Explanation Given information: Each pound of Senior Citizen’s Feast needs 0.6 lb of wheat, 0.2 lb of syrup and each pound of Kids Go needs 0.4 lb of wheat, 0.2 lb of syrup and 0.2 lb of sugar. Suppliers can send at most 2800 lb of wheat, at most 800 lb of sugar and at least 1000 lb of syrup. There is a profit of $ 0.90 on each pound of Senior Citizen’s Feast and $ 1 on each pound of Kids Go. Formula used: The Simplex Method for mixed Constraints: 1. If the provided problem is to minimize f , then maximize − f . 2. Make all constraints ≤ by multiplying both sides of any ≥ constraints by − 1 . 3. In the simplex matrix, examine the upper portion of the last column for any negative entries. a. If there are no negative entries, apply the simplex method. b. If there are no negative entries, go to step 4. 4. When there is a negative value in the upper portion of the last column, proceed as follows: a. Select any negative entry in the same row and use its column as the pivot column. b. In the pivot column, compute all quotients for the entries above the line and determine the pivot form the smaller positive quotient. c. After completing the pivot operations, return to step 3. Calculation: Consider the cost function, C = 0.9 x + 1 y And the constraints, 0.6 x + 0.4 y ≤ 2800 0.2 x + 0.2 y ≥ 1000 0.2 y ≤ 800 Simplifying the constraints, 3 x + 2 y ≤ 14000 x + y ≥ 5000 y ≤ 4000 It can be written as, 3 x + 2 y ≤ 14000 − x − y ≤ 5000 y ≤ 4000 Slack variables are used to write the inequalities as equations. The equations after adding slack variables s 1 , s 2 , s 3 are, 3 x + 2 y + s 1 = 14000 − x − y + s 2 = − 5000 y + s 3 = 4000 The constraints can be written in matrix form as follows: [ x y s 1 s 2 s 3 C 3 2 1 0 0 0 14000 − 1 − 1 0 1 0 0 − 5000 0 1 0 0 1 0 4000 − 0.9 − 1 0 0 0 1 0 ] The upper portion of the last column has negative values. If values for the variables x , y , z , s 1 , s 2 , s 3 are computed then s 2 = − 5000 is obtained. This violates the condition of simplex method that all the variables be non-negative. Choosing − 5000 which is row 2. In this row, there are negative entries in column 1 and column 2. Choosing the pivot column as column 2. The pivot is determined from the quotients as, [ x y s 1 s 2 s 3 C 3 2 1 0 0 0 14000 − 1 − 1 0 1 0 0 − 5000 0 1 0 0 1 0 4000 − 0.9 − 1 0 0 0 1 0 ] 14000 / 2 = 7000 − 5000 / − 1 = 5000 4000 / 1 = 4000 The least positive quotient is 4000 . Hence the pivot row is row 3 and the pivot column is column 2. The intersection is 1 which the pivot as shown is, [ x y s 1 s 2 s 3 C 3 2 1 0 0 0 14000 − 1 − 1 0 1 0 0 − 5000 0 1 0 0 1 0 4000 − 0.9 − 1 0 0 0 1 0 ] Pivoting with this entry gives the following matrix, [ x y s 1 s 2 s 3 C 3 0 1 0 − 2 0 6000 − 1 0 0 1 1 0 − 1000 0 1 0 0 1 0 4000 − 0.9 0 0 0 1 1 4000 ] The upper portion of the last column has one negative value

Mathematical Applications for the Management, Life, and Social Sciences

12th Edition

ISBN: 9781337625340

Author: Ronald J. Harshbarger, James J. Reynolds

Publisher: Cengage Learning

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Chapter 4.5, Problem 28E

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Chapter 4.5, Problem 27E

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The number of pounds of cereals Senior Citizen’s Feast and Kids Go to obtain maximum profit and the maximum profit.

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