. The cost is now 2( 2) 1( 3) 1( 4) 5( 1) 4 3 4 5 16.

118 Chapter 4 58. (a) (b) The cost is given by 1( 3) 2( 1) 1( 6) 5( 3) 1( 5) 3 2 6 15 5 31. Continued on next page

Linear Programming 119 58. continued (c) The indicator value for cell ( II,1 ) is 5 3 6 3 5. The indicator value for cell ( II, 2 ) is 2 3 6 1 4. The indicator value for cell ( III,1 ) is 1 5 6 3 1. Continued on next page

120 Chapter 4 58. (c) continued The indicator value for cell ( III, 2 ) is 7 5 6 1 7. Since one of these cells has an indicator value, which is negative, the initial solution we found is not optimal. Word Search Solution