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Math Help - quadratic attachment?

  1. #1
    Member sluggerbroth's Avatar
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    quadratic attachment?

    I have attached the problem. Help!
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  2. #2
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    Re: quadratic attachment?

    Express the total length of fence needed in terms of x and y. Equate this length to 1500 and express y through x. The total area is xy, so using the obtained expression y(x) the area is xy(x). Find the point of maximum of this function using its derivative.
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  3. #3
    Member sluggerbroth's Avatar
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    Re: quadratic attachment?

    I need more detail please.
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    Re: quadratic attachment?

    Quote Originally Posted by sluggerbroth View Post
    I need more detail please.
    Well, it's like in hostage situation: you have to give us something in exchange for your demands Disclaimer: it's just a joke to make you smile; don't take it seriously.

    However, if you are not able to express the total required fence length in terms of x and y, I am not sure we can help...
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  5. #5
    Member sluggerbroth's Avatar
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    Re: quadratic attachment?

    I hear what you are saying. Are we talking 3x2y=Area? and 1500=9x+8y???
    Book has answer A(x)=1/4x(1500-3x) Max value 46,875 dimension 250x187.5
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    Re: quadratic attachment?

    Quote Originally Posted by sluggerbroth View Post
    I hear what you are saying. Are we talking 3x2y=Area? and 1500=9x+8y???
    Yes, though from the picture and from the answer I think x and y denote the whole width and height, respectively. Thus, 3x + 4y = 1500, from where y = (1500 - 3x) / 4 and the area is indeed x(1500 - 3x) / 4. The maximum point is the same as for x(1500 - 3x) = -3x≤ + 1500x. The vertex of a parabola ax≤ + bx + c is at x = -b / 2a, so the maximum point is x = 250.
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