Explanation Given: The given problem is, Maximize P = x + 4 y subject to x + 3 y ≤ 6 − 2 x + 3 y ≤ − 6 x ≥ 0 , y ≥ 0 Approach: 1. In the standard maximization problem the objective function is to be maximized. 2. All variable involved in the problem should be non negative. 3. Each linear constraint may be written so that the expression involving the variables is less than or equal to a non negative constraint. 4. The optimal solution is reached when all the entries of bottom row of the simplex tableau are non negative. Calculations: Introduce the slack variable u and v in the given linear programming problem, x + 3 y + u = 6 − 2 x + 3 y + v = − 6 − x − 4 y + P = 0 The initial simplex tableau is, x y u v P Constant 1 ( 3 ) 1 0 0 6 − 2 3 0 1 0 − 6 − 1 − 4 0 0 1 0 Table ( 1 ) At x = 0 , y = 0 we have u = 6 v = − 6 , P = 0 In the initial simplex tableau the minimum value for bottom row is − 4 so pivot this column, The pivoted element is 3 so now apply 1 3 R 1

Finite Mathematics for the Managerial, Life, and Social Sciences, 11th Edition

11th Edition

ISBN: 9781285464657

Author: Soo T. Tan

Publisher: Brooks Cole

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Chapter 4.3, Problem 9E

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The linear programming problem.

Maximize P=x+4ysubject to x+3y≤6 −2x+3y≤−6 x≥0,y≥0

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Chapter 4.3, Problem 8E

Chapter 4.3, Problem 10E

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Chapter 4 Solutions

Finite Mathematics for the Managerial, Life, and Social Sciences, 11th Edition

Ch. 4.1 - Give the three characteristics of a standard...Ch. 4.1 - a. When the initial simplex tableau is set up, how...Ch. 4.1 - In the simplex method, how is a pivoted column...Ch. 4.1 - In Exercises 1-6, a write the linear programming...Ch. 4.1 - In Exercises 1-6, a write the linear programming...Ch. 4.1 - In Exercises 1-6, a write the linear programming...Ch. 4.1 - In Exercises 1-6, a write the linear programming...Ch. 4.1 - In Exercises 1-6, a write the linear programming...Ch. 4.1 - Prob. 6E Ch. 4.1 - In Exercises 7-16, determine whether the given...

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The linear programming problem. Maximize P = x + 4 y subject to x + 3 y ≤ 6 − 2 x + 3 y ≤ − 6 x ≥ 0 , y ≥ 0

Solution Summary: The author explains that the optimal solution for the linear programming problem is x=4,y=23.

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To solve:

The linear programming problem.

Maximize P=x+4ysubject to x+3y≤6 −2x+3y≤−6 x≥0,y≥0

Want to see the full answer?

Chapter 4 Solutions

The linear programming problem. Maximize P = x + 4 y subject to x + 3 y ≤ 6 − 2 x + 3 y ≤ − 6 x ≥ 0 , y ≥ 0

Solution Summary: The author explains that the optimal solution for the linear programming problem is x=4,y=23.

Videos

To solve: The linear programming problem. Maximize P=x+4ysubject to x+3y≤6 −2x+3y≤−6 x≥0,y≥0

Want to see the full answer?

Chapter 4 Solutions

To solve:

The linear programming problem.

Maximize P=x+4ysubject to x+3y≤6 −2x+3y≤−6 x≥0,y≥0