Explanation Assume the number of interviews limited to each of the two categories be represented by y 1 and y 2 , respectively. Construct a table in order to fetch the data from the given information as follows. Political Interview Market Interview Limit Total interviews y 1 y 2 8 Earning $8 $10 $60 Bonus points 6 5 40 Cost $45 $55 Then obtain the first constraint as y 1 + y 2 ≥ 8 since she must complete atleast 8 interviews each week. Obtain the second constraint as 8 y 1 + 10 y 2 ≥ 60 since she earns $8 for each political interview and $10 for each market interview and she must earn $60 per week. Obtain the third constraint as 6 y 1 + 5 y 2 ≥ 40 since she earns 6 bonus points for each political interview and 5 bonus points for each market interview and she must earn atleast 40 bonus points per week. Obtain the objective function which represents the minimized cost is, w = 45 y 1 + 55 y 2 . The linear programming problem can be stated as follows. Minimize w = 45 y 1 + 55 y 2 subject to y 1 + y 2 ≥ 8 8 y 1 + 10 y 2 ≥ 60 6 y 1 + 5 y 2 ≥ 40 with y 1 ≥ 0 , y 2 ≥ 0 Obtain the dual of the above problem as follows. Maximize z = 8 x 1 + 60 x 2 + 40 x 3 subject to x 1 + 8 x 2 + 6 x 3 ≤ 45 x 1 + 10 x 2 + 5 x 3 ≤ 55 with x 1 ≥ 0 , x 2 ≥ 0 , x 3 ≥ 0 Add slack variables to the above constraints and obtain the linear programming problem as follows. Maximize z = 8 x 1 + 60 x 2 + 40 x 3 subject to x 1 + 8 x 2 + 6 x 3 + s 1 = 45 x 1 + 10 x 2 + 5 x 3 + s 2 = 55 with x 1 ≥ 0 , x 2 ≥ 0 , x 3 ≥ 0 , s 1 ≥ 0 , s 2 ≥ 0 Rewrite the objective function as z − 8 x 1 − 60 x 2 − 40 x 3 = 0 . Obtain the simplex tableau, x 1 x 2 x 3 s 1 s 2 z [ 1 8 6 1 0 0 45 1 10 5 0 1 0 55 − 8 − 60 − 40 0 0 1 0 ] . Here, the basic variables are s 1 and s 2 . Thus, the basic solution is x 1 = 0 , x 2 = 0 , x 3 = 0 , s 1 = 45 , s 2 = 55 and z = 0 . This is not the optimal solution as only the slack variables are nonzero. Look in the bottom row, the coefficient of x 2 is most negative, so choose second column as a pivot column and bring x 2 into the solution. Divide the entries of the last column by the respective entries in the pivot column as follows. x 1 x 2 x 3 s 1 s 2 z [ 1 8 6 1 0 0 45 1 10 5 0 1 0 55 − 8 − 60 − 40 0 0 1 0 ] 45 / 8 = 5.625 55 / 10 = 5.5 Observe that the least ratio is 5.5. Hence, the second element of the pivot column is the pivot element. x 1 x 2 x 3 s 1 s 2 z [ 1 8 6 1 0 0 45 1 10 5 0 1 0 55 − 8 − 60 − 40 0 0 1 0 ] Perform the row operations on the simplex tableau as follows. x 1 x 2 x 3 s 1 s 2 z [ 1 8 6 1 0 0 45 1 10 5 0 1 0 55 − 8 − 60 − 40 0 0 1 0 ] 10 R 1 − 8 R 2 → R 1 R 3 + 6 R 2 → R 3 The simplex tableau becomes, x 1 x 2 x 3 s 1 s 2 z [ 2 0 20 10 − 8 0 10 1 10 5 0 1 0 55 − 2 0 − 10 0 6 1 330 ] From the above tableau, the bottom row has two negative entries. So, this is not an optimal solution. Look in the bottom row, the coefficient of x 3 is most negative, so choose third column as a pivot column and bring x 3 into the solution. Thus, divide the entries of the last column by the respective entries of the pivot column as follows. x 1 x 2 x 3 s 1 s 2 z [ 2 0 20 10 − 8 0 10 1 10 5 0 1 0 55 − 2 0 − 10 0 6 1 330 ] 10 / 20 = 0.5 55 / 5 = 11 Observe that the least ratio is 0.5. Hence, the first element of the pivot column is the pivot element. x 1 x 2 x 3 s 1 s 2 z [ 2 0 20 10 − 8 0 10 1 10 5 0 1 0 55 − 2 0 − 10 0 6 1 330 ] Perform the row operations on the simplex tableau as follows. x 1 x 2 x 3 s 1 s 2 z [ 2 0 20 10 − 8 0 10 2 40 0 − 10 12 0 210 − 2 0 0 10 4 2 670 ] − R 1 + 4 R 2 → R 2 R 1 + 2 R 3 → R 3 Thus the simplex tableau becomes, x 1 x 2 x 3 s 1 s 2 z [ 2 0 20 10 − 8 0 10 2 40 0 − 10 12 0 210 − 2 0 0 10 4 2 670 ] From the above tableau, the bottom row has one negative entry

Finite Mathematics and Calculus with Applications (10th Edition)

10th Edition

ISBN: 9780321979407

Author: Margaret L. Lial, Raymond N. Greenwell, Nathan P. Ritchey

Publisher: PEARSON

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Chapter 4.3, Problem 23E

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The number of each interview that should be done so as to minimize the time spent.

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Chapter 4.3, Problem 22E

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Chapter 4 Solutions

Finite Mathematics and Calculus with Applications (10th Edition)

Ch. 4.1 - Prob. 1YT Ch. 4.1 - Prob. 2YT Ch. 4.1 - Prob. 1E Ch. 4.1 - Prob. 2E Ch. 4.1 - Prob. 3E Ch. 4.1 - Convert each inequality into an equation by adding...Ch. 4.1 - For Exercises 58, (a) determine the number of...Ch. 4.1 - Prob. 6E Ch. 4.1 - Prob. 7E Ch. 4.1 - Prob. 8E

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The number of each interview that should be done so as to minimize the time spent.

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