Explanation Consider the linear programming problem as follows. Maximize z = 5 x 1 + 3 x 2 subject to 5 x 1 + 2 x 2 ≤ 20 3 x 1 + 6 x 2 ≤ 18 with x 1 ≥ 0 , x 2 ≥ 0 Add slack variables to the above constraints and obtain the linear programming problem as follows. Maximize z = 5 x 1 + 3 x 2 subject to 5 x 1 + 2 x 2 + s 1 = 20 3 x 1 + 6 x 2 + s 2 = 18 with x 1 ≥ 0 , x 2 ≥ 0 , s 1 ≥ 0 , s 2 ≥ 0 Rewrite the objective function as, z − 5 x 1 − 3 x 2 = 0 . The rewritten objective function occupies the last row of the simplex tableau whereas the two linear constraints fill the upper two row as follows. x 1 x 2 s 1 s 2 z [ 5 2 1 0 0 20 3 6 0 1 0 18 − 5 − 3 0 0 1 0 ] Here, the basic variables are s 1 and s 2 . Thus, the basic solution is x 1 = 0 , x 2 = 0 , s 1 = 20 , s 2 = 18 and z = 0 . This is not the optimal solution as only the slack variables are nonzero. Look in the bottom row, the coefficient of x 1 is most negative, so choose first column as a pivot column and bring x 1 into the solution. Divide the entries of the last column by the respective entries in the pivot column as follows. x 1 x 2 s 1 s 2 z [ 5 2 1 0 0 20 3 6 0 1 0 18 − 5 − 3 0 0 1 0 ] → 20 / 5 = 4 → 18 / 3 = 6 Observe that the ratio 4 is the least. Hence, the pivot element is chosen as 5 in the pivot column of the above tableau. x 1 x 2 s 1 s 2 z [ 5 2 1 0 0 20 3 6 0 1 0 18 − 5 − 3 0 0 1 0 ] Simplify the simplex tableau with the help of pivot method until the last row is completely filled with non-negative entries. Apply the operation − 3 R 1 + 5 R 2 → R 2 in the above tableau as follows. x 1 x 2 s 1 s 2 z [ 5 2 1 0 0 20 0 24 − 3 5 0 30 − 5 − 3 0 0 1 0 ] Apply the operation R 3 + R 1 → R 3 in the above tableau, x 1 x 2 s 1 s 2 z [ 5 2 1 0 0 20 0 24 − 3 5 0 30 0 − 1 1 0 1 20 ] From the above tableau, the bottom row has one negative entry that indicates the above tableau does not correspond to optimal solution. Look in the bottom row, the coefficient of x 2 is most negative, so choose second column as a pivot column and bring x 2 into the solution

Finite Mathematics and Calculus with Applications

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ISBN: 9781323188361

Author: Margaret Lial

Publisher: Pearson Education

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Finite Mathematics and Calculus with Applications

Ch. 4.1 - Prob. 1YT Ch. 4.1 - Prob. 2YT Ch. 4.1 - Prob. 1E Ch. 4.1 - Prob. 2E Ch. 4.1 - Prob. 3E Ch. 4.1 - Convert each inequality into an equation by adding...Ch. 4.1 - For Exercises 58, (a) determine the number of...Ch. 4.1 - Prob. 6E Ch. 4.1 - Prob. 7E Ch. 4.1 - Prob. 8E

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The maximization problem using the simplex method.

Solution Summary: The author explains the maximization problem using the simplex method. The maximum value is z=854.

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