Set of optimal solutions being convex set

The set of optimal solution suffices to show that for 0<=k<=1, kx₁ + (1‑k)x₂ is also an optimal solution.

First, kx₁ +(1‑k)x₂ is feasible because A(kx₁ + (1‑k)x₂) = kAx₁ + (1‑k)Ax₂ = kb + (1‑k)b = b and kx₁ + (1‑k)x2>=0 follows from the non‑negativity of x₁ and x₂...

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Chapter 4.7, Problem 6P

Chapter 4.7, Problem 8P

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Chapter 4 Solutions

Operations Research : Applications and Algorithms

Ch. 4.1 - Prob. 1P Ch. 4.1 - Prob. 2P Ch. 4.1 - Prob. 3P Ch. 4.4 - Prob. 1P Ch. 4.4 - Prob. 2P Ch. 4.4 - Prob. 3P Ch. 4.4 - Prob. 4P Ch. 4.4 - Prob. 5P Ch. 4.4 - Prob. 6P Ch. 4.4 - Prob. 7P

Ch. 4.5 - Prob. 1P Ch. 4.5 - Prob. 2P Ch. 4.5 - Prob. 3P Ch. 4.5 - Prob. 4P Ch. 4.5 - Prob. 5P Ch. 4.5 - Prob. 6P Ch. 4.5 - Prob. 7P Ch. 4.6 - Prob. 1P Ch. 4.6 - Prob. 2P Ch. 4.6 - Prob. 3P Ch. 4.6 - Prob. 4P Ch. 4.7 - Prob. 1P Ch. 4.7 - Prob. 2P Ch. 4.7 - Prob. 3P Ch. 4.7 - Prob. 4P Ch. 4.7 - Prob. 5P Ch. 4.7 - Prob. 6P Ch. 4.7 - Prob. 7P Ch. 4.7 - Prob. 8P Ch. 4.7 - Prob. 9P Ch. 4.8 - Prob. 1P Ch. 4.8 - Prob. 2P Ch. 4.8 - Prob. 3P Ch. 4.8 - Prob. 4P Ch. 4.8 - Prob. 5P Ch. 4.8 - Prob. 6P Ch. 4.10 - Prob. 1P Ch. 4.10 - Prob. 2P Ch. 4.10 - Prob. 3P Ch. 4.10 - Prob. 4P Ch. 4.10 - Prob. 5P Ch. 4.11 - Prob. 1P Ch. 4.11 - Prob. 2P Ch. 4.11 - Prob. 3P Ch. 4.11 - Prob. 4P Ch. 4.11 - Prob. 5P Ch. 4.11 - Prob. 6P Ch. 4.12 - Prob. 1P Ch. 4.12 - Prob. 2P Ch. 4.12 - Prob. 3P Ch. 4.12 - Prob. 4P Ch. 4.12 - Prob. 5P Ch. 4.12 - Prob. 6P Ch. 4.13 - Prob. 2P Ch. 4.14 - Prob. 1P Ch. 4.14 - Prob. 2P Ch. 4.14 - Prob. 3P Ch. 4.14 - Prob. 4P Ch. 4.14 - Prob. 5P Ch. 4.14 - Prob. 6P Ch. 4.14 - Prob. 7P Ch. 4.16 - Prob. 1P Ch. 4.16 - Prob. 2P Ch. 4.16 - Prob. 3P Ch. 4.16 - Prob. 5P Ch. 4.16 - Prob. 7P Ch. 4.16 - Prob. 8P Ch. 4.16 - Prob. 9P Ch. 4.16 - Prob. 10P Ch. 4.16 - Prob. 11P Ch. 4.16 - Prob. 12P Ch. 4.16 - Prob. 13P Ch. 4.16 - Prob. 14P Ch. 4.17 - Prob. 1P Ch. 4.17 - Prob. 2P Ch. 4.17 - Prob. 3P Ch. 4.17 - Prob. 4P Ch. 4.17 - Prob. 5P Ch. 4.17 - Prob. 7P Ch. 4.17 - Prob. 8P Ch. 4 - Prob. 1RP Ch. 4 - Prob. 2RP Ch. 4 - Prob. 3RP Ch. 4 - Prob. 4RP Ch. 4 - Prob. 5RP Ch. 4 - Prob. 6RP Ch. 4 - Prob. 7RP Ch. 4 - Prob. 8RP Ch. 4 - Prob. 9RP Ch. 4 - Prob. 10RP Ch. 4 - Prob. 12RP Ch. 4 - Prob. 13RP Ch. 4 - Prob. 14RP Ch. 4 - Prob. 16RP Ch. 4 - Prob. 17RP Ch. 4 - Prob. 18RP Ch. 4 - Prob. 19RP Ch. 4 - Prob. 20RP Ch. 4 - Prob. 21RP Ch. 4 - Prob. 22RP Ch. 4 - Prob. 23RP Ch. 4 - Prob. 24RP Ch. 4 - Prob. 26RP Ch. 4 - Prob. 27RP Ch. 4 - Prob. 28RP

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Solution Summary: The author explains that kx 1 + (1k)x 2 is also an optimal solution to the original LP.

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