Explanation Given Information: The provided constraints are; x + y ≤ 1 y + z ≤ 2 z + w ≤ 3 w + v ≤ 4 And, x ≥ 0 , y ≥ 0 , z ≥ 0 , w ≥ 0 , v ≥ 0 Formula used: Step 1: Use slack variables and write the inequalities in equation form. Step 2: Write the equations in an augmented matrix. Step 3: Choose the smallest negative number on the left side of the bottom row and pivot the column. Step 4: 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. Step 5: Clear the pivot column using the pivot and label the pivot row with the label of the pivot column. Step 6: Repeat the above steps till all the entries in the bottom row are non negative. Use row operations to clear the column of the augmented matrix. Calculation: Consider the constraints, x + y ≤ 1 y + z ≤ 2 z + w ≤ 3 w + v ≤ 4 And, x ≥ 0 , y ≥ 0 , z ≥ 0 , w ≥ 0 , v ≥ 0 Step 1: Use slack variables and write the inequalities in equation form. Add the slack variables s , t , q and r as; x + y + s = 1 y + z + t = 2 z + w + q = 3 w + v + r = 4 Step 2: Write the equations in an augmented matrix. The equations are; x + y + s = 1 y + z + t = 2 z + w + q = 3 w + v + r = 4 And, − x − 2 y − z − 2 w − v + p = 0 Write the equations in augmented matrix as; x y z w v s t r q p s 1 1 0 0 0 1 0 0 0 0 1 t 0 1 1 0 0 0 1 0 0 0 2 r 0 0 1 1 0 0 0 1 0 0 3 q 0 0 0 1 1 0 0 0 1 0 4 p − 1 − 2 − 1 − 2 − 1 0 0 0 0 1 0 Step 3: Choose the smallest negative number on the left side of the bottom row and pivot the column. Since, the smallest negative number on the left side of the bottom row is − 2 . Thus, choose the left most entry and pivot the y column. Step 4: 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. Since, the test ratios are neither 0 nor negative. Thus, the test ratios are 1 1 = 1 and 2 1 = 2 . Thus, the pivot is 1 in column y as; x y z w v s t r q p s 1 1 0 0 0 1 0 0 0 0 1 t 0 1 1 0 0 0 1 0 0 0 2 r 0 0 1 1 0 0 0 1 0 0 3 q 0 0 0 1 1 0 0 0 1 0 4 p − 1 − 2 − 1 − 2 − 1 0 0 0 0 1 0 Step 5: Clear the pivot column using the pivot and label the pivot row with the label of the pivot column. Since, the pivot is t in column y and replace t with y . Thus, y is the active variable. Use row operations to clear the pivot column of the augmented matrix. Use the row operations R 2 − R 1 and R 5 + 2 R 1 as; x y z w v s t r q p y 1 1 0 0 0 1 0 0 0 0 1 t − 1 0 1 0 0 − 1 1 0 0 0 2 r 0 0 1 1 0 0 0 1 0 0 3 q 0 0 0 1 1 0 0 0 1 0 4 p 1 0 − 1 − 2 − 1 2 0 0 0 1 2 Step 6: Repeat the above steps till all the entries in the bottom row are non negative. Since, one entry is still negative in the last row. Choose the smallest negative number on the left side of the bottom row and pivot the column. Since, the smallest negative number on the left side of the bottom row is − 2 . Thus, pivot the w 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

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 6.3, Problem 14E

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To calculate: The maximum value of p=x+2y+z+2w+v subject to constraints x+y≤1,y+z≤2,z+w≤3,w+v≤4 and x≥0,y≥0,z≥0,w≥0,v≥0.

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Chapter 6.3, Problem 13E

Chapter 6.3, Problem 15E

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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 maximum value of p = x + 2 y + z + 2 w + v subject to constraints x + y ≤ 1 , y + z ≤ 2 , z + w ≤ 3 , w + v ≤ 4 and x ≥ 0 , y ≥ 0 , z ≥ 0 , w ≥ 0 , v ≥ 0 .

Solution Summary: The author calculates the maximum value of p=x+2y+z+ 2w+v subject to constraints.

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To calculate: The maximum value of p=x+2y+z+2w+v subject to constraints x+y≤1,y+z≤2,z+w≤3,w+v≤4 and x≥0,y≥0,z≥0,w≥0,v≥0.

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Chapter 6 Solutions