11) Formulate the dual problem for the linear programming problem: 11) C= 3x1 + x2 Minimize subject to 2x1 + 3x2 2 60 X1 + 4x2 2 40 x1, x2 2 0 A) Maximize P=60y1 + 40y2 B) Maximize P= 3y1 + Y2 subject to subject to 2y1 + y2 2 3 3y1 + 4y2 1 y1- y2 × 0 D) Maximize P=60y1 + 40y2 2y1 + y2 s 3 3y1 + 4y2 s1 y1- y2 > 0 C) Maximize P= 3y1 + y2 subject to subject to 2y1 + y2 s3 3y1 + 4y2 s 1 2y1 + y2 3 3y1 + 4y2 1 y1- y2 20 y1- y2 0
11) Formulate the dual problem for the linear programming problem: 11) C= 3x1 + x2 Minimize subject to 2x1 + 3x2 2 60 X1 + 4x2 2 40 x1, x2 2 0 A) Maximize P=60y1 + 40y2 B) Maximize P= 3y1 + Y2 subject to subject to 2y1 + y2 2 3 3y1 + 4y2 1 y1- y2 × 0 D) Maximize P=60y1 + 40y2 2y1 + y2 s 3 3y1 + 4y2 s1 y1- y2 > 0 C) Maximize P= 3y1 + y2 subject to subject to 2y1 + y2 s3 3y1 + 4y2 s 1 2y1 + y2 3 3y1 + 4y2 1 y1- y2 20 y1- y2 0
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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10) Solve the following linear programming probiem using the simpiex method:
Maximize P =x1 - x2
subject to
X1 + x2 s 4
2x1 + 7x2 s 14
x1, x2 2 0
A) Max P = 14 at x1= 4 and x2 = 0
B) Max P = 4 at x1= 4 and x2 = 4
C) Max P = 4 at x1= 4 and x2 = 0
D) Max P = 4 at x1= 14 and x2 = 0
11) Formulate the dual problem for the linear programming problem:
11)
Minimize
C= 3x1 + x2
subject to
2x1 + 3x2 2 60
X1 + 4x2 2 40
X1, x2 2 0
A) Maximize
P = 60y1 + 40y2
B) Maximize
P = 3y1 + y2
subject to
subject to
2y1 + y2 s3
2y1 + y2 2 3
Зу1 + 4y2 <1
y1, y2 2 0
P = 3y1 + y2
Зу1 + 4y2 > 1
y1, y2 2 0
C) Maximize
D) Maximize
P = 60y1 + 40y2
subject to
subject to
2y1 + y2 s3
Зу1 + 4у2 <1
2y1 + y2 2 3
3y1 + 4y2 > 1
y1, y2 - 0
y1, y2 2 0
PAGE 4 REVIEW_SANJAC
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