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Math Help - quadratic equations word problems geometry (math 20 pure)

  1. #1
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    quadratic equations word problems geometry (math 20 pure)

    Hey! I'm having trouble with a certain question.

    A rectangular field is to be enclosed by 600 m of fence.
    a) What dimensions will give a maximum area?
    b) What is the maximum area?

    This is what I have so far:

    x= 2 sides of rectangular field
    y= 2 other sides of rectangular field

    I know that:
    2x+2y=600
    and for the area I need to do (x)(y) = area.

    Now what???

    Thank you : )
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  2. #2
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    Where's that function notation you studied earlier? You need it now.

    Area is a function of x and y

    Area(x,y) = x*y

    We also know that x and y are related, so y can be expressed in terms of x.

    2x + 2y = 600

    x + y = 300

    y = 300 - x

    After substitution, this gives

    Area(x) = x*(300-x) = 300x - x^2

    This should look like a nice quadratic or parabola. You should have studied how to find the maximum of such things.

    Show us what you get.
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  3. #3
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    When I graphed " 300x - x^2 " in my calculator, it gave me a straight line.
    Am I doing something wrong?
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  4. #4
    MHF Contributor kalagota's Avatar
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    no, it is a parabola. i will look like a "line" in small intervals..

    anyways, the point is, you have to find the maximum area right? so, could you show us what you did?
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  5. #5
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    Haha, you were right, it did work on my calculator. I found the maximum on my calculator and it gave me x = 150 and y = 22500, but I have no idea how to use those results to answer the two questions that go along with the word problem.
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  6. #6
    MHF Contributor kalagota's Avatar
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    Quote Originally Posted by dancingqueen9 View Post
    Haha, you were right, it did work on my calculator. I found the maximum on my calculator and it gave me x = 150 and y = 22500, but I have no idea how to use those results to answer the two questions that go along with the word problem.
    take note of the y=22500. this is your A(150) or the maximum area..

    now, you have already have one dimension, what is the other dimension? use the constraints..
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