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Maximize P = x1 +52 subject to the constraints A = D = After a few rounds of pivoting, the final simplex tableau can have the following form. Provide the missing numbers as well as values for the solution: B = E = C = The maximum value of the objective function is P = BV $1 X2 P 2x1 x1 + + x2 ≤10 2x2 10 *1,2 > 0 found when x₁ = P x1 X2 $1 03/2 0 1 01/2 $2 -1/2 1 0 1/2 D 1 A B C and 2 = RHS 5 5 E

Maximize P = x1 +52 subject to the constraints A = D = After a few rounds of pivoting, the final simplex tableau can have the following form. Provide the missing numbers as well as values for the solution: B = E = C = The maximum value of the objective function is P = BV $1 X2 P 2x1 x1 + + x2 ≤10 2x2 10 *1,2 > 0 found when x₁ = P x1 X2 $1 03/2 0 1 01/2 $2 -1/2 1 0 1/2 D 1 A B C and 2 = RHS 5 5 E

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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Maximize P = x₁ + 5x2 subject to the constraints A = D = B = E = C = After a few rounds of pivoting, the final simplex tableau can have the following form. Provide the missing numbers as well as values for the solution: The maximum value of the objective function is P = BV $1 X2 P 2x1 ++ found when x1 = x1 + X2 2x2 1,2 P x 1 x2 $1 0 3/2 0 1 01/2 1 0 1 A B с < 10 ≤ 10 > 0 $2 RHS 5 5 E -1/2 1/2 D and 2
Transcribed Image Text:Maximize P = x₁ + 5x2 subject to the constraints A = D = B = E = C = After a few rounds of pivoting, the final simplex tableau can have the following form. Provide the missing numbers as well as values for the solution: The maximum value of the objective function is P = BV $1 X2 P 2x1 ++ found when x1 = x1 + X2 2x2 1,2 P x 1 x2 $1 0 3/2 0 1 01/2 1 0 1 A B с < 10 ≤ 10 > 0 $2 RHS 5 5 E -1/2 1/2 D and 2
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