Explanation Assume the number of pills limited to each of the two categories be represented by y 1 and y 2 , respectively. Construct a table in order to fetch the data from the given information as follows. Pill 1 Pill 2 Total Vitamin A 1600 IU 400IU 3200IU Vitamin B 1 1 mg 1 mg 5 mg Vitamin C 20 mg 70 mg 200 mg Cost $0.1 $0.2 Then obtain the first constraint as 1600 y 1 + 400 y 2 ≥ 3200 since he must have atleast 3200IU of vitamin A where pill 1 consists 1600IU and pill 2 contains 400IU of vitamin A. Obtain the second constraint as y 1 + y 2 ≥ 5 since he must have atleast 5mg of vitamin B 1 where pill 1 consists 1mg and pill 2 contains 1mg of vitamin B 1 . Obtain the third constraint as 20 y 1 + 70 y 2 ≥ 200 since he must have atleast 200mg of vitamin C where pill 1 consists 20mg and pill 2 contains 70mg of vitamin C . Obtain the objective function which represents the minimized cost is, w = 0.1 y 1 + 0.2 y 2 The linear programming problem can be stated as follows. Minimize w = 0.1 y 1 + 0.2 y 2 subject to 1600 y 1 + 400 y 2 ≥ 3200 y 1 + y 2 ≥ 5 2 0 y 1 + 70 y 2 ≥ 200 with y 1 ≥ 0 , y 2 ≥ 0 Obtain the dual of the above problem as follows. Maximize z = 3200 x 1 + 5 x 2 + 200 x 3 subject to 1600 x 1 + 1 x 2 + 20 x 3 ≤ 0.1 400 x 1 + 1 x 2 + 70 x 3 ≤ 0.2 with x 1 ≥ 0 , x 2 ≥ 0 , x 3 ≥ 0 Add slack variables to the above constraints and obtain the linear programming problem as follows. Maximize z = 3200 x 1 + 5 x 2 + 200 x 3 subject to 1600 x 1 + 1 x 2 + 20 x 3 + s 1 = 0.1 400 x 1 + 1 x 2 + 70 x 3 + s 2 = 0.2 with x 1 ≥ 0 , x 2 ≥ 0 , x 3 ≥ 0 , s 1 ≥ 0 , s 2 ≥ 0 Rewrite the objective function as z − 3200 x 1 − 5 x 2 − 200 x 3 = 0 . Obtain the simplex tableau, x 1 x 2 x 3 s 1 s 2 z [ 1600 1 20 1 0 0 0.1 400 1 70 0 1 0 0.2 − 3200 − 5 − 200 0 0 1 0 ] . Here, the basic variables are s 1 and s 2 . Thus, the basic solution is s 1 = 0.1 , s 2 = 0.2 , 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 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 x 3 s 1 s 2 z [ 1600 1 20 1 0 0 0.1 400 1 70 0 1 0 0.2 − 3200 − 5 − 200 0 0 1 0 ] 0.1 / 1600 0.2 / 400 Observe that the least ratio is 0.1 1600 . Hence, the second element of the pivot column is the pivot element. x 1 x 2 x 3 s 1 s 2 z [ 1600 1 20 1 0 0 0.1 400 1 70 0 1 0 0.2 − 3200 − 5 − 200 0 0 1 0 ] Perform the row operations and obtain the simplified tableau as follows. x 1 x 2 x 3 s 1 s 2 z [ 1600 1 20 1 0 0 .1 400 1 70 0 1 0 .2 − 3200 − 5 − 200 0 0 1 0 ] − R 1 + 4 R 2 → R 2 2 R 1 + R 3 → R 3 Thus, the simplex tableau becomes, x 1 x 2 x 3 s 1 s 2 z [ 1600 1 20 1 0 0 0.1 0 3 260 − 1 4 0 0.7 0 − 3 − 160 2 0 1 0.2 ] . From the above tableau, the bottom row has two negative entries. So, this is not an optimal solution. Look in the bottom row, the coefficient of x 3 is most negative, so choose third column as a pivot column and bring x 3 into the solution. Thus, divide the entries of the last column by the respective entries of the pivot column as follows. x 1 x 2 x 3 s 1 s 2 z [ 1600 1 20 1 0 0 0

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 28E

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Chapter 4 Solutions

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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