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Math Help - Area of Rectangle inside graph

  1. #1
    Member OnMyWayToBeAMathProffesor's Avatar
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    Area of Rectangle inside graph

    So I got the answer to this question but it seems a little iffy. If someone could just double check my work, that would be wonderful.

    I am not sure if this matters, but the level of math in this question is AP Calc-AB.

    Find the area of the largest rectangle (with sides parallel to the coordinate axis) that cna be inscribed in the region enclosed by the graphs f(x)=18-x^2 and g(x)=2x^2-9.

    there is a picture, but it isn't that helpful.

    x.doc

    so this is what i did. since we are looking for the largest area, i started with the area formula. A=bh. I labeled each the distance from the y axis x. and since the rectangle is touching each graph, these are the x coordinates. x, 2x^2-9 for the bottom right hand corner and x, 18-x^2 for the top right hand corner. So I know b=2x and h=the distance between x, 2x^2-9 and x, 18-x^2. So i used the distance formula; d=\sqrt{(x-x)^2+(18-x^2-2x^2+9)^2}
    d^2=(18-x^2-2x^2+9)^2
    d^2=(27-3x^2)^2
    d=27-3x^2

    so then i did A=(2x)(27-3x^2) i got the derivative and A'=54-18x^2 i found the zeros and one was negative so the only possibility was x=1.732.

    so x=1.732 will optimize the rectangle, correct?
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  2. #2
    MHF Contributor
    skeeter's Avatar
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    correct decimal value, but you're making this way too difficult.

    base of rectangle = 2x

    height of rectangle = f(x) - g(x) = 27 - 3x^2

    A = 2x(27 - 3x^2) = 54x - 6x^3

    \frac{dA}{dx} = 54 - 18x^2

    54 - 18x^2 = 0

    3 - x^2 = 0

    x = \sqrt{3}

    now sub in \sqrt{3} for x in the area equation and ATQ.
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  3. #3
    Member OnMyWayToBeAMathProffesor's Avatar
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    o wow, thanks you so much, that seems much easier.
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