Explanation Given: Minimize w = 3 y 1 + 2 y 2 + 3 y 3 Subject to: 2 y 1 + 3 y 2 + 6 y 3 ≤ 60 y 1 + 4 y 2 + 5 y 3 ≥ 40 with y 1 ≥ 0 , y 2 ≥ 0 , y 3 ≥ 0 Explanation: Consider the problem: Minimize w = 3 y 1 + 2 y 2 + 3 y 3 Subject to: 2 y 1 + 3 y 2 + 6 y 3 ≤ 60 y 1 + 4 y 2 + 5 y 3 ≥ 40 with y 1 ≥ 0 , y 2 ≥ 0 , y 3 ≥ 0 Since this is a nonstandard minimization problem, convert it into a maximization problem by letting z equal to negative of the objective function: z = − w and then find the maximum value of : z = − 3 y 1 − 2 y 2 − 3 y 3 The problem is as follows: Maximize z = − 3 y 1 − 2 y 2 − 3 y 3 Subject to: 2 y 1 + 3 y 2 + 6 y 3 ≤ 60 y 1 + 4 y 2 + 5 y 3 ≥ 40 with y 1 ≥ 0 , y 2 ≥ 0 , y 3 ≥ 0 Now adding slack and surplus variables in to the constraints gives: Maximize z = − 3 y 1 − 2 y 2 − 3 y 3 Subject to: 2 y 1 + 3 y 2 + 6 y 3 + s 1 + 0 s 2 = 60 y 1 + 4 y 2 + 5 y 3 + 0 s 1 − s 2 = 40 3 y 1 + 2 y 2 + 3 y 3 + z = 0 The initial simplex tableau is: y 1 y 2 y 3 s 1 s 2 z [ 2 3 6 1 0 0 60 1 4 5 0 − 1 0 40 3 2 3 0 0 1 0 ] _ The solution is y 1 = 0 , y 2 = 0 , y 3 = 0 , s 1 = 60 , and s 2 = − 40 , since s 2 is negative, solution is not feasible. Since s 2 has a − 1 in row 2 and the positive entry farthest to the left in row 2 is 1 in column 1. Now: min { 60 2 , 40 1 } = min { 30 , 40 } = 30 Since 2 gives the smallest quotient, so 2 becomes the pivot in column 1. Perform row operation to change the other number in the pivot column to 0. R 1 → 1 2 R 1 gives, y 1 y 2 y 3 s 1 s 2 z [ 1 3 2 3 1 2 0 0 30 1 4 5 0 − 1 0 40 3 2 3 0 0 1 0 ] _ R 2 → R 2 + ( − 1 ) R 1 , R 3 → R 3 + ( − 1 ) R 1 gives, y 1 y 2 y 3 s 1 s 2 z [ 1 3 2 3 1 3 0 0 30 0 5 2 2 − 1 2 − 1 0 10 0 − 5 2 − 6 − 3 2 0 1 − 90 ] _ The basic solution is y 1 = 30 , y 2 = 0 , y 3 = 0 , s 1 = 0 , and s 2 = − 10 .Again, s 2 is negative, solution is not feasible. s 1 and s 2 have a − 1 2 , − 1 in row 2, respectively, and the positive entry farthest to the left in row 2 is 5 2 in column 2. Now min { 30 3 2 , 10 5 2 } = min { 20 , 4 } = 4 Since 5 2 gives the smallest quotient, so 5 2 becomes the pivot. Perform row operation to change the other number in the pivot column to 0. R 2 → 2 5 R 2 gives, y 1 y 2 y 3 s 1 s 2 z [ 1 3 2 3 1 3 0 0 30 0 1 4 5 − 1 5 − 2 5 0 4 0 − 5 2 − 6 − 3 2 0 1 − 90 ] _ R 1 → R 1 + ( − 3 2 ) R 2 , R 3 → R 3 + ( 5 2 ) R 2 gives, y 1 y 2 y 3 s 1 s 2 z [ 1 0 9 5 4 5 3 5 0 24 0 1 4 5 − 1 5 − 2 5 0 4 0 0 − 4 − 2 − 1 1 − 80 ] The basic solution is y 1 = 24 , y 2 = 4 , y 3 = 0 , s 1 = 0 , and s 2 = 0 , a feasible solution but not optimal solution because some indicators are negative

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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