SOLUTION: Please help. I need to graph this inequalities and fine the feasible region. The inequalities are; −3≤y≤5
4x+y≤5
−2x+y≤5.
and says E = 4x - 3
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Question 810282: Please help. I need to graph this inequalities and fine the feasible region. The inequalities are; −3≤y≤5
4x+y≤5
−2x+y≤5.
and says E = 4x - 3y.
My teacher gets mad it homework is wrong but mainly if not done. Thanks.
Answer by KMST(5328) (Show Source): You can put this solution on YOUR website!
I will show you how to graph those inequalities and find the feasible region first.
After that, I will tell you what I think of .
THE FEASIBLE REGION:
The borders of your feasible region may be part of the lines
, ,
, and
Each of those lines is part of the solution to one of the inequalities.
The lines represented by and are easy to figure out.
They are "horizontal" lines, where all the points have the same y-coordinate:
The other two lines are slanted lines that are a little harder to figure out.
However, since two points determine a line, all we need is two find 2 points for each one.
Also, since they have to cross the horizontal lines and ,
we might as well find the point where they cross and .
To find each intersection point we solve a system of equations, but these are very simple.
Where does the horizontal line cross the slanted lines?
--> --> --> gives us point {0,5) .
--> --> --> gives us point {0,5) too.
So, it turns out that the 3 lines represented by
, , and all cross at point .
Where does the horizontal line cross the slanted lines?
--> --> --> --> --> gives us point .
--> --> --> --> --> gives us point .
Now we can mark those points and draw the slanted lines:
, passing through (0,5) and {2,-3) , and
, passing through (0,5) and {-4,-3) .
Is triangle ABC the feasible region?
(Feasible regions are often polygons, like triangles, quadrilaterals, maybe even pentagons).
If triangle ABC is the feasible region, the origin, point should be a solution of all the inequalities.
, , and (((-2*0+0=0<5}}} ,
so point is a solution of all the inequalities.
For each of the 4 inequalities that determine the feasible region,
, , , and ,
the solution is the corresponding boundary line, plus the whole side of the line that contains the origin, point .
So the points that satisfy all 4 inequalities are the points in triangle ABC, including the sides of the triangle.
WHAT ABOUT ?
Usually you are asked to find out where in the feasible region the function has a maximum or minimum value.
(E is a linear function of x and y).
For a linear function in a polygon-shaped feasible region,
the maximum must happen in one of the vertices or all along one tof the sides/edges of the feasible region.
The sames goes for the minimum.
All we have to do is calculate the value of at points A, B, and C
At ,
At ,
At ,
So the maximum of E happens at , where ,
and the minimum happens at , where .
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