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Math Help - Largest Rectangle

  1. #1
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    Largest Rectangle

    A rectangle has its base on the x-axis and its upper two vertices on the parabola y=12-x^2. What is the largest area the rectangle can have, and what are its dimensions?

    I wasn't at school for the lesson so I don't know where to begin. Any help would be nice. Thanks.
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  2. #2
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    Quote Originally Posted by lancelot854 View Post
    A rectangle has its base on the x-axis and its upper two vertices on the parabola y=12-x^2. What is the largest area the rectangle can have, and what are its dimensions?

    I wasn't at school for the lesson so I don't know where to begin. Any help would be nice. Thanks.
    12-x^2=(\sqrt{12}-x)(\sqrt{12}+x)

    This crosses the x-axis at \pm\sqrt{12}

    It is symmetrical about the y-axis so we can maximise the half of the rectangle that lies on the positive side of the x-axis.

    The entire rectangle area A is 2x\left(12-x^2\right)=24x-2x^3

    This is a maximum when the tangent to the graph is horizontal, hence differentiate A and set the derivative to zero to find x.
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  3. #3
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    You know that the length of this rectangle is 2x (simply from drawing it), and the height of this rectangle is y, but for consistency's sake, rewrite it as height = 12 - x^2. So, the area of our rectangle is,
    A(x) = 2x(12 - x^2) = 24x - 2x^3. You should use the first derivative test to find out when this maximizes.
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  4. #4
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    Thanks for the help guys, however, I just don't understand where the 2x in this problem comes from. How could I find this? Thanks again.
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  5. #5
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    The picture of a rectangle tangent to a parabola is here, (generically):
    http://www.ms.uky.edu/~carl/ma123/ko...chap14e115.gif
    We want the horizontal length of this rectangle. Well, the rectangle starts (horizontally at some point on the graph (-x, y), and ends at the point (x, y). So the length is simply  x - (-x) = 2x.
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  6. #6
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    Ok, it all makes sense now, thanks a lot.
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