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Thread: "feasible solutions"?

  1. #1 "feasible solutions"? 
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    Hey there,

    I've one question on this problem. So, I'm working on word problems based on graphing inequalities.

    And there's this one question I don't get. It says "Explain why the feasible solutions of this problem are ordered pairs of whole numbers".

    Actually, I'll put down full c) part from this word problem.

    c) A feasible solution is one that satisfies all the conditions of the problem, including the obvious ones that are not stated. (<--- "obvious ones that are not stated", what??) Explain why the feasible solutions of this problem are ordered pairs of whole numbers. Describe two possible solutions.

    I'll copy down the full word problem for you.

    "A cycle dealer wishes to purchase touring cycles for $300 each and racing cycles for $400 each. She plans to spend not more than $8400. She buys x touring models and y racing models. An inequality that represents this is 300x+400y\< (that's "≤", if you can see the symbol) 8400, which can be simplified to y \< (that's "≤", if you can see the symbol) 21-0.75x."

    So, I graphed the inequality and here's what I've got - I picked two possible solutions which would be (11,12) and (20,10).

    (11,12) is located in the shaded area and (20,10) is located on the line, by the way.

    Thanks for any help.......


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  3. #2  
    Forum Professor serpicojr's Avatar
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    If I just gave you the inequality 300x + 400y ≤ 8400 and asked you for solutions, you could come up with some answers which don't make sense in the context of your problem. For example, consider the solution (x,y) = (-1,-1). This is a solution: -700 ≤ 8400. However, I can't buy negative numbers of bikes. So this cannot be a solution in the context of the problem. Similarly, I could take (x,y) = (1/2,1/2) as a solution to the inequality, since 350 ≤ 8400. But I cannot buy half a bike, and so this isn't a solution to the problem, either. So these are some obvious conditions which are not stated: you can only buy bikes in whole number quantities. The problem is basically asking why not every pair of real numbers (x,y) satisfying 300x + 400y ≤ 8400 is a solution to the problem.


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  4. #3  
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    Ahhh yes, aye I think I got you.

    I see, if I pick a solution which actually exists but does not match the answer, that would be a feasible solution, correct?

    Also, if I pick an existing solution that's a (10, 11.5) - that would be a feasible solution?

    Again, thank you for your reply.
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  5. #4  
    Forum Professor serpicojr's Avatar
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    Maybe you should think of "feasible" as meaning "possible" in this context.

    (10, 11.5) wouldn't be a feasible solution--it's not possible to buy 11 and a half bikes. (Well, I guess it is, but you know what I mean...)
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  6. #5  
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    Well, there are two words involved, "feasible" and "possible solutions" in part c. but okay.

    Besides, 11.5 wouldn't count as a whole number, right? So...
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  7. #6  
    Forum Professor serpicojr's Avatar
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    I'm pretty sure they use feasible and possible to mean the same thing. Did they ever define what they meant by a possible solution?

    But, yeah, that's the point: 11.5 isn't a whole number, so it can't be a feasible solution.
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