Linear Programming: Min 2A + 3B s.t. 1A+3B<=12, 3A+1B>=13, ...  TOPIC_SOLVED

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Linear Programming: Min 2A + 3B s.t. 1A+3B<=12, 3A+1B>=13, ...

Postby needhelp2 on Thu Mar 24, 2011 6:49 am

I am stucktrying to solve for the following:
Min 2A + 2B
s.t.
A. 1A + 3B <= 12
B. 3A + 1B >= 13
C. 1A - 1B = 3

I need to
A. Show the Feasibility Region
B. What is the extreem point of the feasible region
C. Find the optimal solution using the graphical solution procedure.

I found my lines:
A.(0,4) (12,0)
B. (0,13)(4.333,0) This one I am unsure
C. (0,-3)(3,0)

I believe my extreme points of feasibility have to be on the line because of equation C, however dont know how to get to it.
I can add lines B and C (4,1)
but when I add lines A and C I get confused.

Please help.
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Postby stapel_eliz on Thu Mar 24, 2011 11:24 am

needhelp2 wrote:I found my lines:
A.(0,4) (12,0)
B. (0,13)(4.333,0) This one I am unsure
C. (0,-3)(3,0)

What you have posted are pairs of points. Perhaps you mean that you have found points so that you can now draw the lines...?

needhelp2 wrote:I believe my extreme points of feasibility have to be on the line...

I'm not sure what you mean by the above...? As your book and instructor should have mentioned (and as the lesson provided to you in your previous thread explained), the points to test in the optimization equation are the corner points of the feasibility region. So you need to find the intersections of the various pairs of lines (or look on the graph, if the points are "neat" whole numbers).

needhelp2 wrote:dont know how to get to it.
I can add lines B and C (4,1)
but when I add lines A and C I get confused.

When putting two equations equal (for instance, B = -(1/3)A + 4 and B = -3A + 13/3), you are, in effect, solving a system of equations. To learn how that works, please try here.
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