Question 461306
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In general, to graph a two-variable inequality:


1.  Replace the inequality sign with an equals sign, then graph the resulting straight line.  If the original inequality was inclusive of equals, then make the graph using a solid line indicating that the boundary line is included in the solution set.  If the original inequality was strict, graph the line with a dashed line to indicate the line as being excluded from the solution set.


2.  Select an ordered pair representing any point NOT on the boundary line you just graphed.  Whenever the boundary line does NOT pass through the origin, the origin is an outstanding choice for this step.


3.  Substitute the coordinate values from the ordered pair from step 2 into the original inequality.  If the result is a true statement, then shade in the half plane that contains the test point selected.  Otherwise, shade in the other side of the boundary line.


In the case of graphing a feasibility area, graph all of your constraints onto one set of axes.  The feasibility area is the area where all of the shaded areas overlap.  (sometimes, particularly when you have several constraints, and/or when you are using graphing software, it is helpful to do the shading on the opposite side of all of your constraint inequalities.  That is to say, shade in the EXCLUDED part of the coordinate plane.  If you do that, the feasible area is that area that ends up with no shading at all -- and it is easier to visualize the shape and extent of the feasible area that way.


As for the problem you answered incorrectly, if you follow the above steps to graph the feasible area, you will note that only one of the given vertices is actually IN the feasible area.  If you write back, I'll send you a file with the feasible area graphed.


John
*[tex \LARGE e^{i\pi} + 1 = 0]
My calculator said it, I believe it, that settles it
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