1. 6  Graphing Systems of Linear Inequalities in Two Variables  Linear Programming Problems  Graphical Solutions of Linear Programming Problems  The Simplex Method: Standard Maximization Problems  The Simplex Method: Standard Minimization Problems Linear Programming: A Geometric Approach

2. 6.1 Graphing Systems of Linear Inequalities in Two Variables

3. Graphing Linear Inequalities  We’ve seen that a linear equation in two variables x and y has a solution set that may be exhibited graphically as points on a straight line in the xy-plane.  There is also a simple graphical representation for linear inequalities of two variables:

4. Procedure for Graphing Linear Inequalities 1.Draw the graph of the equation obtained for the given inequality by replacing the inequality sign with an equal sign. ✦ Use a dashed or dotted line if the problem involves a strict inequality,. ✦ Otherwise, use a solid line to indicate that the line itself constitutes part of the solution. 2.Pick a test point lying in one of the half-planes determined by the line sketched in step 1 and substitute the values of x and y into the given inequality. ✦ Use the origin whenever possible. 3.If the inequality is satisfied, the graph of the inequality includes the half-plane containing the test point. ✦ Otherwise, the solution includes the half-plane not containing the test point.

5. Examples  Determine the solution set for the inequality 2x + 3y  6. Solution  Replacing the inequality  with an equality =, we obtain the equation 2x + 3y = 6, whose graph is: x y 7531 77553311– 1– 177553311– 1– 1 – 5– 3 – 1135– 5– 3 – 1135– 5– 3 – 1135– 5– 3 – 1135 2x + 3y = 6

6. Examples  Determine the solution set for the inequality 2x + 3y  6. Solution  Picking the origin as a test point, we find 2(0) + 3(0)  6, or 0  6, which is false.  Thus, the solution set is: x y 7531 77553311– 1– 177553311– 1– 1 – 5– 3 – 1135– 5– 3 – 1135– 5– 3 – 1135– 5– 3 – 1135 2x + 3y = 6 2x + 3y  6 (0, 0)

7. Graphing Systems of Linear Inequalities  The solution set of a system of linear inequalities in two variables x and y is the set of all points (x, y) that satisfy each inequality of the system.  The graphical solution of such a system may be obtained by graphing the solution set for each inequality independently and then determining the region in common with each solution set.

8. – 5 – 3 13 5 Examples  Graph x – 3y > 0. Solution  Replacing the inequality > with an equality =, we obtain the equation x – 3y = 0, whose graph is: x y 31 3311– 1– 1– 3– 33311– 1– 1– 3– 3 x – 3y = 0

9. Examples  Graph x – 3y > 0. Solution  We use a dashed line to indicate the line itself will not be part of the solution, since we are dealing with a strict inequality >. x y x – 3y = 0 – 5 – 3 13 5 31 3311– 1– 1– 3– 33311– 1– 1– 3– 3

10. 31 3311– 1– 1– 3– 33311– 1– 1– 3– 3Examples  Graph x – 3y > 0. Solution  Since the origin lies on the line, we cannot use the origin as a testing point: x y x – 3y = 0 (0, 0)

11. Examples  Graph x – 3y > 0. Solution  Picking instead (3, 0) as a test point, we find (3) – 2(0) > 0, or 3 > 0, which is true.  Thus, the solution set is: y x – 3y = 0 x – 3y > 0 – 5 – 3 13 5 31 3311– 1– 1– 3– 33311– 1– 1– 3– 3 x (3, 0)

12. Graphing Systems of Linear Inequalities  The solution set of a system of linear inequalities in two variables x and y is the set of all points (x, y) that satisfy each inequality of the system.  The graphical solution of such a system may be obtained by graphing the solution set for each inequality independently and then determining the region in common with each solution set.

13. Example  Determine the solution set for the system Solution  The intersection of the solution regions of the two inequalities represents the solution to the system: x y 4321 4x + 3y  12 4x + 3y = 12 – 1 1 23 – 1 1 23 – 1 1 23 – 1 1 23

14. Example  Determine the solution set for the system Solution  The intersection of the solution regions of the two inequalities represents the solution to the system: x y x – y  0 x – y = 0 4321 – 1 1 23 – 1 1 23 – 1 1 23 – 1 1 23

15. Example  Determine the solution set for the system Solution  The intersection of the solution regions of the two inequalities represents the solution to the system: x y 4x + 3y = 12 x – y = 0 4321 – 1 1 23 – 1 1 23 – 1 1 23 – 1 1 23

16. Bounded and Unbounded Sets  The solution set of a system of linear inequalities is bounded if it can be enclosed by a circle.  Otherwise, it is unbounded.

