Solve the following LP (noting that x2 is urs) x2 being urs reflects that the business can must purchase units if >0 but is able to sell units x2 < 0 Max z = 30x1 - 4x2 s.t. 5x1 <= 30 + x2 x1 <= 5 x1 >= 0; x2 urs after introducing x2' and x2" as called for, we have a standard max so there is no Big-M or two Phase needed. 1. in the first tableau the entries in the x2" column are what number times the entries in the x2' column 2. in the next tableau, the entries in the x2" column are what number times the entries in the x2¹ column 3. The optimal solution for z is 4. the optimal value for x2" is 5. the optimal value for x2 is

Linear Algebra: A Modern Introduction
4th Edition
ISBN:9781285463247
Author:David Poole
Publisher:David Poole
Chapter2: Systems Of Linear Equations
Section2.4: Applications
Problem 28EQ
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Solve the following LP (noting that x2 is urs)
x2 being urs reflects that the business can must purchase units if >0
but is able to sell units x2 < 0
Max z 30x1 - 4x2
s.t. 5x1 <= 30 + x2
x1 <= 5
x1 >= 0; x2 urs
after introducing x2' and x2" as called for, we have a standard max
so there is no Big-M or two Phase needed.
1. in the first tableau the entries in the x2" column are what number
times the entries in the x2¹ column
2. in the next tableau, the entries in the x2" column are what number
times the entries in the x2' column
3. The optimal solution for z is
4. the optimal value for x2" is
5. the optimal value for x2 is
Transcribed Image Text:Solve the following LP (noting that x2 is urs) x2 being urs reflects that the business can must purchase units if >0 but is able to sell units x2 < 0 Max z 30x1 - 4x2 s.t. 5x1 <= 30 + x2 x1 <= 5 x1 >= 0; x2 urs after introducing x2' and x2" as called for, we have a standard max so there is no Big-M or two Phase needed. 1. in the first tableau the entries in the x2" column are what number times the entries in the x2¹ column 2. in the next tableau, the entries in the x2" column are what number times the entries in the x2' column 3. The optimal solution for z is 4. the optimal value for x2" is 5. the optimal value for x2 is
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