Consider the LP maximize $ \large z = cx \ $ subject to $ \ Ax ≤ b \ , x ≥ 0 \ $ where $ \ c \ $ is a nonzero vector. Suppose that the point $ \ x_ 0 \ $ is such that $ \ Ax_ 0 < b \ $ and $ \ x_ 0 > 0 \ $.
Show that $ \ x_ 0 \ $ cannot be an optimal solution.
Answer:
Max $ \ Z=cx \ $ subject to $ \ Ax \leq b , \ x \geq 0 \ $
Given $ \ x_0 \ $ satisfying $ \ Ax_0 <b \ $.
Then for $ \ \epsilon >0 \ $ , there exists $ \ x'=x_0+\epsilon \ $ such that $ Ax' \leq b \ , \ x' \geq 0 $
Also,
$ cx_0 \leq cx' \ $
$ \Rightarrow x_0 \ $ is not optimal solution.
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