Explanation Given: The provided problem is: 3 x 1 + 5 x 2 ≤ 47 x 1 + x 2 ≤ 25 5 x 1 + 2 x 2 ≤ 35 2 x 1 + x 2 ≤ 30 Calculation: From the above information the constraints are as, 3 x 1 + 5 x 2 ≤ 47 x 1 + x 2 ≤ 25 5 x 1 + 2 x 2 ≤ 35 2 x 1 + x 2 ≤ 30 where x 1 , x 2 are all non-negative. The first of these constraints says that 3 x 1 + 5 x 2 ≤ 47 is less than or perhaps equal to 47. Suppose s 1 is the slack variable. Therefore, the first equation is, 3 x 1 + 5 x 2 + s 1 = 47 The second constraints x 1 + x 2 ≤ 25 can be expressed by adding slack variable s 2 . Therefore, the second equation is, x 1 + x 2 + s 2 = 25 The third constraints 5 x 1 + 2 x 2 ≤ 35 can be expressed by adding slack variable s 3 . Therefore, the third equation is, 5 x 1 + 2 x 2 + s 3 = 35 The fourth constraints 2 x 1 + x 2 ≤ 30 can be expressed by adding slack variable s 3 . Therefore, the third equation is, 2 x 1 + x 2 + s 4 = 30 The maximum profit function, z = 2 x 1 + 9 x 2 . Now linear programming problem can be written as follows, The maximum profit z = 2 x 1 + 9 x 2 Subjected to: 3 x 1 + 5 x 2 ≤ 47 x 1 + x 2 ≤ 25 5 x 1 + 2 x 2 ≤ 35 2 x 1 + x 2 ≤ 30 Here, x 1 ≥ 0 , x 2 ≥ 0 , s 1 ≥ 0 , s 2 ≥ 0 , s 3 ≥ 0 , s 4 ≥ 0 Thus, the initial simplex tableau is: x 1 x 2 s 1 s 2 s 3 s 4 [ 3 5 1 0 0 0 47 1 2 0 1 0 0 25 5 2 0 0 1 0 35 2 1 0 0 0 1 30 − 2 − 9 0 0 0 0 0 ] Locate the most negative indicator

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ISBN: 8220101336323

Author: MULLINS

Publisher: PEARSON

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Chapter 7, Problem 30RE

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To calculate: The maximum value of the provided linear programming problem.

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Chapter 7, Problem 29RE

Chapter 7, Problem 31RE

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