17. Example  The solution to the problem we just discussed is unbounded, since the solution set cannot be enclosed in a circle: x y 4x + 3y = 12 x – y = 0 4321 – 1 1 23 – 1 1 23 – 1 1 23 – 1 1 23

18. 7531 –1 1359 –1 1359Example  Determine the solution set for the system Solution  The intersection of the solution regions of the four inequalities represents the solution to the system: x y

19. Example  Determine the solution set for the system Solution  Note that the solution to this problem is bounded, since it can be enclosed by a circle: –1 1359 –1 1359 x y 7531

20. 6.2 Linear Programming Problems Maximize Subject to

21. Linear Programming Problem  A linear programming problem consists of a linear objective function to be maximized or minimized subject to certain constraints in the form of linear equations or inequalities.

22. Applied Example 1: A Production Problem  Ace Novelty wishes to produce two types of souvenirs: type-A will result in a profit of $1.00, and type-B in a profit of $1.20.  To manufacture a type-A souvenir requires 2 minutes on machine I and 1 minute on machine II.  A type-B souvenir requires 1 minute on machine I and 3 minutes on machine II.  There are 3 hours available on machine I and 5 hours available on machine II.  How many souvenirs of each type should Ace make in order to maximize its profit?

23. Applied Example 1: A Production Problem Solution  Let’s first tabulate the given information:  Let x be the number of type-A souvenirs and y the number of type-B souvenirs to be made. Type-AType-B Time Available Profit/Unit Profit/Unit$1.00$1.20 Machine I Machine I 2 min 1 min 180 min Machine II Machine II 1 min 3 min 300 min

24. Applied Example 1: A Production Problem Solution  Let’s first tabulate the given information:  Then, the total profit (in dollars) is given by which is the objective function to be maximized. Type-AType-B Time Available Profit/Unit Profit/Unit$1.00$1.20 Machine I Machine I 2 min 1 min 180 min Machine II Machine II 1 min 3 min 300 min

25. Applied Example 1: A Production Problem Solution  Let’s first tabulate the given information:  The total amount of time that machine I is used is and must not exceed 180 minutes.  Thus, we have the inequality Type-AType-B Time Available Profit/Unit Profit/Unit$1.00$1.20 Machine I Machine I 2 min 1 min 180 min Machine II Machine II 1 min 3 min 300 min

26. Applied Example 1: A Production Problem Solution  Let’s first tabulate the given information:  The total amount of time that machine II is used is and must not exceed 300 minutes.  Thus, we have the inequality Type-AType-B Time Available Profit/Unit Profit/Unit$1.00$1.20 Machine I Machine I 2 min 1 min 180 min Machine II Machine II 1 min 3 min 300 min

27. Applied Example 1: A Production Problem Solution  Let’s first tabulate the given information:  Finally, neither x nor y can be negative, so Type-AType-B Time Available Profit/Unit Profit/Unit$1.00$1.20 Machine I Machine I 2 min 1 min 180 min Machine II Machine II 1 min 3 min 300 min

28. Applied Example 1: A Production Problem Solution  In short, we want to maximize the objective function subject to the system of inequalities  We will discuss the solution to this problem in section 6.4.

29. Applied Example 2: A Nutrition Problem  A nutritionist advises an individual who is suffering from iron and vitamin B deficiency to take at least 2400 milligrams (mg) of iron, 2100 mg of vitamin B 1, and 1500 mg of vitamin B 2 over a period of time.  Two vitamin pills are suitable, brand-A and brand-B.  Each brand-A pill costs 6 cents and contains 40 mg of iron, 10 mg of vitamin B 1, and 5 mg of vitamin B 2.  Each brand-B pill costs 8 cents and contains 10 mg of iron and 15 mg each of vitamins B 1 and B 2.  What combination of pills should the individual purchase in order to meet the minimum iron and vitamin requirements at the lowest cost?

