Question 145904
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) 
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(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.
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Notice that two of the inequalities are above their boundary lines
and the 3rd is below it boundary line.
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The vertices you want are (6,0),(9,0),(0,3),(0,9), and (3/2,3/2)
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{{{graph(400,300,-10,10,-10,10,(-1/3)x+2,-x+3,-x+9)}}}
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Cheers,
Stan H.