SOLUTION: Help Needed; Use graphical methods to solve the linear programming problem. Maximize z = 6x + 7y subject to: 2x + 3y ≤ 12 2x + y ≤ 8 x ≥ 0 y ≥ 0

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Question 201961: Help Needed;
Use graphical methods to solve the linear programming problem.
Maximize z = 6x + 7y
subject to: 2x + 3y ≤ 12
2x + y ≤ 8
x ≥ 0
y ≥ 0

A) Maximum of 24 when x = 4 and y = 0
B) Maximum of 32 when x = 2 and y = 3
C) Maximum of 32 when x = 3 and y = 2
D) Maximum of 52 when x = 4 and y = 4
Answer by jim_thompson5910(35256)   (Show Source): You can put this solution on YOUR website!
Simply graph each inequality and find the shaded area. Now locate each of the vertices that fall on the border of this shaded area. With each vertex, plug in the x and y coordinates into . The pair of coordinates that maximize "z" will be the answer.


If what I'm saying doesn't help, then...



Start with the given system of inequalities





In order to graph this system of inequalities, we need to graph each inequality one at a time.


First lets graph the first inequality
In order to graph , we need to graph the equation (just replace the inequality sign with an equal sign).
So lets graph the line (note: if you need help with graphing, check out this solver)
graph of
Now lets pick a test point, say (0,0). Any point will work, (just make sure the point doesn't lie on the line) but this point is the easiest to work with. Now evaluate the inequality with the test point

Substitute (0,0) into the inequality
Plug in and
Simplify



(note: for some reason, some of the following images do not display correctly in Internet Explorer. So I recommend the use of Firefox to see these images.)


Since this inequality is true, we simply shade the entire region that contains (0,0)
Graph of with the boundary (which is the line in red) and the shaded region (in green)

---------------------------------------------------------------


Now lets graph the second inequality
In order to graph , we need to graph the equation (just replace the inequality sign with an equal sign).
So lets graph the line (note: if you need help with graphing, check out this solver)
graph of
Now lets pick a test point, say (0,0). Any point will work, (just make sure the point doesn't lie on the line) but this point is the easiest to work with. Now evaluate the inequality with the test point

Substitute (0,0) into the inequality
Plug in and
Simplify



Since this inequality is true, we simply shade the entire region that contains (0,0)
Graph of with the boundary (which is the line in red) and the shaded region (in green)

---------------------------------------------------------------


Now lets graph the third inequality
In order to graph , we need to graph the equation (just replace the inequality sign with an equal sign).
So lets graph the line (simply draw a vertical line through )
graph of (note:the graph is the line that is overlapping the y-axis. So it may be hard to see)
Now lets pick a test point, say (1,0). Any point will work, (just make sure the point doesn't lie on the line) but this point is the easiest to work with. Now evaluate the inequality with the test point

Substitute (1,0) into the inequality
Plug in and
Simplify



Since this inequality is true, we simply shade the entire region that contains (1,0)
Graph of with the boundary (which is the line in red) and the shaded region (in green)

---------------------------------------------------------------


Now lets graph the fourth inequality
In order to graph , we need to graph the equation (just replace the inequality sign with an equal sign).
So lets graph the line (simply draw a horizontal line through )
graph of (note:the graph is the line that is overlapping the x-axis. So it may be hard to see)
Now lets pick a test point, say (0,1). Any point will work, (just make sure the point doesn't lie on the line) but this point is the easiest to work with. Now evaluate the inequality with the test point

Substitute (0,1) into the inequality
Plug in and
Simplify



Since this inequality is true, we simply shade the entire region that contains (0,1)
Graph of with the boundary (which is the line in red) and the shaded region (in green)

---------------------------------------------------------------


So we essentially have these 4 regions:

Region #1
Graph of


Region #2
Graph of


Region #3
Graph of


Region #4
Graph of




When these inequalities are graphed on the same coordinate system, the regions overlap to produce this region. It's a little hard to see, but after evenly shading each region, the intersecting region will be the most shaded in.







Here is a cleaner look at the intersection of regions




Here is the intersection of the 4 regions represented by the series of dots



Take note that the vertices of the shaded region are: (0,4), (4,0) and (3,2)


It turns out that if we plug in and , we'll get the largest possible "z" value of


Note: you should plug in the other vertices to verify that this is indeed the max.


So the answer is C) Maximum of 32 when x = 3 and y = 2
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