30. Applied Example 2: A Nutrition Problem Solution  Let’s first tabulate the given information:  Let x be the number of brand-A pills and y the number of brand-B pills to be purchased. Brand-ABrand-B Minimum Requirement Cost/Pill Cost/Pill 6¢6¢6¢6¢ 8¢8¢8¢8¢ Iron Iron 40 mg 10 mg 2400 mg Vitamin B 1 Vitamin B 1 10 mg 15 mg 2100 mg Vitamin B 2 Vitamin B 2 5mg 5mg 15 mg 1500 mg

31. Applied Example 2: A Nutrition Problem Solution  Let’s first tabulate the given information:  The cost C (in cents) is given by and is the objective function to be minimized. Brand-ABrand-B Minimum Requirement Cost/Pill Cost/Pill 6¢6¢6¢6¢ 8¢8¢8¢8¢ Iron Iron 40 mg 10 mg 2400 mg Vitamin B 1 Vitamin B 1 10 mg 15 mg 2100 mg Vitamin B 2 Vitamin B 2 5mg 5mg 15 mg 1500 mg

32. Applied Example 2: A Nutrition Problem Solution  Let’s first tabulate the given information:  The amount of iron contained in x brand-A pills and y brand-B pills is given by 40x + 10y mg, and this must be greater than or equal to 2400 mg.  This translates into the inequality Brand-ABrand-B Minimum Requirement Cost/Pill Cost/Pill 6¢6¢6¢6¢ 8¢8¢8¢8¢ Iron Iron 40 mg 10 mg 2400 mg Vitamin B 1 Vitamin B 1 10 mg 15 mg 2100 mg Vitamin B 2 Vitamin B 2 5mg 5mg 15 mg 1500 mg

33. Applied Example 2: A Nutrition Problem Solution  Let’s first tabulate the given information:  The amount of vitamin B 1 contained in x brand-A pills and y brand-B pills is given by 10x + 15y mg, and this must be greater or equal to 2100 mg.  This translates into the inequality Brand-ABrand-B Minimum Requirement Cost/Pill Cost/Pill 6¢6¢6¢6¢ 8¢8¢8¢8¢ Iron Iron 40 mg 10 mg 2400 mg Vitamin B 1 Vitamin B 1 10 mg 15 mg 2100 mg Vitamin B 2 Vitamin B 2 5mg 5mg 15 mg 1500 mg

34. Applied Example 2: A Nutrition Problem Solution  Let’s first tabulate the given information:  The amount of vitamin B 2 contained in x brand-A pills and y brand-B pills is given by 5x + 15y mg, and this must be greater or equal to 1500 mg.  This translates into the inequality Brand-ABrand-B Minimum Requirement Cost/Pill Cost/Pill 6¢6¢6¢6¢ 8¢8¢8¢8¢ Iron Iron 40 mg 10 mg 2400 mg Vitamin B 1 Vitamin B 1 10 mg 15 mg 2100 mg Vitamin B 2 Vitamin B 2 5mg 5mg 15 mg 1500 mg

35. Applied Example 2: A Nutrition Problem Solution  In short, we want to minimize the objective function subject to the system of inequalities  We will discuss the solution to this problem in section 6.4.

36. 6.3 Graphical Solutions of Linear Programming Problems

37. Feasible Solution Set and Optimal Solution  The constraints in a linear programming problem form a system of linear inequalities, which have a solution set S.  Each point in S is a candidate for the solution of the linear programming problem and is referred to as a feasible solution.  The set S itself is referred to as a feasible set.  Among all the points in the set S, the point(s) that optimizes the objective function of the linear programming problem is called an optimal solution.

38. Theorem 1 Linear Programming  If a linear programming problem has a solution, then it must occur at a vertex, or corner point, of the feasible set S associated with the problem.  If the objective function P is optimized at two adjacent vertices of S, then it is optimized at every point on the line segment joining these vertices, in which case there are infinitely many solutions to the problem.

39. Theorem 2 Existence of a Solution  Suppose we are given a linear programming problem with a feasible set S and an objective function P = ax + by. a.If S is bounded, then P has both a maximum and a minimum value on S. b.If S is unbounded and both a and b are nonnegative, then P has a minimum value on S provided that the constraints defining S include the inequalities x  0 and y  0. c.If S is the empty set, then the linear programming problem has no solution: that is, P has neither a maximum nor a minimum value.

