Question 197946: Value of c=2x-y subject to the constraints:
x>=1
x<=4
y<=8
x-3y<=-2
The last constraint really confuses me. Thanks for your time. Mike
Found 2 solutions by jim_thompson5910, solver91311:
Answer by jim_thompson5910(35256)   (Show Source):
You can put this solution on YOUR website! Simply graph every inequality on the xy plane. These graphs will form a closed polygon. It turns out that the values that will optimize "c" will be the x and y values that are at the vertices. So simply plug in the x and y values at the vertices to see which value yields the largest/smallest value of "c". Let me know if this makes sense.
Answer by solver91311(24713)  (Show Source):
You can put this solution on YOUR website!
I'm presuming that you are doing a linear programming problem. You don't say what you want to do with your objective function, that is whether you want to maximize or minimize (oddly enough, there are solutions both ways for this function and set of constraints).
I also presume, because of the simplicity of the problem -- there are only two variables and four constraints after all, that you are probably working this one graphically.
Given that, so far you should have an area of feasibility bounded by y = 8, x = 1, and x = 4 with no lower bound. You need to further define your area of feasibility with the
constraint.
Step 1: Graph the line  . You could go through the process of creating the slope intercept form, but quickest is to just pick a couple values for x and substitute then solve.
Step 2: Once you have your boundary drawn, decide which side to shade. You need a test point that is NOT on the line. Since this line does not pass through the origin, the origin is an excellent test point. Substitute the coordinates of the test point into the original inequality. If you get a true statement, shade the side with the test point in it. If you get a false statement, shade the other side.
 \leq -2)
is clearly false. (0,0) is below the line, so shade above the line.
Where that shading overlaps your previously defined area of feasibility is your new area of feasibility. In other words, the trapezoid with vertices (1,8), (4,8), (1,1), and (4,2). It looks like those vertices are your points of interest.
Using the Solver add-in in Excel confirms that (4,2) is the point where the function is maximized, and (1,8) is the point where the function is minimized.
John
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