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Math Help - optimization problem

  1. #1
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    optimization problem

    A farmer wishes to fence an area next to his barn. He needs a wire fence that costs $1 per linear foot in front of the barn and and wooden fencing that costs $2 per foot on the other sides.
    Find x and y so that he can enclose the maximum area if his budget for materials is $4400.


    A=2xy

    c=(2*2x) + (1*y)=4400

    4x=4400-y

    x=1100-\frac{y}{4}

    A=2(1100-\frac{y}{4})y=2200y-\frac{y^2}{2}

    A'=2200-y

    y=2200<br />

    Y is supposed to be \frac{2200}{3}.
    So either my area function or my constraint is wrong?
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  2. #2
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    Does the problem say rectangular area? If so,

    You're area should just be xy, if you draw a picture you will see this, you need the 2 in there when considering cost, as you have done
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  3. #3
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    why do you think Y is incorrect? because I am actually getting 2200 and 550
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  4. #4
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    They had a picture like this:

    where W was labled " x", and L was labeled " y". The top horizontal piece opposite the "L" was labeled " barn(wire)."

    I still ended up getting y=2200, when I did A=xy though.

    A=(1100-\frac{y}{4})y
    1100y-\frac{y^2}{4})
    A'=1100-\frac{y}{2})
    1100=\frac{y}{2}
    y=2200
    ?
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  5. #5
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    Quote Originally Posted by artvandalay11 View Post
    why do you think Y is incorrect? because I am actually getting 2200 and 550
    Here are the solutions: http://www.math.ufl.edu/~jysmith/MAC...0answerf09.pdf

    This is #22.
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  6. #6
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    Yes I'm sorry about the xy comment, I realized it wouldnt make a difference for your Y value since the 2 goes away when you differentiate, but that answer has to be incorrect


    We both agree on Cost:

    C=2(2x)+1(y)=4x+y=4400

    Now let's try the "answer"

    4(550)+\frac{2200}{3}=2200+\frac{2200}{3}\not =4400
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  7. #7
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    Y would equal \frac{2200}{3} if the cost for that side of fencing was $3 per foot
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  8. #8
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    That is true. I guess I'll have to straighten that out with my professor.Well, anyways .
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