40. The Method of Corners 1.Graph the feasible set. 2.Find the coordinates of all corner points (vertices) of the feasible set. 3.Evaluate the objective function at each corner point. 4.Find the vertex that renders the objective function a maximum or a minimum. ✦ If there is only one such vertex, it constitutes a unique solution to the problem. ✦ If there are two such adjacent vertices, there are infinitely many optimal solutions given by the points on the line segment determined by these vertices.

41. Applied Example 1: A Production Problem  Recall Applied Example 1 from the last section (3.2), which required us to find the optimal quantities to produce of type-A and type-B souvenirs in order to maximize profits.  We restated the problem as a linear programming problem in which we wanted to maximize the objective function subject to the system of inequalities  We can now solve the problem graphically.

42. 200 100 100200300 Applied Example 1: A Production Problem  We first graph the feasible set S for the problem. ✦ Graph the solution for the inequality considering only positive values for x and y: x y (90, 0) (0, 180)

43. 200 100 Applied Example 1: A Production Problem  We first graph the feasible set S for the problem. ✦ Graph the solution for the inequality considering only positive values for x and y: 100200300 x y (0, 100) (300, 0)

44. 200 100 Applied Example 1: A Production Problem  We first graph the feasible set S for the problem. ✦ Graph the intersection of the solutions to the inequalities, yielding the feasible set S. (Note that the feasible set S is bounded) 100200300 x y S

45. 200 100 Applied Example 1: A Production Problem  Next, find the vertices of the feasible set S. ✦ The vertices are A(0, 0), B(90, 0), C(48, 84), and D(0, 100). 100200300 x y S C(48, 84) D(0, 100) B(90, 0) A(0, 0)

46. 200 100 Applied Example 1: A Production Problem  Now, find the values of P at the vertices and tabulate them: 100200300 x y S C(48, 84) D(0, 100) B(90, 0) A(0, 0) Vertex Vertex P = x + 1.2 y A(0, 0) A(0, 0) 0 B(90, 0) B(90, 0)90 C(48, 84) C(48, 84)148.8 D(0, 100) D(0, 100)120

47. 200 100 Applied Example 1: A Production Problem  Finally, identify the vertex with the highest value for P: ✦ We can see that P is maximized at the vertex C(48, 84) and has a value of 148.8. 100200300 x y S D(0, 100) B(90, 0) A(0, 0) Vertex Vertex P = x + 1.2 y A(0, 0) A(0, 0) 0 B(90, 0) B(90, 0)90 C(48, 84) C(48, 84)148.8 D(0, 100) D(0, 100)120 C(48, 84)

48. Applied Example 1: A Production Problem  Finally, identify the vertex with the highest value for P: ✦ We can see that P is maximized at the vertex C(48, 84) and has a value of 148.8. ✦ Recalling what the symbols x, y, and P represent, we conclude that ACE Novelty would maximize its profit at $148.80 by producing 48 type-A souvenirs and 84 type-B souvenirs.

49. Applied Example 2: A Nutrition Problem  Recall Applied Example 2 from the last section (3.2), which asked us to determine the optimal combination of pills to be purchased in order to meet the minimum iron and vitamin requirements at the lowest cost.  We restated the problem as a linear programming problem in which we wanted to minimize the objective function subject to the system of inequalities  We can now solve the problem graphically.

50. 200 100 Applied Example 2: A Nutrition Problem  We first graph the feasible set S for the problem. ✦ Graph the solution for the inequality considering only positive values for x and y: 100200300 x y (60, 0) (0, 240)

51. 200 100 Applied Example 2: A Nutrition Problem  We first graph the feasible set S for the problem. ✦ Graph the solution for the inequality considering only positive values for x and y: 100200300 x y (210, 0) (0, 140)

52. 200 100 Applied Example 2: A Nutrition Problem  We first graph the feasible set S for the problem. ✦ Graph the solution for the inequality considering only positive values for x and y: 100200300 x y (300, 0) (0, 100)

53. 200 100 Applied Example 2: A Nutrition Problem  We first graph the feasible set S for the problem. ✦ Graph the intersection of the solutions to the inequalities, yielding the feasible set S. (Note that the feasible set S is unbounded) 100200300 x y S

