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 + r = 3 w + v + q = 4 Step 2: Write the equations in an augmented matrix. The equations are x + y + s = 1 y + z + t = 2 z + w + r = 3 w + v + q = 4 − x + y − z − 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 − 1 − 1 − 1 − 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 − 1 . Thus, choose the left most entry and pivot the x 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 only test ratio is 1 1 = 1 . Thus, the pivot is 1 in column x 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 − 1 − 1 − 1 − 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 s in column x and replace s with x . Thus, x is the active variable. Use row operations to clear the pivot column of the augmented matrix. Use the row operations R 5 + R 1 as; x y z w v s t r q p x 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 0 0 − 1 − 1 − 1 1 0 0 0 1 1 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 − 1 . Thus, pivot the z 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. Since, the test ratios are neither 0 nor negative. Thus, the test ratios are 2 1 = 2 and 3 1 = 3 . Thus, the pivot is 1 in column z as; x y z w v s t r q p x 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 0 0 − 1 − 1 − 1 1 0 0 0 1 1 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 z and replace t with z . Thus, z is the active variable. Use row operations to clear the pivot column of the augmented matrix. Use the row operations R 3 − R 2 and R 5 + R 2 as; x y z w v s t r q p x 1 1 0 0 0 1 0 0 0 0 1 z 0 1 1 0 0 0 1 0 0 0 2 r 0 − 1 0 1 0 0 − 1 1 0 0 1 q 0 0 0 1 1 0 0 0 1 0 4 p 0 1 0 − 1 − 1 1 1 0 0 1 3 Since, one entry is still negative in the last row

Bundle: Finite Mathematics, 7th + WebAssign Printed Access Card for Waner/Costenoble's Finite Mathematics, 7th Edition, Single-Term

7th Edition

ISBN: 9781337604956

Author: Waner

Publisher: CENGAGE L

See similar textbooks

Videos

Question

Chapter 5.3, Problem 13E

To determine

To calculate: The maximum value of p=x+y+z+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.

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Chapter 5.3, Problem 12E

Chapter 5.3, Problem 14E

Students have asked these similar questions

π sin2x sin3x sin5x sin30x sinx - sin6x - sin10x - sin15x) 2 dx

f(x+T) = f(x) f(x) = sin² (x) + cos² (x) T=?

HW: The frame shown in the figure is pinned at A and C. Use moment distribution method, with and without modifications, to draw NFD, SFD, and BMD. B I I 40 kN/m A 3 m 4 m

Chapter 5 Solutions

Bundle: Finite Mathematics, 7th + WebAssign Printed Access Card for Waner/Costenoble's Finite Mathematics, 7th Edition, Single-Term

Ch. 5.1 - In Exercises 1-26, sketch the region that...Ch. 5.1 - In Exercises 1-26, sketch the region that...Ch. 5.1 - In Exercises 1-26, sketch the region that...Ch. 5.1 - In Exercises 1-26, sketch the region that...Ch. 5.1 - In Exercises 1-26, sketch the region that...Ch. 5.1 - Prob. 6E Ch. 5.1 - In Exercises 1-26, sketch the region that...Ch. 5.1 - In Exercises 1-26, sketch the region that...Ch. 5.1 - Prob. 9E Ch. 5.1 - In Exercises 1-26, sketch the region that...

Knowledge Booster

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, subject and related others by exploring similar questions and additional content below.

The maximum value of p = x + y + z + 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+y+z+w+v subject to constraints.

Videos

To calculate: The maximum value of p=x+y+z+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.

Want to see the full answer?

Chapter 5 Solutions

The maximum value of p = x + y + z + 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+y+z+w+v subject to constraints.

Videos

To calculate: The maximum value of p=x+y+z+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.

Want to see the full answer?

Chapter 5 Solutions

To calculate: The maximum value of p=x+y+z+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.