Explanation Given: Maximize z = 10 x 1 + 12 x 2 Subject to: 2 x 1 + 2 x 2 = 17 2 x 1 + 5 x 2 ≥ 22 4 x 1 + 3 x 2 ≥ 28 With x 1 ≥ 0 , x 2 ≥ 0 Explanation: The problem has three constraints for the objective function. We need 2 slack variables and 1 artificial variable to make them linear equation. Adding artificial variable a 1 , 2 x 1 + 2 x 2 = 17 2 x 1 + 2 x 2 + a 1 = 17 Subtracting surplus variable s 1 , 2 x 1 + 5 x 2 ≥ 22 2 x 1 + 5 x 2 − s 1 = 22 Adding surplus variable s 2 , 4 x 1 + 3 x 2 ≥ 28 4 x 1 + 3 x 2 + s 2 = 28 Before building tableau, all the variables should be on left side of the equation and constants on right. Rewriting the objective function, − 10 x 1 − 12 x 2 + z = 0 Initial simplex tableau is as follows, [ x 1 x 2 a 1 s 1 s 2 z 2 2 1 0 0 0 17 2 5 0 − 1 0 0 22 4 3 0 0 1 0 28 − 10 − 12 0 0 0 1 0 ] To get maximum of the function, pivot procedure is applied till all the indictors are nonnegative and solutions are read from the tableau. Step 1: Artificial variable a 1 should be zero. Pivot column is farthest nonnegative value in the a 1 row and the smallest quotient formed by dividing each number in the pivot column from the right side of the tableau of the corresponding row, becomes pivot. Corresponding quotients, 17 2 = 8.5 22 2 = 11 28 4 = 7 [ x 1 x 2 a 1 s 1 s 2 z 2 2 1 0 0 0 17 2 5 0 − 1 0 0 22 4 3 0 0 1 0 28 − 10 − 12 0 0 0 1 0 ] Perform row operations to make all zeros except pivot in x 1 column. Row 1: 2 R 1 − R 3 Row 2: 2 R 2 − R 3 Row 4: 2 R 4 + 5 R 3 [ x 1 x 2 a 1 s 1 s 2 z 0 1 2 0 − 1 0 6 0 7 0 − 2 − 1 0 16 4 3 0 0 1 0 28 0 − 9 0 0 5 2 140 ] From the above tableau a 1 = 3 .Repeating step 2 to make it zero. Step 2: The smallest quotient formed by dividing each number in the pivot column from the right side of the tableau of the corresponding row, becomes pivot row. Corresponding quotients: 6 1 = 6 16 7 = 2.28 28 3 = 9.33 Quotient 2.28 is smallest, specifying 7 is the pivot in the 2 nd row, 2 nd column. [ x 1 x 2 a 1 s 1 s 2 z 0 1 2 0 − 1 0 6 0 7 0 − 2 − 1 0 16 4 3 0 0 1 0 28 0 − 9 0 0 5 2 140 ] Perform row operations to make all zeros except pivot in x 2 column. Row 1: 7 R 1 − R 2 Row 3: 7 R 3 − 3 R 2 Row 4: 7 R 4 + 9 R 2 [ x 1 x 2 a 1 s 1 s 2 z 0 0 14 2 − 6 0 26 0 7 0 − 2 − 1 0 16 28 0 0 6 10 0 148 0 0 0 − 18 26 14 1124 ] From the above tableau a 1 = 13 7

Finite Mathematics, Books a la Carte Plus MyLab Math Access Card Package (11th Edition)

11th Edition

ISBN: 9780133886818

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

Publisher: PEARSON

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Finite Mathematics, Books a la Carte Plus MyLab Math Access Card Package (11th Edition)

Ch. 4.1 - Convert each inequality into an equation by adding...Ch. 4.1 - Prob. 2E Ch. 4.1 - Convert each inequality into an equation by adding...Ch. 4.1 - Prob. 4E Ch. 4.1 - For Exercises 5-8. (a) determine the number of...Ch. 4.1 - Prob. 6E Ch. 4.1 - For Exercises 5-8, (a) determine the number of...Ch. 4.1 - For Exercises 5-8, (a) determine the number of...Ch. 4.1 - Introduce slack variables as necessary, then...Ch. 4.1 - Introduce slack variables as necessary, then write...

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Solution Summary: The author explains that the problem has three constraints for the objective function: we need 2 slack variables and 1 artificial variable.

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