54. 200 100 Applied Example 2: A Nutrition Problem  Next, find the vertices of the feasible set S. ✦ The vertices are A(0, 240), B(30, 120), C(120, 60), and D(300, 0). 100200300 x y S C(120, 60) D(300, 0) A(0, 240) B(30, 120)

55. Applied Example 2: A Nutrition Problem  Now, find the values of C at the vertices and tabulate them: 200 100 100200300 x y S C(120, 60) D(300, 0) A(0, 240) B(30, 120) Vertex Vertex C = 6x + 8y A(0, 240) A(0, 240) 1920 1920 B(30, 120) B(30, 120)1140 C(120, 60) C(120, 60)1200 D(300, 0) D(300, 0)1800

56. Applied Example 2: A Nutrition Problem  Finally, identify the vertex with the lowest value for C: ✦ We can see that C is minimized at the vertex B(30, 120) and has a value of 1140. 200 100 100200300 x y S C(120, 60) D(300, 0) A(0, 240) Vertex Vertex C = 6x + 8y A(0, 240) A(0, 240) 1920 1920 B(30, 120) B(30, 120)1140 C(120, 60) C(120, 60)1200 D(300, 0) D(300, 0)1800 B(30, 120)

57. Applied Example 2: A Nutrition Problem  Finally, identify the vertex with the lowest value for C: ✦ We can see that C is minimized at the vertex B(30, 120) and has a value of 1140. ✦ Recalling what the symbols x, y, and C represent, we conclude that the individual should purchase 30 brand-A pills and 120 brand-B pills at a minimum cost of $11.40.

58. 6.4 The Simplex Method: Standard Maximization Problems xyuvPConstant103/5–1/50 48 48 01–1/52/50 84 84 009/257/251 148 4/5 148 4/5

59. The Simplex Method  The simplex method is an iterative procedure.  Beginning at a vertex of the feasible region S, each iteration brings us to another vertex of S with an improved value of the objective function.  The iteration ends when the optimal solution is reached.

60. A Standard Linear Programming Problem  A standard maximization problem is one in which 1.The objective function is to be maximized. 2.All the variables involved in the problem are nonnegative. 3.All other linear constraints may be written so that the expression involving the variables is less than or equal to a nonnegative constant.

61. Setting Up the Initial Simplex Tableau 1.Transform the system of linear inequalities into a system of linear equations by introducing slack variables. 2.Rewrite the objective function in the form where all the variables are on the left and the coefficient of P is +1. Write this equation below the equations in step 1. 3.Write the augmented matrix associated with this system of linear equations.

62. Applied Example 1: A Production Problem  Recall the production problem discussed in section 6.3, which required us to maximize the objective function subject to the system of inequalities  This is a standard maximization problem and may be solved by the simplex method.  Set up the initial simplex tableau for this linear programming problem.

63. Applied Example 1: A Production Problem Solution  First, introduce the slack variables u and v into the inequalities and turn these into equations, getting  Next, rewrite the objective function in the form

64. Applied Example 1: A Production Problem Solution  Placing the restated objective function below the system of equations of the constraints we get  Thus, the initial tableau associated with this system is xyuvPConstant21100180 13010300 –1 – 6/5 001 0

65. The Simplex Method 1.Set up the initial simplex tableau. 2.Determine whether the optimal solution has been reached by examining all entries in the last row to the left of the vertical line. a.If all the entries are nonnegative, the optimal solution has been reached. Proceed to step 4. b.If there are one or more negative entries, the optimal solution has not been reached. Proceed to step 3. 3.Perform the pivot operation. Return to step 2. 4.Determine the optimal solution(s).

