Explanation Given: The company produces and ships three different plates. The details can have reproduced in the following table, Production and Shipment of three plates Cake plate Wheat plate Dinner plate Per week Variables x 1 x 2 x 3 Material Cost 15 10 8 ≤ 1500 Labor in hours 5 4 4 ≤ 2700 Shipment 6 5 5 ≤ 1200 Profit 15 12 5 z Explanation: The table summarizes the information of the company. Production and Shipment of three plates Cake plate Wheat plate Dinner plate Per week Variables x 1 x 2 x 3 Material Cost 15 10 8 ≤ 1500 Labor in hours 5 4 4 ≤ 2700 Shipment 6 5 5 ≤ 1200 Profit 15 12 5 z Constraints are Material Cost, Labor and Shipment with profit as the objective function. The maximization problem can be stated as Maximize z = 15 x 1 + 12 x 2 + 5 x 3 subject to: 15 x 1 + 10 x 2 + 8 x 3 ≤ 1500 5 x 1 + 4 x 2 + 4 x 3 ≤ 2700 6 x 1 + 5 x 2 + 5 x 3 ≤ 1200 With x 1 ≥ 0 , x 2 ≥ 0 , x 3 ≥ 0 The problem has 3 constraints, with “at most” ( ≤ ) values, for the objective function. We need 3 slack variables to make them linear equation. Adding slack variable s 1 to 15 x 1 + 10 x 2 + 8 x 3 ≤ 1500 , 15 x 1 + 10 x 2 + 8 x 3 + s 1 = 1500 Adding slack variable s 2 to 5 x 1 + 4 x 2 + 4 x 3 ≤ 2700 , 5 x 1 + 4 x 2 + 4 x 3 + s 2 = 2700 Adding slack variable s 3 to 5 x 1 + 4 x 2 + 4 x 3 ≤ 2700 , 6 x 1 + 5 x 2 + 5 x 3 + s 3 = 1200 Rewriting the objective function, − 15 x 1 − 12 x 2 − 5 x 3 + z = 0 Initial simplex tableau is as follows, [ x 1 x 2 x 3 s 1 s 2 s 3 z 15 10 8 1 0 0 0 1500 5 4 4 0 1 0 0 2700 6 5 5 0 0 1 0 1200 − 15 − 12 − 5 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 nonnegative 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: 1500 15 = 100 2700 5 = 540 1200 6 = 200 Quotient 100 is smallest, specifying 15 is the pivot in the 1 st row, 1 st column. [ x 1 x 2 x 3 s 1 s 2 s 3 z 15 10 8 1 0 0 0 1500 5 4 4 0 1 0 0 2700 6 5 5 0 0 1 0 1200 − 15 − 12 − 5 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 41RE

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Solve the maximization problem for the production problem in Exercise 37.

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Chapter 4, Problem 40RE

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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 production problem in Exercise 37.

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Solve the maximization problem for the production problem in Exercise 37.

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Chapter 4 Solutions