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

Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
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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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