66. Applied Example 1: A Production Problem  Recall again the production problem discussed previously.  We have already performed step 1 obtaining the initial simplex tableau:  Now, complete the solution to the problem. xyuvPConstant21100180 13010300 –1 – 6/5 001 0

67. Applied Example 1: A Production Problem Solution Step 2.Determine whether the optimal solution has been reached. ✦ Since there are negative entries in the last row of the tableau, the initial solution is not optimal. xyuvPConstant21100180 13010300 –1 – 6/5 001 0

68. Applied Example 1: A Production Problem Solution Step 3.Perform the pivot operation. ✦ Since the entry – 6/5 is the most negative entry to the left of the vertical line in the last row of the tableau, the second column in the tableau is the pivot column. xyuvPConstant21100180 13010300 –1 – 6/5 001 0

69. Applied Example 1: A Production Problem Solution Step 3.Perform the pivot operation. ✦ Divide each positive number of the pivot column into the corresponding entry in the column of constants and compare the ratios thus obtained. ✦ We see that the ratio 300/3 = 100 is less than the ratio 180/1 = 180, so row 2 is the pivot row. xyuvPConstant21100180 13010300 –1 – 6/5 001 0

70. Applied Example 1: A Production Problem Solution Step 3.Perform the pivot operation. ✦ The entry 3 lying in the pivot column and the pivot row is the pivot element. xyuvPConstant21100180 13010300 –1 – 6/5 001 0

71. Applied Example 1: A Production Problem Solution Step 3.Perform the pivot operation. ✦ Convert the pivot element into a 1. xyuvPConstant21100180 13010300 –1 – 6/5 001 0

72. Applied Example 1: A Production Problem Solution Step 3.Perform the pivot operation. ✦ Convert the pivot element into a 1. xyuvPConstant21100180 1/3101/30100 –1 – 6/5 001 0

73. Applied Example 1: A Production Problem Solution Step 3.Perform the pivot operation. ✦ Use elementary row operations to convert the pivot column into a unit column. xyuvPConstant21100180 1/3101/30100 –1 – 6/5 001 0

74. Applied Example 1: A Production Problem Solution Step 3.Perform the pivot operation. ✦ Use elementary row operations to convert the pivot column into a unit column. xyuvPConstant5/301–1/30 80 80 1/3101/30100 –3/5002/51120

75. Applied Example 1: A Production Problem Solution Step 3.Perform the pivot operation. ✦ This completes an iteration. ✦ The last row of the tableau contains a negative number, so an optimal solution has not been reached. ✦ Therefore, we repeat the iteration step. xyuvPConstant5/301–1/30 80 80 1/3101/30100 –3/5002/51120

76. Applied Example 1: A Production Problem Solution Step 3.Perform the pivot operation again. ✦ Since the entry – 3/5 is the most negative entry to the left of the vertical line in the last row of the tableau, the first column in the tableau is now the pivot column. xyuvPConstant5/301–1/30 80 80 1/3101/30100 –3/5002/51120

77. Applied Example 1: A Production Problem Solution Step 3.Perform the pivot operation. ✦ Divide each positive number of the pivot column into the corresponding entry in the column of constants and compare the ratios thus obtained. ✦ We see that the ratio 80/(5/3) = 48 is less than the ratio 100/(1/3) = 300, so row 1 is the pivot row now. xyuvPConstant5/301–1/30 80 80 1/3101/30100 –3/5002/51120 Ratio

78. Applied Example 1: A Production Problem Solution Step 3.Perform the pivot operation. ✦ The entry 5/3 lying in the pivot column and the pivot row is the pivot element. xyuvPConstant5/301–1/30 80 80 1/3101/30100 –3/5002/51120

79. Applied Example 1: A Production Problem Solution Step 3.Perform the pivot operation. ✦ Convert the pivot element into a 1. xyuvPConstant5/301–1/30 80 80 1/3101/30100 –3/5002/51120

80. Applied Example 1: A Production Problem Solution Step 3.Perform the pivot operation. ✦ Convert the pivot element into a 1. xyuvPConstant103/5–1/50 48 48 1/3101/30100 –3/5002/51120

81. Applied Example 1: A Production Problem Solution Step 3.Perform the pivot operation. ✦ Use elementary row operations to convert the pivot column into a unit column. xyuvPConstant103/5–1/50 48 48 1/3101/30100 –3/5002/51120

82. Applied Example 1: A Production Problem Solution Step 3.Perform the pivot operation. ✦ Use elementary row operations to convert the pivot column into a unit column. xyuvPConstant103/5–1/50 48 48 01–1/52/50 84 84 009/257/251 148 4/5 148 4/5

83. Applied Example 1: A Production Problem Solution Step 3.Perform the pivot operation. ✦ The last row of the tableau contains no negative numbers, so an optimal solution has been reached. xyuvPConstant103/5–1/50 48 48 01–1/52/50 84 84 009/257/251 148 4/5 148 4/5

