Explanation Assume that the number of items limited to each of the three categories be represented by y 1 , y 2 and y 3 respectively. Large bowl Small bowl Pot Total Clay 3 lb 2 lb 4 lb 72 lb Glaze 6 fl 6 fl 2 fl 108 fl Hours 5 6 4 Then obtain the first constraint as 3 y 1 + 2 y 2 + 4 y 3 ≥ 72 since large bowl requires 3 lb of clay, small bowl requires 2 lb of clay and pot requires 4 lb of clay whereas total clay must be 72 lb. Obtain the first constraint as 6 y 1 + 6 y 2 + 2 y 3 ≥ 108 since large bowl requires 6 fl oz of glaze, small bowl requires 6 fl oz of glaze and pot requires 2 fl oz of glaze whereas total glaze must be 108 fl oz of glaze. Obtain the objective function which represents the minimized time is, w = 5 y 1 + 6 y 2 + 4 y 3 . The linear programming problem can be stated as follows. Minimize w = 5 y 1 + 6 y 2 + 4 y 3 subject to 3 y 1 + 2 y 2 + 4 y 3 ≥ 72 6 y 1 + 6 y 2 + 2 y 3 ≥ 108 with y 1 ≥ 0 , y 2 ≥ 0 , y 3 ≥ 0 Obtain the dual of the above problem as follows. Maximize z = 72 x 1 + 108 x 2 subject to 3 x 1 + 6 x 2 ≤ 5 2 x 1 + 6 x 2 ≤ 6 4 x 1 + 2 x 2 ≤ 4 with x 1 ≥ 0 , x 2 ≥ 0 Add slack variables to the above constraints and obtain the linear programming problem as follows. Maximize z = 72 x 1 + 108 x 2 subject to 3 x 1 + 6 x 2 + s 1 = 5 2 x 1 + 6 x 2 + s 2 = 6 4 x 1 + 2 x 2 + s 3 = 4 with x 1 ≥ 0 , x 2 ≥ 0 , s 1 ≥ 0 , s 2 ≥ 0 , s 3 ≥ 0 Rewrite the objective function as z − 72 x 1 − 108 x 2 = 0 . Obtain the simplex tableau, x 1 x 2 s 1 s 2 s 3 z [ 3 6 1 0 0 0 5 2 6 0 1 0 0 6 4 2 0 0 1 0 4 − 72 − 108 0 0 0 1 0 ] . Here, the basic variables are s 1 , s 2 and s 3 . Thus, the basic solution is s 1 = 5 , s 2 = 6 , s 3 = 4 , x 1 = 0 , x 2 = 0 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 2 is most negative, so choose second column as a pivot column and bring x 2 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 s 3 z [ 3 6 1 0 0 0 5 2 6 0 1 0 0 6 4 2 0 0 1 0 4 − 72 − 108 0 0 0 1 0 ] 5 / 6 = 0.83 6 / 6 = 1 4 / 2 = 2 Observe that the least ratio is 0.83. Hence, the first element of the pivot column is chosen as the pivot element. x 1 x 2 s 1 s 2 s 3 z [ 3 6 1 0 0 0 5 2 6 0 1 0 0 6 4 2 0 0 1 0 4 − 72 − 108 0 0 0 1 0 ] Perform the row operations and obtain the simplified tableau as follows. x 1 x 2 s 1 s 2 s 3 z [ 3 6 1 0 0 0 5 2 6 0 1 0 0 6 4 2 0 0 1 0 4 − 72 − 108 0 0 0 1 0 ] R 2 − R 1 → R 2 − R 2 + 3 R 3 → R 3 18 R 1 + R 4 → R 4 That is, x 1 x 2 s 1 s 2 s 3 z [ 3 6 1 0 0 0 5 − 1 0 − 1 1 0 0 1 9 0 − 1 0 3 0 7 − 18 0 18 0 0 1 90 ] From the above tableau, the bottom row has one negative entry

Finite Mathematics and Calculus with Applications (10th Edition)

10th Edition

ISBN: 9780321979407

Author: Margaret L. Lial, Raymond N. Greenwell, Nathan P. Ritchey

Publisher: PEARSON

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Chapter 4.3, Problem 26E

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Chapter 4.3, Problem 25E

Chapter 4.3, Problem 27E

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Finite Mathematics and Calculus with Applications (10th Edition)

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 number of items that should be made to minimize the time and the minimum time.

Solution Summary: The author explains the number of items that should be made to minimize the time and the minimum time. The linear programming problem can be stated as follows.

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Chapter 4 Solutions