Explanation Given: Objective function z = 5 x 1 + 7 x 2 subjected to x 1 + x 2 = 15 2 x 1 + 4 x 2 ≥ 30 3 x 1 + 5 x 2 ≥ 10 where x 1 ≥ 0 , x 2 ≥ 0 Explanation: The corresponding system of equations requires surplus variables s 1 , s 2 and an artificial variable a 1 . The system becomes x 1 + x 2 + a 1 = 15 2 x 1 + 4 x 2 − s 1 = 30 3 x 1 + 5 x 2 − s 2 = 10 Also x 1 ≥ 0 , x 2 ≥ 0 Produce a tableau, given by: [ x 1 x 2 a 1 s 1 s 2 z 1 1 1 0 0 0 15 2 4 0 − 1 0 0 30 3 5 0 0 − 1 0 10 − 5 − 7 0 0 0 1 0 ] _ In this case, s 1 , s 2 , and a 1 are the basic variables, but since s 1 = − 30 , choose the positive entry farthest on the left in the second row. Apply test ratio for pivot: 15 1 = 15 , 30 2 = 15 , 10 3 = 3.33 Therefore, 3 is the pivot as the least positive ratio is obtained for this element. Use the first element in the third row as the pivot. Use row operation 1 3 R 3 → R 3 to change the pivot to 1. Use row operations R 1 − R 3 → R 1 , R 2 − 2 R 3 → R 2 , R 4 + 5 R 3 → R 4 to change all the remaining elements in this column to 0. [ x 1 x 2 a 1 s 1 s 2 z 1 1 1 0 0 0 15 2 4 0 − 1 0 0 30 1 5 3 0 0 − 1 3 0 10 3 − 5 − 7 0 0 0 1 0 ] [ x 1 x 2 a 1 s 1 s 2 z 0 − 2 3 1 0 1 3 0 35 3 0 2 3 0 − 1 2 3 0 70 3 1 5 3 0 0 − 1 3 0 10 3 0 4 3 0 0 − 5 3 1 50 3 ] _ Since the basic variable s 1 is negative, choose an entry farthest on the left in the second row. Now to find the pivot element in the second column, use ratio test. 35 3 − 2 3 = − 17.5 , 70 3 2 3 = 35 , 10 3 5 3 = 2 Therefore, 5 3 is the pivot, as the least positive ratio is obtained from this element. Use the third row as the pivot row and the second column as the pivot column. Use row operation 3 5 R 1 → R 1 to change the pivot to 1. Use row operations R 1 + 2 3 R 2 → R 2 , R 2 − 2 3 R 3 → R 2 , R 4 − 4 3 R 3 → R 4 to change all the remaining elements in this column to 0. [ x 1 x 2 a 1 s 1 s 2 z 0 − 2 3 1 0 1 3 0 35 3 0 2 3 0 − 1 2 3 0 70 3 3 5 1 0 0 − 1 5 0 2 0 4 3 0 0 − 5 3 1 50 3 ] [ x 1 x 2 a 1 s 1 s 2 z 2 5 0 1 0 1 5 0 13 − 2 5 0 0 − 1 4 5 0 22 3 5 1 0 0 − 1 5 0 2 − 4 5 0 0 0 − 7 5 1 14 ] _ Since the last row consists of negative numbers, take the number that is the smallest (greatest number when the sign gets reversed)

11th Edition

ISBN: 9780321979438

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

Publisher: PEARSON

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Chapter 4 Solutions

Finite Mathematics (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 = 5 x 1 + 7 x 2 .

Solution Summary: The author explains that the objective function is underset_z=105, which attains its value when the variables are

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