84. Applied Example 1: A Production Problem Solution Step 4.Determine the optimal solution. ✦ Locate the basic variables in the final tableau. In this case, the basic variables are x, y, and P.  The optimal value for x is 48.  The optimal value for y is 84.  The optimal value for P is 148.8. ✦ Thus, the firm will maximize profits at $148.80 by producing 48 type-A souvenirs and 84 type-B souvenirs. This agrees with the results obtained in section 6.3. xyuvPConstant103/5–1/50 48 48 01–1/52/50 84 84 009/257/251 148 4/5 148 4/5

85. 6.5 The Simplex Method: Standard Minimization Problems

86. Minimization with  Constraints  In the last section we developed the simplex method to solve linear programming problems that satisfy three conditions: 1.The objective function is to be maximized. 2.All the variables involved are nonnegative. 3.Each linear constraint may be written so that the expression involving the variables is less than or equal to a nonnegative constant.  We will now see how the simplex method can be used to solve minimization problems that meet the second and third conditions listed above.

87. Example  Solve the following linear programming problem:  This problem involves the minimization of the objective function and so is not a standard maximization problem.  Note, however, that all the other conditions for a standard maximization hold true.

88. Example  We can use the simplex method to solve this problem by converting the objective function from minimizing C to its equivalent of maximizing P = – C.  Thus, the restated linear programming problem is  This problem can now be solved using the simplex method as discussed in section 6.4.

89. ExampleSolution Step 1.Set up the initial simplex tableau. ✦ Turn the constraints into equations adding to them the slack variables u and v. Also rearrange the objective function and place it below the constraints: ✦ Write the coefficients of the system in a tableau: xyuvPConstant54100 32 32 12010 10 10 –2–3001 0

90. ExampleSolution Step 2.Determine whether the optimal solution has been reached. ✦ Since there are negative entries in the last row of the tableau, the initial solution is not optimal. xyuvPConstant5410032 1201010 –2–3001 0

91. ExampleSolution Step 3.Perform the pivot operation. ✦ Since the entry – 3 is the most negative entry to the left of the vertical line in the last row of the tableau, the second column in the tableau is the pivot column. xyuvPConstant5410032 1201010 –2–3001 0

92. ExampleSolution Step 3.Perform the pivot operation. ✦ Divide each positive number of the pivot column into the corresponding entry in the column of constants and compare the ratios thus obtained. ✦ We see that the ratio 10/2 = 5 is less than the ratio 32/4 = 8, so row 2 is the pivot row. xyuvPConstant5410032 1201010 –2–3001 0 Ratio

93. ExampleSolution Step 3.Perform the pivot operation. ✦ The entry 2 lying in the pivot column and the pivot row is the pivot element. xyuvPConstant5410032 1201010 –2–3001 0

94. ExampleSolution Step 3.Perform the pivot operation. ✦ Convert the pivot element into a 1. xyuvPConstant5410032 1201010 –2–3001 0

95. ExampleSolution Step 3.Perform the pivot operation. ✦ Convert the pivot element into a 1. xyuvPConstant5410032 1/2101/20 5 –2–3001 0

96. ExampleSolution Step 3.Perform the pivot operation. ✦ Use elementary row operations to convert the pivot column into a unit column. xyuvPConstant5410032 1/2101/20 5 –2–3001 0

97. ExampleSolution Step 3.Perform the pivot operation. ✦ Use elementary row operations to convert the pivot column into a unit column. xyuvPConstant301–2012 1/2101/20 5 –1/2003/2115

98. ExampleSolution Step 3.Perform the pivot operation. ✦ This completes an iteration. ✦ The last row of the tableau contains a negative number, so an optimal solution has not been reached. ✦ Therefore, we repeat the iteration step. xyuvPConstant301–2012 1/2101/20 5 –1/2003/2115

99. ExampleSolution Step 3.Perform the pivot operation. ✦ Since the entry –1/2 is the most negative entry to the left of the vertical line in the last row of the tableau, the first column in the tableau is now the pivot column. xyuvPConstant301–2012 1/2101/20 5 –1/2003/2115

