SOLUTION: z=2x+4y
x>=0, y>=0 Linear programming problem, graphing a solution, and maximum.
x+3y>=6
x+y>=3
x+y<=9
I come up with serveral vertices. Even when using the example. I kn
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Question 145904This question is from textbook Blitzer College Algebra
: z=2x+4y
x>=0, y>=0 Linear programming problem, graphing a solution, and maximum.
x+3y>=6
x+y>=3
x+y<=9
I come up with serveral vertices. Even when using the example. I know how to put the vertices back into the equations, but I need to figure out how to limit or pick the correct vertices; (0,0), (0,2), (6,0), (0,3), (3,0), (5,4), (6,3), (0,9) My graphe looks like a octagon kind of including 0. I am missing something and trying desperately to understand this. Can someone show me how I conclude which vertices are correct?
Heidi
This question is from textbook Blitzer College Algebra
Found 2 solutions by scott8148, stanbon:
Answer by scott8148(6628) (Show Source): You can put this solution on YOUR website!
you are trying to maximize z
test the vertices by substituting into z=2x+4y to find the maximum
some points can be eliminated by inspection __ (0,0) is obviously not the maximum
__ (6,3) must be greater than each of the first five points in the list
__ the maximum appears to be one of the last three points in the list
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
z=2x+4y
x>=0, y>=0 Linear programming problem, graphing a solution, and maximum.
x+3y>=6
x+y>=3
x+y<=9
I come up with serveral vertices. Even when using the example. I know how to put the vertices back into the equations, but I need to figure out how to limit or pick the correct vertices; (0,0), (0,2), (6,0), (0,3), (3,0), (5,4), (6,3), (0,9)
--------------
(0,0),(0,2),(3,0),(5,4) and (6,3) are not vertices of the
area that is enclose by y>=-x+3, y>=(-1/3)x+2, and y<=-3+9.
------------
Notice that two of the inequalities are above their boundary lines
and the 3rd is below it boundary line.
-------------
The vertices you want are (6,0),(9,0),(0,3),(0,9), and (3/2,3/2)
-------------
================
Cheers,
Stan H.
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