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The following intermediate tableau for a dual maximum problem was obtained using the simplex method for optimizing a minimum problem. (see image) Perform all the necessary pivot and row operations to obtain the final tableau. Then, using the final tableau, answer the following questions:               The minimum function value is:  The value of  x1 in the minimum problem is :              The value of  x2  in the minimum problem is:

The following intermediate tableau for a dual maximum problem was obtained using the simplex method for optimizing a minimum problem. (see image) Perform all the necessary pivot and row operations to obtain the final tableau. Then, using the final tableau, answer the following questions:               The minimum function value is:  The value of  x1 in the minimum problem is :              The value of  x2  in the minimum problem is:

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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The following intermediate tableau for a dual maximum problem was obtained using the simplex method for optimizing a minimum problem. (see image)

Perform all the necessary pivot and row operations to obtain the final tableau. Then, using the final tableau, answer the following questions:              

The minimum function value is: 

The value of  x1 in the minimum problem is :             

The value of  x2  in the minimum problem is: 

Py₁ 00 0 1 Y2 $1 4 1 3 0 $2 1 2 2 1 0 8 0 2 RHS 4 10 40
Transcribed Image Text:Py₁ 00 0 1 Y2 $1 4 1 3 0 $2 1 2 2 1 0 8 0 2 RHS 4 10 40
Expert Solution
Step 1: Analysis and Introduction

Intermediate Dual simplex tableau of a LPP is provided as:

Error converting from MathML to accessible text.

To find:

The solution to the LPP and find the value of x subscript 1 and x subscript 2 in the minimization problem.

Concept used:

Enter the variable which has most non-negative constraint and make the column as pivot by taking away the column where the minimum ratio persists.

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