Explanation Given: A popular soft drink called Sugarlo, which advertised as having a sugar content of no more than 10%, is blended form five ingredients each of which has some sugar content. The sugar content of the ingredients and their costs per gallon are given below in table. Ingredients 1 2 3 4 5 water Sugar Content (%) 0.28 0.19 0.43 0.57 0.22 0 Cost ($/gal) 0.48 0.32 0.53 0.28 0.43 0.04 Where 1, 2, 3, 4, 5, and water are ingredients. At least 0.01 of the content of Sugarlo must come from ingredients 3 or 4, 0.01 must come from ingredients 2 or 5, 0.01 from ingredients 1 or 4. Method used: Solving a Nonstandard Problem: 1. If necessary, convert the problem to a maximization problem. (If the minimization problem has objective function w , rewrite as a maximization problem with the same constraints and objective function z = − w ). 2. Add slack variables and subtract surplus variables as needed. 3. Write the initial simplex tableau. 4. If any basic variable has a negative value, locate the nonzero number in that variable`s column, and note what row it is in. 5. In the row located in Step 4, find the positive entry that is farthest to the left, and note what column it is in. 6. In the column found in Step 5, choose a pivot by investigating quotients. 7. Use the row operations to change the other numbers in the pivot column to 0. 8. Continue Step 4through 7 until all basic variables are nonnegative. If it never becomes impossible to continue, then the problem has no feasible solution. 9. Once, a feasible solution has been found, continue to use the simplex method until the optimal solution is not found. Calculation: Solution of the model using the simplex method is as follows. Let us assume that y 1 be the number of gallons of ingredient 1, y 2 be the number of gallons of ingredient 2, y 3 be the number of gallons of ingredient 3, y 4 be the number of gallons of ingredient 4, y 5 be the number of gallons of ingredient 5, and y 6 be the number of gallons of water. Construct the objective function and the corresponding constraints as follows. The sugar content of the ingredients and their costs per gallon are given below in table. Ingredients y 1 y 2 y 3 y 4 y 5 y 6 Sugar Content (%) 0.28 0.19 0.43 0.57 0.22 0 Cost ($/gal) 0.48 0.32 0.53 0.28 0.43 0.04 Hence the objective function that minimize the total cost is, Minimize : w = 0.48 y 1 + 0.32 y 2 + 0.53 y 3 + 0.28 y 4 + 0.43 y 5 + 0.04 y 6 . Observe that the popular soft drink called Sugarlo, which advertised as having a sugar content of no more than 10%, is blended form five ingredients. That is the 10% of the 15000 is 15000 × 10 100 = 1500 . Similarly, 0.01 of the 15000 is 15000 × 0.01 = 150 . Hence, the constraint that gives 10% of the 15000 blended sugar is 0.28 y 1 + 0.19 y 2 + 0.43 y 3 + 0.57 y 4 + 0.22 y 5 ≤ 1500 . Also given that at least 0.01 of the content of Sugarlo must come from ingredient s 3 or 4, 0.01 must come from ingredients 2 or 5, 0.01 from ingredients 1 or 4. Hence, the constraints are y 3 + y 4 ≥ 150 , y 2 + y 5 ≥ 150 , and y 1 + y 4 ≥ 150 . The amount of each ingredient should be used in preparing 15000 gal of Sugarlo. That is y 1 + y 2 + y 3 + y 4 + y 5 + y 6 = 15000 . Combine the objective function and the constraints; the mathematical form of the linear programming problem is, Minimize : w = 0.48 y 1 + 0.32 y 2 + 0.53 y 3 + 0.28 y 4 + 0.43 y 5 + 0.04 y 6 subject t o : 0.28 y 1 + 0.19 y 2 + 0.43 y 3 + 0.57 y 4 + 0.22 y 5 ≤ 1500 y 3 + y 4 ≥ 150 y 2 + y 5 ≥ 150 y 1 + y 4 ≥ 150 y 1 + y 2 + y 3 + y 4 + y 5 + y 6 = 15 , 000 with y 1 ≥ 0 , y 2 ≥ 0 , y 3 ≥ 0. y 4 ≥ 0 , y 5 ≥ 0 , y 6 ≥ 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 = − 0.48 y 1 − 0.32 y 2 − 0.53 y 3 − 0.28 y 4 − 0.43 y 5 − 0.04 y 6 . Note that the constraints in the maximization problem are same as minimization problem. Thus, the equivalent maximization problem is as follows. Maximize : z = − 0.48 y 1 − 0.32 y 2 − 0.53 y 3 − 0.28 y 4 − 0.43 y 5 − 0.04 y 6 subject t o : 0.28 y 1 + 0.19 y 2 + 0.43 y 3 + 0.57 y 4 + 0.22 y 5 ≤ 1500 y 3 + y 4 ≥ 150 y 2 + y 5 ≥ 150 y 1 + y 4 ≥ 150 y 1 + y 2 + y 3 + y 4 + y 5 + y 6 = 15 , 000 with y 1 ≥ 0 , y 2 ≥ 0 , y 3 ≥ 0. y 4 ≥ 0 , y 5 ≥ 0 , y 6 ≥ 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 30E

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Chapter 4.4, Problem 29E

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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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