Explanation Method used: Solving a Nonstandard Problem: 1. If necessary, convert the problem to a maximization problem. (If the minimization problem has objective function w , rewrite as a maximization problem with the same constraints and objective function z = − w ). 2. Add slack variables and subtract surplus variables as needed. 3. Write the initial simplex tableau. 4. If any basic variable has a negative value, locate the nonzero number in that variable`s column, and note what row it is in. 5. In the row located in Step 4, find the positive entry that is farthest to the left, and note what column it is in. 6. In the column found in Step 5, choose a pivot by investigating quotients. 7. Use the row operations to change the other numbers in the pivot column to 0. 8. Continue Step 4 through 7 until all basic variables are nonnegative. If it never becomes impossible to continue, then the problem has no feasible solution. 9. Once a feasible solution has been found, continue to use the simplex method until the optimal solution is not found. Calculation: Solution of the model using the simplex method is as follows. Let us assume that y 1 be number of ounces of ingredient I, y 2 be the number of ounces of ingredients II, and y 3 be the number of ounces of ingredients III. Observe that the cost of three ingredient I, II, and III are $0.30, $0.09, and $0.27, respectively. So, the cost of three ingredient I, II, and III in 100 parts are $30, $9, and $27 Hence, the objective function cab be written as, Minimize : w = 30 y 1 + 9 y 2 + 27 y 3 . The constraints for the given model obtained as follows. Observe that for the proper performance, the total amount of the additive must be at least 10 oz per gal of gasoline. Therefore, the first constraint is y 1 + y 2 + y 3 ≥ 10 . And for the safety reason, the amount of additive should not exceed 15 oz per gal of gasoline. That is, this amount given by the constraint y 1 + y 2 + y 3 ≤ 15 . Notice that at least 1 4 oz of ingredient I must be used for every ounce of ingredient II, and at least 1 oz of ingredient III must be used for every ounce of ingredient I. That is, the above statement given by the constraints y 1 ≥ 1 4 y 2 and y 3 ≥ y 1 , or the two constraints can be written as − 4 y 1 + y 2 ≤ 0 and y 1 − y 3 ≤ 0 . Combine the objective function and the constraints; the mathematical form of the linear programming problem is, Minimize : w = 30 y 1 + 9 y 2 + 27 y 3 subject t o : y 1 + y 2 + y 3 ≥ 10 y 1 + y 2 + y 3 ≤ 15 − 4 y 1 + y 2 ≤ 0 y 1 − y 3 ≤ 0 with y 1 ≥ 0 , y 2 ≥ 0 , y 3 ≥ 0. Here, the given linear programming is of minimization type. Change the minimization objective function into maximization by taking z = − w . So, the objective function become, Maximize : z = − 30 y 1 − 9 y 2 − 27 y 3 . Note that the constraints in the maximization problem are same as minimization problem. Thus, the equivalent maximization problem is as follows. Maximize : z = − 30 y 1 − 9 y 2 − 27 y 3 subject t o : y 1 + y 2 + y 3 ≥ 10 y 1 + y 2 + y 3 ≤ 15 − 4 y 1 + y 2 ≤ 0 y 1 − y 3 ≤ 0 with y 1 ≥ 0 , y 2 ≥ 0 , y 3 ≥ 0. Note that the above linear programming problem is maximization type. So it is in standard form. Subtract surplus variables and add slack variables the above linear programming can be written as, Maximize : z = − 30 y 1 − 9 y 2 − 27 y 3 subject t o : y 1 + y 2 + y 3 − s 1 = 10 y 1 + y 2 + y 3 + s 2 = 15 − 4 y 1 + y 2 + s 3 = 0 y 1 − y 3 + s 4 = 0 with y 1 ≥ 0 , y 2 ≥ 0 , y 3 ≥ 0 , s i ≥ 0 for i = 1 , 2 , 3 , 4. (1) Multiplied by 100 in second constraint, the system (1) can be written in the system of equation as, y 1 + y 2 + y 3 − s 1 = 10 y 1 + y 2 + y 3 + s 2 = 15 − 4 y 1 + y 2 + s 3 = 0 y 1 − y 3 + s 4 = 0 30 y 1 + 9 y 2 + 27 y 3 + z = 0 . Construct the initial simplex table form the above linear system as follows. [ y 1 y 2 y 3 s 1 s 2 s 3 s 4 z 1 1 1 − 1 0 0 0 0 10 1 1 1 0 1 0 0 0 15 − 4 1 0 0 0 1 0 0 0 1 0 − 1 0 0 0 1 0 0 30 9 27 0 0 0 0 1 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 4.4, Problem 32E

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To find: The mixture of three ingredients that produces the minimum cost of the additive, and how much of the additive should be used per gal of gasoline

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Chapter 4.4, Problem 31E

Chapter 4.4, Problem 33E

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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 mixture of three ingredients that produces the minimum cost of the additive, and how much of the additive should be used per gal of gasoline

Solution Summary: The author explains the simplex method of solving a nonstandard problem by adding slack variables and subtracting surplus variables.

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To find: The mixture of three ingredients that produces the minimum cost of the additive, and how much of the additive should be used per gal of gasoline

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Chapter 4 Solutions