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One corner point of the region enclosed by the inequalities x ≥ 0, x + 2y ≥ 0, and 2x + y ≤ 4 is

1. (1, 1)
2. (0, 1)
3. (3, 0)
4. (0, 0)

Correct answer: (0, 0)

Solution

Region: x>=0, x+2y>=0, 2x+y<=4. Vertices are intersections of the boundaries: x=0 & x+2y=0 -> (0,0); x=0 & 2x+y=4 -> (0,4); x+2y=0 & 2x+y=4 -> (8/3,-4/3). Among the options only (0,0) is a corner point. (3,0) gives 2x+y=6>4, outside the region.

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