Explanation Given: The company makes two white wines. The details are reproduced in the following table: Production and profit of two wines Fruity Wine Crystal Wine Total Variables x 1 x 2 Grape A (bushels) 2 1 ≤ 110 Grape B (bushels) 2 3 ≤ 125 Sugar (lb) 2 1 ≤ 90 Profit 12 15 z Explanation: The table summarizes the information of the company. Production and profit of two wines Fruity Wine Crystal Wine Total Variables x 1 x 2 Grape A (bushels) 2 1 ≤ 110 Grape B (bushels) 2 3 ≤ 125 Sugar (lb) 2 1 ≤ 90 Profit 12 15 z From the table above, three constraints are there, each for grape A, grape B, and sugar. The maximization problem can be stated as Maximize z = 12 x 1 + 15 x 2 subject to: 2 x 1 + x 2 ≤ 110 2 x 1 + 3 x 2 ≤ 125 2 x 1 + x 2 ≤ 90 with x 1 ≥ 0 , x 2 ≥ 0 , x 3 ≥ 0 The problem has three constraints, with “at most” ( ≤ ) values, for the objective function. We need three slack variables to make them linear equation. Add slack variable s 1 to 2 x 1 + x 2 ≤ 110 : 2 x 1 + x 2 + s 1 = 110 Add slack variable s 2 to 2 x 1 + 3 x 2 ≤ 125 : 2 x 1 + 3 x 2 + s 2 = 125 Add slack variable s 3 to 2 x 1 + x 2 ≤ 90 : 2 x 1 + x 2 + s 3 = 90 Rewriting the objective function: − 12 x 1 − 15 x 2 + z = 0 Initial simplex tableau is as follows: [ x 1 x 2 s 1 s 2 s 3 z 2 1 1 0 0 0 110 2 3 0 1 0 0 125 2 1 0 0 1 0 90 − 12 − 15 0 0 0 1 0 ] The solutions from the simplex tableau are not optimum. To get maximum of the function, pivot procedure is applied till all the indictors are non-negative and solutions are read from the tableau. Step 1 Find the largest negative indicator (number in the last row), because it indicates the variable with the largest coefficient in the objective function. It becomes the pivot column. From the above tableau − 15 is largest negative number. 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 are 110 1 = 110 125 3 = 41.6 90 1 = 90 Quotient 41.6 is smallest, specifying 3 is the pivot in the2nd row, 2nd column. [ x 1 x 2 s 1 s 2 s 3 z 2 1 1 0 0 0 110 2 3 0 1 0 0 125 2 1 0 0 1 0 90 − 12 − 15 0 0 0 1 0 ] Perform row operations to make all zeros except pivot in x 1 column

11th Edition

ISBN: 9780321979438

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

Publisher: PEARSON

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Chapter 4, Problem 43RE

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Solve the maximization problem for the winery problem in Exercise 39.

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Chapter 4, Problem 42RE

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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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Solve the maximization problem for the winery problem in Exercise 39.

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Chapter 4 Solutions