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Thread: Area question

  1. #1
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    Area question

    I have to find the maximum area of a rectangle that already has a 100m fence in the ground, and I am given 200m to add onto it. However, my answer doesn't make sense:

    A=b x h
    P=300=2b + 2h
    300=2(100+x)+2h
    300=200+2x+2h
    h=50-x
    A=b(50-x)
    A=50b-xb
    dA/db = 50-x
    0=50-x
    x=50
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  2. #2
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    Quote Originally Posted by Dudealadude View Post
    I have to find the maximum area of a rectangle that already has a 100m fence in the ground, and I am given 200m to add onto it. However, my answer doesn't make sense:

    A=b x h
    P=300=2b + 2h
    300=2(100+x)+2h
    300=200+2x+2h
    h=50-x
    A=b(50-x)

    b=100+x

    A=(100+x)(50-x)=5000-50x-x^2

    dA/dx=-2x-50

    x=-25


    A=50b-xb
    dA/db = 50-x
    0=50-x
    x=50
    Hence, you must create a kink in the 100m side so that it is 75m long after.
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  3. #3
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    I think the purpose of the question is that one side has to be atleast 100m, in your answer they are both 75. No kinks allowed. Thanks though!
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  4. #4
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    Quote Originally Posted by Dudealadude View Post
    I think the purpose of the question is that one side has to be atleast 100m, in your answer they are both 75. No kinks allowed. Thanks though!
    In that case x=0

    This is because the graph is an inverted U-shape with an absolute maximum at x=-25.
    Hence for x non-negative, the area will be maximised at x=0,
    since any positive x will give a lower value for area.

    therefore, h=50.
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