Explanation Given Information: The given linear programming problem is: Maximize 2 x 1 + 5 x 2 + x 3 Subject to 3 x 1 + x 2 + x 3 ≤ 9 2 x 1 + x 2 + 3 x 3 ≤ 100 2 x 1 + x 2 ≤ 5 Formula used: Following steps of the simplex method, 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 maximisation problem, Maximize 2 x 1 + 5 x 2 + x 3 Subject to 3 x 1 + x 2 + x 3 ≤ 9 2 x 1 + x 2 + 3 x 3 ≤ 100 2 x 1 + x 2 ≤ 5 x ≥ 0 , y ≥ 0 Step 1: Use slack variables and write the inequalities in equation form. Add the slack variables s , t and u as below, 3 x 1 + x 2 + x 3 + s = 9 2 x 1 + x 2 + 3 x 3 + t = 100 2 x 1 + x 2 + u = 5 Step 2: Write the equations in an augmented matrix. The equations are, 3 x 1 + x 2 + x 3 + s = 9 2 x 1 + x 2 + 3 x 3 + t = 100 2 x 1 + x 2 + u = 5 And, f − 2 x 1 − 5 x 2 − x 3 = 0 Write the equations in augmented matrix, x 1 x 2 x 3 s t u f s 3 1 1 1 0 0 0 9 t 2 1 3 0 1 0 0 8 u 2 1 0 0 0 1 0 5 f − 2 − 5 − 1 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 − 5 . Thus, choose the second entry in last row and pivot the x 2 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 9 1 = 9 , 8 1 = 8 and 5 1 = 5 . Since, 5 is smallest. Thus, the pivot is 1 in column x 2 . x 1 x 2 x 3 s t u f s 3 1 1 1 0 0 0 9 t 2 1 3 0 1 0 0 8 u 2 1 0 0 0 1 0 5 f − 2 − 5 − 1 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 u in column x 2 and replace u with x 2 . Thu, x 2 is the active variable. Use row operations to clear the pivot column of the augmented matrix. Use the row operations R 1 − R 3 → R 1 , R 2 − R 3 → R 2 and R 4 + 5 R 3 → R 4

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 4.3, Problem 6CP

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To calculate: The linear programming problem:

Maximize 2x1+5x2+x3

Subject to

3x1+x2+x3≤92x1+x2+3x3≤1002x1+x2≤5

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Chapter 4.3, Problem 5CP

Chapter 4.3, Problem 1E

Chapter 4 Solutions

Mathematical Applications for the Management, Life, and Social Sciences

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The linear programming problem: Maximize 2 x 1 + 5 x 2 + x 3 Subject to 3 x 1 + x 2 + x 3 ≤ 9 2 x 1 + x 2 + 3 x 3 ≤ 100 2 x 1 + x 2 ≤ 5

Solution Summary: The author explains how to calculate the linear programming problem by using slack variables and an augmented matrix.

Videos

To calculate: The linear programming problem:

Maximize 2x1+5x2+x3

Subject to

3x1+x2+x3≤92x1+x2+3x3≤1002x1+x2≤5

Want to see the full answer?

Chapter 4 Solutions

The linear programming problem: Maximize 2 x 1 + 5 x 2 + x 3 Subject to 3 x 1 + x 2 + x 3 ≤ 9 2 x 1 + x 2 + 3 x 3 ≤ 100 2 x 1 + x 2 ≤ 5

Solution Summary: The author explains how to calculate the linear programming problem by using slack variables and an augmented matrix.

Videos

To calculate: The linear programming problem: Maximize 2x1+5x2+x3 Subject to 3x1+x2+x3≤92x1+x2+3x3≤1002x1+x2≤5

Want to see the full answer?

Chapter 4 Solutions

To calculate: The linear programming problem:

Maximize 2x1+5x2+x3

Subject to

3x1+x2+x3≤92x1+x2+3x3≤1002x1+x2≤5