Explanation Given Information: The provided constraints are x + 2 y ≥ 20 2 x + y ≥ 10 x ≥ 0 , y ≥ 0 Formula used: To convert the minimization problem to the maximization problem, multiply the objective function by − 1 . To compute solution of general linear programming problems, use slack and surplus variables to covert inequalities into equations and star the rows for which the active variables have negative value. If there are starred rows, use Phase I to proceed. Phase I: To remove the stars that is to make active variables have positive values. Start from the first starred row and locate the largest positive number and pivot the column, use test ratios and the smallest test ratio corresponds to a pivot and clear the pivot column. Check again and star the rows for which the active variables have negative value and repeat the process until there are no starred rows and move to Phase II. Phase II: Use the simplex method to solve the obtained standard maximization tableau. Choose the smallest negative number on the left side of the bottom row and pivot the column. Select the pivot which is the smallest of the test ratios a b , where a is the entry in the right most column and b is the corresponding entry in the pivot column. Clear the pivot column using the pivot and label the pivot row with the label of the pivot column. Repeat till all the entries in the bottom row are non negative. If there are no negative entries in the bottom row of the standard maximization tableau, then the obtained tableau represents the solution to the problem. Use row operations to clear the column of the augmented matrix. Calculation: Consider the objective function c = 6 x + 6 y of the minimization problem. Recall that to convert the minimization problem to the maximization problem, multiply the objective function by − 1 . − c = − ( 3 x + 2 y ) p = − 3 x − 2 y Thus, the corresponding maximization problem is Maximize p = − 3 x − 2 y subject to x + 2 y ≥ 20 2 x + y ≥ 10 x ≥ 0 , y ≥ 0 Consider the constraints x + 2 y ≥ 20 2 x + y ≥ 10 x ≥ 0 , y ≥ 0 Recall that to compute solution of general linear programming problems, use slack and surplus variables to covert inequalities into equations and star the rows for which the active variables have negative value. Subtract the surplus variable s and t . x + 2 y − s = 20 2 x + y − t = 10 Thus, the equations are x + 2 y − s = 20 2 x + y − t = 10 3 x + 2 y + p = 0 Write the equations in augmented matrix. x y s t p s 1 2 − 1 0 0 20 t 2 1 0 − 1 0 10 p 3 2 0 0 1 0 Check the values of the active variables s , t . s = 20 − 1 = − 20 t = 10 − 1 = − 10 Since, the value of s and t is negative, star the s and t row. Thus, the tableau is x y s t p * s 1 2 − 1 0 0 20 * t 2 1 0 − 1 0 10 p 3 2 0 0 1 0 Since, there are starred rows, use Phase I to proceed. Phase I: To remove the stars that is to make active variables have positive values. Start from the first starred row and locate the largest positive number and pivot the column, use test ratios and the smallest test ratio corresponds to a pivot and clear the pivot column. Check again and star the rows for which the active variables have negative value and repeat the process until there are no starred rows and move to Phase II. The first starred row is s and the largest positive number in the t row is 2. Thus, pivot the y column. The test ratios are 20 2 = 10 and 20 1 = 20 . Since, 10 is the smallest, pivot the entry 2 in y column. x y s t p * s 1 2 − 1 0 0 20 * t 2 1 0 − 1 0 10 p 3 2 0 0 1 0 Use row operations 2 R 2 − R 1 and R 3 − R 1 to clear the pivot column of the augmented matrix

Finite Mathematics and Applied Calculus (MindTap Course List)

7th Edition

ISBN: 9781337274203

Author: Stefan Waner, Steven Costenoble

Publisher: Cengage Learning

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Chapter 6.4, Problem 12E

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To calculate: The minimum value of c=3x+2y subject to

x+2y≥202x+y≥10x≥0,y≥0

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Chapter 6.4, Problem 11E

Chapter 6.4, Problem 13E

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Finite Mathematics and Applied Calculus (MindTap Course List)

Ch. 6.1 - In Exercises 1-26, sketch the region that...Ch. 6.1 - In Exercises 1-26, sketch the region that...Ch. 6.1 - Prob. 3E Ch. 6.1 - Prob. 4E Ch. 6.1 - Prob. 5E Ch. 6.1 - Prob. 6E Ch. 6.1 - Prob. 7E Ch. 6.1 - In Exercises 1-26, sketch the region that...Ch. 6.1 - Prob. 9E Ch. 6.1 - Prob. 10E

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The minimum value of c = 3 x + 2 y subject to x + 2 y ≥ 20 2 x + y ≥ 10 x ≥ 0 , y ≥ 0

Solution Summary: The author explains the simplex method to solve the obtained standard maximization tableau.

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To calculate: The minimum value of c=3x+2y subject to

x+2y≥202x+y≥10x≥0,y≥0

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Chapter 6 Solutions

The minimum value of c = 3 x + 2 y subject to x + 2 y ≥ 20 2 x + y ≥ 10 x ≥ 0 , y ≥ 0

Solution Summary: The author explains the simplex method to solve the obtained standard maximization tableau.

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To calculate: The minimum value of c=3x+2y subject to x+2y≥202x+y≥10x≥0,y≥0

Want to see the full answer?

Chapter 6 Solutions

To calculate: The minimum value of c=3x+2y subject to

x+2y≥202x+y≥10x≥0,y≥0