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Which of these sets of points are the corner points for the feasible region satisfying the system of linear inequalities 2x + 8y < or = 32, 5x + 2y > or = 30, x> or =0, y > or =0?

a) (5,9), (4,0), (1/2, 4/9)
b) (0,4), (15, 0), (0, 0)
c) (0,4), (0,15), (4, 0), (4, 9)
d) (0, 4), (0, 15), (44/9, 25/9)

2007-11-05 06:35:54 · 2 answers · asked by brownk_1999 1 in Science & Mathematics Mathematics

2 answers

Are you sure the inequalities aren't 2x+8y >= 32 and 5x+2y<=30? This would lead to answer d. But they problem you stated (2x+8y <= 32 and 5x+2y>=30) leads to another answer not listed: (6,0), (16,0), and (44/9, 25/9).

Assuming the inequalities are 2x+8y >= 32 and 5x+2y<=30 we can show that:

y>= 4-x/4 and y<=15-5x/2

So at x = 0, y must be between 4 and 15.

Setting the two equations equal yields a third corner point:
4-x/4 = 15-5x/2

which is true at x=44/9. Plugging this x value into either equation above yields y = 25/9.

2007-11-05 07:00:51 · answer #1 · answered by RJ Hunt 2 · 1 0

Go with R. J.

2007-11-05 07:18:00 · answer #2 · answered by Tony 7 · 0 0

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