Posted by **Anonymous** on Wednesday, September 30, 2009 at 12:39am.

Given the system of inequalities below, find the vertices of the feasible region. Report your vertices starting with the one which has the smallest x-value. If more than one vertex has the same, smallest x-value, start with the one that has the smallest y-value. Proceed clockwise from the first vertex.

y<-x+7

y>-7x+11

x>0 y> 0

if someone could explain it to me that would be wonderful!

- math -
**Anonymous**, Wednesday, September 30, 2009 at 12:45am
where did you get 10?

- math -
**Shavaleir**, Wednesday, September 30, 2009 at 12:52am
x and y equals 10

- math -
**MathMate**, Thursday, October 1, 2009 at 8:23am
y<-x+7

y>-7x+11

x>0 y> 0

Graphically, draw the four inequalities as through they are lines,

y=-x+7

y=-7x+11

x=0

y= 0

They should intersect at six distinct points.

Create the figure that represents the region of feasibility.

The goal is to find the boundaries of the region of feasibility bounded by four intersection points, and the correspoinding segments of straight line representing the inequalities.

This can be achieved (as described in the question) by starting at the origin (intersection of x>0 and y>0) and proceeding along the x-axis to the next intersection point. If this point satisfies all constraints, or at the border of it, proceed along the straight line. If not, go back to the previous point and continue with the intersecting line.

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