a. Explanation Suppose the number of units limited to each of the two categories of test tubes be represented by y 1 and y 2 , respectively. Construct a table in order to fetch the data from the given information as follows. Small Test Tubes Large Test Tubes Total Stock 900 600 Requirement 2700 Cost $18 $15 Then obtain the first and second constraint as y 1 ≥ 900 and y 2 ≥ 600 since small test tube to be bought by the department is at least 900 and large to be bought is 600 at least. Obtain the third constraint as y 1 + y 2 ≥ 2700 since total test tubes that are to be bought by the department is 2700. Obtain the fourth constraint as y 1 − 2 y 2 ≥ 0 since small test tubes are to be bought twice more than the large test tubes. Obtain the objective function which represents the minimized cost is w = 18 y 1 + 15 y 2 . The linear programming problem can be stated as follows. Minimize w = 18 y 1 + 15 y 2 subject to y 1 ≥ 900 y 2 ≥ 600 y 1 + y 2 ≥ 2700 y 1 − 2 y 2 ≥ 0 with y 1 ≥ 0 , y 2 ≥ 0 Obtain the dual of the above problem as follows. Maximize z = 900 x 1 + 600 x 2 + 2700 x 3 + 0 x 4 subject to x 1 + x 3 + x 4 ≤ 18 x 2 + x 3 − 2 x 4 ≤ 15 with x 1 ≥ 0 , x 2 ≥ 0 , x 3 ≥ 0 , x 4 ≥ 0 Add slack variables to the above constraints and obtain the linear programming problem as follows. Maximize z = 900 x 1 + 600 x 2 + 2700 x 3 + 0 x 4 subject to x 1 + x 3 + x 4 + s 1 = 18 x 2 + x 3 − 2 x 4 + s 2 = 15 with x 1 ≥ 0 , x 2 ≥ 0 , x 3 ≥ 0 , x 4 ≥ 0 , s 1 ≥ 0 , s 2 ≥ 0 Rewrite the objective function is, z − 900 x 1 − 600 x 2 − 2700 x 3 = 0 Obtain the simplex tableau as follows. x 1 x 2 x 3 x 4 s 1 s 2 z [ 1 0 1 1 1 0 0 18 0 1 1 − 2 0 1 0 15 − 900 − 600 − 2700 0 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 , x 3 = 0 , x 4 = 0 , s 1 = 18 , s 2 = 15 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 3 is most negative, so choose fourth column as a pivot column and bring x 3 into the solution. Divide the entries of the last column by the respective entries in the pivot column as follows. x 1 x 2 x 3 x 4 s 1 s 2 z [ 1 0 1 1 1 0 0 18 0 1 1 − 2 0 1 0 15 − 900 − 600 − 2700 0 0 0 1 0 ] 18 / 1 = 18 15 / 1 = 15 Observe that the least ratio is 15. Hence, the second element of the pivot column is chosen as the pivot element. x 1 x 2 x 3 x 4 s 1 s 2 z [ 1 0 1 1 1 0 0 18 0 1 1 − 2 0 1 0 15 − 900 − 600 − 2700 0 0 0 1 0 ] Apply the operation R 1 − R 2 → R 1 in the above tableau as follows. x 1 x 2 x 3 x 4 s 1 s 2 z [ 1 − 1 0 3 1 − 1 0 3 0 1 1 − 2 0 1 0 15 − 900 − 600 − 2700 0 0 0 1 0 ] Apply the operation R 3 + 2700 R 2 → R 3 in the above tableau as follows To determine To compute: The total costs if the minimum number of test tubes is increased.

Finite Mathematics and Calculus with Applications

1st Edition

ISBN: 9781323188361

Author: Margaret Lial

Publisher: Pearson Education

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Chapter 4.3, Problem 20E

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The number of each size of test tube that should be ordered so as to minimize total cost.

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To compute: The total costs if the minimum number of test tubes is increased.

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Chapter 4.3, Problem 19E

Chapter 4.3, Problem 21E

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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 number of each size of test tube that should be ordered so as to minimize total cost.

Solution Summary: The author explains how to determine the number of units limited to each of the two categories of test tubes.

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The number of each size of test tube that should be ordered so as to minimize total cost.

To compute: The total costs if the minimum number of test tubes is increased.

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Chapter 4 Solutions