Explanation Given: The provided problem is, The maximum profit function z = 5 x 1 + 6 x 2 + 3 x 3 x 1 + x 2 + x 3 ≤ 100 2 x 1 + 3 x 2 ≤ 500 x 1 + 2 x 3 ≤ 350 x 1 ≥ 0 , x 2 ≥ 0 , x 3 ≥ 0 Formula used: Calculation: From the above information, the constrains are as, x 1 + x 2 + x 3 ≤ 100 2 x 1 + 3 x 2 ≤ 500 x 1 + 2 x 3 ≤ 350 Where x 1 , x 2 and x 3 are all nonnegative. The first of these constraints says that x 1 + x 2 + x 3 ≤ 100 is less than or perhaps equal to 100. Suppose s 1 is the slack variable, Therefore, the first equation is: x 1 + x 2 + x 3 + s 1 = 100 The second constraints 2 x 1 + 3 x 2 ≤ 500 can be expressed by adding slack variable s 2 . Therefore, the second equation is, 2 x 1 + 3 x 2 + s 2 = 500 The third constraints x 1 + 2 x 3 ≤ 350 can be expressed by adding slack variable s 3 . Therefore, the third equation is, x 1 + 2 x 3 + s 3 = 350 The maximum profit function z = 5 x 1 + 6 x 2 + 3 x 3 Now linear programming problem can be written as follows, The maximum profit z − 5 x 1 − 6 x 2 − 3 x 3 = 0

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Author: MULLINS

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Chapter 7, Problem 21RE

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To calculate: The slack variable and set up the initial simplex tableau for z=5x1+6x2+3x3.

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Chapter 7, Problem 20RE

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The slack variable and set up the initial simplex tableau for z = 5 x 1 + 6 x 2 + 3 x 3 .

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