Explanation Given: Minimize w = 24 y 1 + 30 y 2 + 36 y 3 Subject to 5 y 1 + 10 y 2 + 15 y 3 ≥ 1200 y 1 + y 2 + y 3 ≤ 50 with y 1 ≥ 0 , y 2 ≥ 0 , y 3 ≥ 0 Explanation: The problem has 2 constraints for the objective function. We need 1 surplus variable and 1 slack variable to make them linear equation. Convert objective function to maximize z = − w = − 24 y 1 − 30 y 2 − 36 y 3 The problem can be restated in maximum as Maximize z = − 24 y 1 − 30 y 2 − 36 y 3 Subject to 5 x 1 + 10 x 2 + 15 x 3 ≥ 1200 x 1 + x 2 + x 3 ≤ 50 with x 1 ≥ 0 , x 2 ≥ 0 , x 3 ≥ 0 Subtracting surplus variable s 1 : 5 x 1 + 10 x 2 + 15 x 3 ≥ 1200 Thus 5 x 1 + 10 x 2 + 15 x 3 − s 1 = 1200 Adding slack variable s 2 : x 1 + x 2 + x 3 ≤ 50 Thus x 1 + x 2 + x 3 + s 2 = 50 Before building tableau, all the variables should be on the left side of the equation and constants on the right. Rewriting the objective function: 24 y 1 + 30 y 2 + 36 y 3 + z = 0 Initial simplex tableau is as follows: [ x 1 x 2 x 3 s 1 s 2 z 5 10 15 − 1 0 0 1200 1 1 1 0 1 0 50 24 30 36 0 0 1 0 ] To get the maximum of the function, variable s 1 should be zero. The farthest left of − 1 is 5, which is the pivot column. The smallest quotient formed by dividing each number in the pivot column from the right side of the tableau of the corresponding row becomes the pivot row. Corresponding quotients: 1200 5 = 240 50 1 = 50 Quotient 50 is the smallest, specifying 1 as the pivot in the 2nd row, 1st column: [ x 1 x 2 x 3 s 1 s 2 z 5 10 15 − 1 0 0 1200 1 1 1 0 1 0 50 24 30 36 0 0 1 0 ] Perform row operations to make all zeros except pivot in the x 1 column. Row 1: R 1 − 5 R 2 Row 3: R 3 − 24 R 2 [ x 1 x 2 x 3 s 1 s 2 z 0 5 10 − 1 − 5 0 950 1 1 1 0 1 0 50 0 6 12 0 − 24 1 − 1200 ] s 1 = − 950

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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Solve the problem using the simplex method

Solution Summary: The author explains that the minimization problem has no feasible solution. The problem needs 1 surplus variable and 1 slack variable to make them linear equations.

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