min 4a1 + 5xз 2x1 + x2 – 53 -3x1 subject to = 1 + 4x3 + x4 = 2 X; 2 0, i = 1, 2, 3, 4.
min 4a1 + 5xз 2x1 + x2 – 53 -3x1 subject to = 1 + 4x3 + x4 = 2 X; 2 0, i = 1, 2, 3, 4.
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
Related questions
Question
Consider the LP Problem (attached image):
(b) Write down the dual problem and find an optimal solution. Is it unique?

Transcribed Image Text:4x1 + 5x3
2x1 + x2 – 5x3
-3x1
subject to
= 1
+ 42з + х4 — 2
Ti 2 0, i = 1, 2, 3, 4.
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