Explanation Given: There are three kinds of sets, namely Royal Flush set, Deluxe Diamond set and the Full House set. There are 1000 poker chips, 4 decks cards, 10 dice and 2 dealer buttons in the Royal Flush set. In the case of Deluxe Diamond there are 600 poker chips, 2 decks of cards, 5 dice and 1 dealer button. For Full House poker set, there are 300 poker chips, 2 decks of cards, 5 dice and 1 dealer button. The company has a total of 2 , 8000 , 000 poker chips, 10 , 000 decks of cards, 25 , 000 dice and 6000 dealer buttons in stock. For set of Royal poker, the profit is $ 38 , for Deluxe Diamond, the profit amount is $ 22 and in the case of Full House poker set, the profit is $ 12 . Calculation: Suppose x 1 is the number of Royal Flush sets, x 2 is the number of Deluxe Diamond sets and x 3 is the number of Full House sets. z is the maximum profit. The summaries information are as follows, Poker chips decks chips dice dealer buttons Maximum profit on each sets Royal Flush sets ( x 1 ) 1000 4 10 2 $ 38 Deluxe Diamond sets ( x 2 ) 600 2 5 1 $ 22 Full House sets ( x 3 ) 300 2 5 1 $ 12 Total number of stocks 2 , 800 , 000 10 , 000 25 , 000 6000 From the above information the constraints are as, 1000 x 1 + 600 x 2 + 300 x 3 ≤ 2 , 800 , 000 4 x 1 + 2 x 2 + 2 x 3 ≤ 10 , 000 10 x 1 + 5 x 2 + 5 x 3 ≤ 25 , 000 2 x 1 + x 2 + x 3 ≤ 6 , 000 Where x 1 , x 2 and x 3 are all nonnegative. The first of these constraints says that 1000 x 1 + 600 x 2 + 300 x 3 ≤ 2 , 800 , 000 is less than or perhaps equal to 2 , 800 , 000 . Suppose s 1 is the slack variable, Therefore, the first equation is 1000 x 1 + 600 x 2 + 300 x 3 + s 1 = 2 , 800 , 000 The second constraints 4 x 1 + 2 x 2 + 2 x 3 ≤ 10 , 000 can be expressed by adding slack variable s 2 . Therefore, the second equation is, 4 x 1 + 2 x 2 + 2 x 3 + s 2 = 10 , 000 The third constraints 10 x 1 + 5 x 2 + 5 x 3 ≤ 25 , 000 can be expressed by adding slack variable s 3 . Therefore, the third equation is, 10 x 1 + 5 x 2 + 5 x 3 + s 3 = 25 , 000 The fourth constraints 2 x 1 + x 2 + x 3 ≤ 6 , 000 can be expressed by adding slack variable s 4 . Therefore, the third equation is, 2 x 1 + x 2 + x 3 + s 4 = 6 , 000 The maximum profit function z = 38 x 1 + 22 x 2 + 12 x 3 Now linear programming problem can be written as follows, The maximum profit z = 38 x 1 + 22 x 2 + 12 x 3 Subjected to: 1000 x 1 + 600 x 2 + 300 x 3 ≤ 2 , 800 , 000 4 x 1 + 2 x 2 + 2 x 3 ≤ 10 , 000 10 x 1 + 5 x 2 + 5 x 3 ≤ 25 , 000 2 x 1 + x 2 + x 3 ≤ 6 , 000 Here, x 1 ≥ 0 , x 2 ≥ 0 , x 3 ≥ 0 , s 1 ≥ 0 , s 2 ≥ 0 , s 3 ≥ 0 , s 4 ≥ 0 Thus, the initial simplex tableau is x 1 x 2 x 3 s 1 s 2 s 3 s 4 [ 1000 600 300 1 0 0 0 2 , 800 , 000 4 2 2 0 1 0 0 10 , 000 10 5 5 0 0 1 0 25 , 000 2 1 1 0 0 0 1 6000 − 38 − 22 − 12 0 0 0 0 0 ] The pivot is 10 , in the third row and first column. Multiply third row by ‘ 1 10 ’ (b) To determine To calculate: The value of slack variable.

Mathematics with Applications In the Management, Natural, and Social Sciences (12th Edition)

12th Edition

ISBN: 9780134767628

Author: Margaret L. Lial, Thomas W. Hungerford, John P. Holcomb, Bernadette Mullins

Publisher: PEARSON

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Chapter 7.5, Problem 10E

To determine

To calculate: Number of each type of set to be assembled to get maximum profit and also amount of maximum profit.

(b)

To determine

To calculate: The value of slack variable.

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Chapter 7.5, Problem 9E

Chapter 7.5, Problem 11E

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Mathematics with Applications In the Management, Natural, and Social Sciences (12th Edition)

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