Linear programming graphical solutions

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sunflower69
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How do you do part e) I get x=5 and y=9, but answer is (6,5)
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sunflower69
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Size:  115.3 KBHere is what graph looks like
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mqb2766
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There are three lines (two constraint, one objective) which are almost parallel so the solution is a bit confusing.
(5,9) does not satisfy the 2nd or 3rd constraints.
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sunflower69
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(Original post by mqb2766)
There are three lines (two constraint, one objective) which are almost parallel so the solution is a bit confusing.
(5,9) does not satisfy the 2nd or 3rd constraints.
Oh yea, so how do I get to (6,5)
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mqb2766
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(Original post by sunflower69)
Oh yea, so how do I get to (6,5)
Note the solutions here are integers, so the maximum of the objective is close to (but not necessarily on) a vertex

I'd start at (8,0) which is feasible and gives £28.00 and check other integer solutions close by, in your feasible space. Note you can slide up the 5x+2y=40 constraint as it's slightly steeper (downwards) than the objective so you get more money and still be feasible. The best point (6,5) occurs here. However, the other two constraints are shallower (gradient) than the objective, so you'll lose money while still being feasible if you slide along them on the boundary of the feasible space.

Tbh, they could have made the gradients of the constraints and objectives a bit more different if they want you to get a graphical solution.
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