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How many corner points will there be for the feasible region bounded by: x>0, y>0, 2x+5y<7?

User Drecker
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1 Answer

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Final answer:

There is 1 corner point for the feasible region bounded by the given inequalities.

Step-by-step explanation:

To find the corner points of the feasible region bounded by the inequalities x>0, y>0, and 2x+5y<7, we need to graph the inequalities and identify the points where the lines intersect.

First, we graph the line 2x+5y=7. To do this, we find two points on the line by setting x=0 and solving for y, and setting y=0 and solving for x. The two points are (0,7/5) and (7/2,0).

Next, we graph the inequalities x>0 and y>0. These are the positive x and y quadrants of the coordinate plane.

The corner point of the feasible region is the intersection of the lines and the positive x and y quadrants. In this case, there is only one intersection point, which is (7/2,0). Therefore, there is 1 corner point for the feasible region.

User Lone Lunatic
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