Maximize z = x₁ + x₂ subject to x₁ + 2x₂ ≤ 2000, x₁ + x₂ ≤ 1500, x₂ ≤ 600, x₁, x₂ ≥ 0. The solution is
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Correct answer: C. x₁ = 1000, x₂ = 500, z = 1500
Explanation
The maximum z = 1500 lies on x₁ + x₂ = 1500, and x₁ + 2x₂ ≤ 2000 then forces x₂ ≤ 500. Only (1000, 500) among the options satisfies every constraint (A and D break x₁ + 2x₂ ≤ 2000; B breaks x₂ ≤ 600).
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