Explanation Given: Consider the linear programming problem as follows. Minimize w = 15 y 1 + 12 y 2 subject to 3 y 1 + 5 y 2 ≥ 20 3 y 1 + y 2 ≥ 18 with y 1 ≥ 0 , y 2 ≥ 0 Calculation: Obtain the augmented matrix for the given linear programming problem. [ 3 5 20 3 1 18 15 12 0 ] Observe that the last row is the coefficients of the objective function w = 15 y 1 + 12 y 2 and the upper two rows are the coefficients of the two linear constraints 3 y 1 + 5 y 2 ≥ 20 and 3 y 1 + y 2 ≥ 18 with their constants. Obtain the transpose of the augmented matrix obtained above as follows. [ 3 3 15 5 1 12 20 18 0 ] Observe that the last row becomes the coefficients of the objective function that is z = 20 x 1 + 18 x 2 and the upper two rows becomes the coefficients as well as constant of the two linear constraints that are 3 x 1 + 3 x 2 ≤ 15 and 5 x 1 + x 2 ≤ 12 . Hence, the dual of the linear programming problem is obtained as follows. Maximize z = 20 x 1 + 18 x 2 subject to 3 x 1 + 3 x 2 ≤ 15 5 x 1 + x 2 ≤ 12 with x 1 ≥ 0 , x 2 ≥ 0 Add slack variable to the above linear programming problem. Maximize z = 20 x 1 + 18 x 2 subject to 3 x 1 + 3 x 2 + s 1 = 15 5 x 1 + x 2 + s 2 = 12 with x 1 ≥ 0 , x 2 ≥ 0 , s 1 ≥ 0 , s 2 ≥ 0 Rewrite the objective function as, z − 20 x 1 − 18 x 2 = 0 . Now, obtain the simplex tableau, x 1 x 2 s 1 s 2 z [ 3 3 1 0 0 15 5 1 0 1 0 12 − 20 − 18 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 , s 1 = 15 , s 2 = 12 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 1 is most negative, so choose first column as a pivot column and bring x 1 into the solution. Divide the entries of the last column by the respective entries in the pivot column as follows. x 1 x 2 s 1 s 2 z [ 3 3 1 0 0 15 5 1 0 1 0 12 − 20 − 18 0 0 1 0 ] → 15 / 3 → 12 / 5 Observe that 12 5 is the least. Hence, the pivot element is chosen as the second element of the pivot column in the above tableau. x 1 x 2 s 1 s 2 z [ 3 3 1 0 0 15 5 1 0 1 0 12 − 20 − 18 0 0 1 0 ] Simplify the simplex tableau with the help of pivot method until the last row is completely filled with non-negative entries. Apply the operation 5 R 1 − 3 R 2 → R 1 in the above tableau as follows. x 1 x 2 s 1 s 2 z [ 0 12 5 − 3 0 39 5 1 0 1 0 12 − 20 − 18 0 0 1 0 ] Apply the operation R 3 + 4 R 2 → R 3 in the above tableau as follows

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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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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Solution Summary: The author explains how to solve the minimization problem by obtaining the augmented matrix for the linear programming problem.

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