Explanation Given: Objective function z = 3 x 1 + 2 x 2 subjected to x 1 + x 2 = 50 4 x 1 + 2 x 2 ≥ 120 5 x 1 + 2 x 2 ≤ 200 where x 1 ≥ 0 , x 2 ≥ 0 Explanation: Corresponding system of equations requires slack variables s 1 , s 2 and an artificial variable a 1 . The system becomes 5 x 1 + 2 x 2 + s 1 = 200 4 x 1 + 2 x 2 − s 2 = 120 x 1 + x 2 + a 1 = 50 z − 3 x 1 − 2 x 2 = 0 Also, x 1 ≥ 0 , x 2 ≥ 0 Produce a tableau given by [ x 1 x 2 s 1 s 2 a 1 z 5 2 1 0 0 0 200 4 2 0 − 1 0 0 120 1 1 0 0 1 0 50 − 3 − 2 0 0 0 1 0 ] _ In this case, s 1 and s 2 are the basic variables, but s 2 = − 200 chose the positive entry farthest on the left in the second row. Apply test ratio for pivot: 200 5 = 40 , 120 4 = 30 , 50 1 = 50 Therefore, 4 is the pivot as the least positive ratio is obtained for this element. Use the first element in the second row as the pivot. Use the row operation 1 4 R 2 → R 2 to change the pivot to 1. Use row operations R 1 − 5 R 2 → R 1 , R 3 − R 2 → R 3 , R 4 + 3 R 2 → R 4 to change all the remaining elements in this column to 0. [ x 1 x 2 s 1 s 2 a 1 z 0 2 1 0 0 0 200 1 1 2 0 − 1 4 0 0 30 0 1 0 0 1 0 50 0 − 2 0 0 0 1 0 ] [ x 1 x 2 s 1 s 2 a 1 z 0 − 1 2 1 5 4 0 0 50 1 1 2 0 − 1 4 0 0 30 0 1 2 0 1 4 1 0 20 0 − 1 2 0 − 3 4 0 1 90 ] _ Since the last row consists of negative numbers, take the number that is the smallest (greatest number when the sign gets reversed). Therefore, the most negative number is − 3 4 ; now, to find the pivot element in this column, use ratio test: 50 5 4 = 40 , 30 − 1 4 = − 120 , 20 1 4 = 80 Therefore, 5 4 is the pivot, as the least positive ratio is obtained from this element. Use the first row as the pivot row and the third column as the pivot column. Use the row operation 4 5 R 1 → R 1 to change the pivot to 1. Use row operations R 2 + 1 4 R 1 → R 2 , R 3 − 1 4 R 1 → R 3 , R 4 + 3 4 R 1 → R 4 to change all the remaining elements in this column to 0

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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Chapter 4.4, Problem 17E

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Chapter 4 Solutions

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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Maximum of z = 3 x 1 + 2 x 2 .

Solution Summary: The author explains that the objective function's maximum value is underset_z=4003.

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Maximum of z=3x1+2x2.

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