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Exercise 2.10 · Q6

Q.x−2y≥0, 2x−y≤−2, x≥0, y≥0x-2y\ge0,\ 2x-y\le-2,\ x\ge0,\ y\ge0.

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Step 1. x−2y≥0  ⟺  y≤x2x-2y\ge0\iff y\le\dfrac x2. 2x−y≤−2  ⟺  y≥2x+22x-y\le-2\iff y\ge2x+2.

Step 2. For x≥0x\ge0: y≥2x+2≥2y\ge2x+2\ge2 (since x≥0x\ge0), so any feasible point needs y≥2y\ge2.

Step 3. But also y≤x2y\le\dfrac x2, so we would need 2x+2≤x2⇒4x+4≤x⇒3x≤−4⇒x≤−432x+2\le\dfrac x2\Rightarrow4x+4\le x\Rightarrow3x\le-4\Rightarrow x\le-\dfrac43, contradicting x≥0x\ge0. …

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