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Quadratic Functions Word Problem

  1. Mar 30, 2013 #1
    1. The problem statement, all variables and given/known data

    Your budget for constructing a rectangular enclosure which consists of a high surrounding fence and a lower inside fence that divides the enclosure in half is 2400 dollars. The high fence costs 8 dollars per foot and the low fence costs 4 dollars per foot. Find the dimensions and the maximum area of each half of the enclosure.

    2. Relevant equations

    -None-

    3. The attempt at a solution

    I know you probably have to figure out how much each type of fencing is going to cost then divide that for the particular fence giving you the dimensions. I don't know how to set up the proper formula for getting it though. I tried 16x+8y=2400 then solving for y and plugging it back in but I keep going in circles. I couldn't get much further than that because I am confused.

    Thanks for any help!
     
  2. jcsd
  3. Mar 30, 2013 #2

    SteamKing

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    First, draw a picture of this problem. Remember, you are trying to maximize area enclosed while staying under budget with the purchase of the fence. You are going to need to write an equation for the area enclosed by the fence and another equation for the cost of the fence.
     
  4. Mar 30, 2013 #3
    there is a picture in the book but I still don't understand it

    Area= xy
    Cost of fence, 8x+4y=2400 ?

    ?

    Image:

    8ZYmfeM.png
     
    Last edited: Mar 30, 2013
  5. Mar 30, 2013 #4
    Let x be length, y be height:

    Low Fence = y.
    2*8*x = Cost of Horizontal Fence
    2*8*y + 4*y = Cost of Vertical Fence + Cost of Low Fence

    16x + 20y = 2400

    Thus, y = (2400-16x)/20

    You might need calculus from here onward. Did you learn calculus?
     
  6. Mar 30, 2013 #5
    I am in precalculus
    After you solve for y do you plug it back into the original equation?
     
  7. Mar 31, 2013 #6
    No, you shouldn't. I'm not very sure how to do this in this case, because I only know the Calculus method, sorry
     
  8. Mar 31, 2013 #7

    ehild

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    The enclosed area A=xy must be maximum. Substitute for y: You get a quadratic function A(x). If you plot it, at what x is the peak of the parabola? (Use completing the square)

    ehild
     
  9. Mar 31, 2013 #8
    Thanks for the help I just used x=-b/2a to find h and subbed it back in for k giving me (h, k) the vertex which are the lengths 75' x 60' of the sides
     
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