SOLUTION: A little help please!? Find all four corner points of the feasible region of the following system of inequalities: x + 4y <= 8 x

SOLUTION: A little help please!? Find all four corner points of the feasible region of the following system of inequalities: x + 4y <= 8 x - y <= 3 x => 0 y => 0 I do not have

Question 427098: A little help please!?
Find all four corner points of the feasible region of the following system of inequalities:
x + 4y <= 8
x - y <= 3
x => 0
y => 0
I do not have a clue...
Thank you in advance!! :)
Answer by ankor@dixie-net.com(22740) (Show Source): You can put this solution on YOUR website!
Find all four corner points of the feasible region of the following system of inequalities:
Put equations in the slope/intercept form to graph
x + 4y <= 8
4y <= -x + 8
y <= + 2
This is plotted as the red line
:
x - y <= 3
-y <= -x + 3
y has to be positive, multiply by -1, this reverses the inequality sign
y >= x - 3
This is plotted as the green line
:
This just means all values are >0
x => 0
y => 0
:
Plot these using x=0 and x=8. Looks something like this:

Note that the area of feasibility is:
Less or at the red line
above or at the green line
At or Above the x axis
At or to the right of the y axis
:
You can see the coordinates of the corner points of this area:
0,0; 0,2; 4,1; 3,0
:
Perhaps this will give you a couple clues to this.
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