100. ExampleSolution Step 3.Perform the pivot operation. ✦ Divide each positive number of the pivot column into the corresponding entry in the column of constants and compare the ratios thus obtained. ✦ We see that the ratio 12/3 = 4 is less than the ratio 5/(1/2) = 10, so row 1 is now the pivot row. xyuvPConstant301–2012 1/2101/20 5 –1/2003/2115

101. ExampleSolution Step 3.Perform the pivot operation. ✦ The entry 3 lying in the pivot column and the pivot row is the pivot element. xyuvPConstant301–2012 1/2101/20 5 –1/2003/2115

102. ExampleSolution Step 3.Perform the pivot operation. ✦ Convert the pivot element into a 1. xyuvPConstant301–2012 1/2101/20 5 –1/2003/2115

103. ExampleSolution Step 3.Perform the pivot operation. ✦ Convert the pivot element into a 1. xyuvPConstant101/3–2/30 4 1/2101/20 5 –1/2003/2115

104. ExampleSolution Step 3.Perform the pivot operation. ✦ Use elementary row operations to convert the pivot column into a unit column. xyuvPConstant101/3–2/30 4 1/2101/20 5 –1/2003/2115

105. ExampleSolution Step 3.Perform the pivot operation. ✦ Use elementary row operations to convert the pivot column into a unit column. xyuvPConstant101/3–2/30 4 01–1/65/60 3 001/67/6117

106. ExampleSolution Step 3.Perform the pivot operation. ✦ The last row of the tableau contains no negative numbers, so an optimal solution has been reached. xyuvPConstant101/3–2/30 4 01–1/65/60 3 001/67/6117

107. ExampleSolution Step 4.Determine the optimal solution. ✦ Locate the basic variables in the final tableau. In this case, the basic variables are x, y, and P.  The optimal value for x is 4.  The optimal value for y is 3.  The optimal value for P is 17, which means that the minimized value for C is –17. xyuvPConstant101/3–2/30 4 01–1/65/60 3 001/67/6117

108. The Dual Problem  Another special class of linear programming problems we encounter in practical applications is characterized by the following conditions: 1.The objective function is to be minimized. 2.All the variables involved are nonnegative. 3.All other linear constraints may be written so that the expression involving the variables is greater than or equal to a nonnegative constant.  Such problems are called standard minimization problems.

109. The Dual Problem  In solving this kind of linear programming problem, it helps to note that each maximization problem is associated with a minimization problem, and vice versa.  The given problem is called the primal problem, and the related problem is called the dual problem.

110. Example  Write the dual problem associated with this problem:  We first write down a tableau for the primal problem: xyConstant40102400 10152100 5151500 68 Primal Problem

111. Example  Next, we interchange the columns and rows of the tableau and head the three columns of the resulting array with the three variables u, v, and w, obtaining xyConstant40102400 10152100 5151500 68 uvwConstant401056 1015158 240021001500

112. Example  Consider the resulting tableau as if it were the initial simplex tableau for a standard maximization problem.  From it we can reconstruct the required dual problem: uvwConstant401056 1015158 240021001500 Dual Problem

113. Theorem 1 The Fundamental Theorem of Duality  A primal problem has a solution if and only if the corresponding dual problem has a solution.  Furthermore, if a solution exists, then: a.The objective functions of both the primal and the dual problem attain the same optimal value. b.The optimal solution to the primal problem appears under the slack variables in the last row of the final simplex tableau associated with the dual problem.

114. Example  Complete the solution of the problem from our last example: Dual Problem

115. ExampleSolution  The dual problem associated with the given primal problem is a standard maximization problem.  Thus, we can proceed with the simplex method.  First, we introduce to the system of equations the slack variables x and y, and restate the inequalities as equations, obtaining

116. ExampleSolution  Next, we transcribe the coefficients of the system of equations into an initial simplex tableau: uvwxyPConstant401051006 1015150108 –2400–2100–15000010

117. ExampleSolution  Continue with the simplex iterative method until a final tableau is obtained with the solution for the problem:  The fundamental theorem of duality tells us that the solution to the primal problem is x = 30 and y = 120, with a minimum value for C of 1140. uvwxyPConstant10 – 3/20 3/100 – 1/50 0 1/50 1/50 0111/10 – 1/50 2/25013/25 004503012011140 Solution for the primal problem

118. End of Chapter