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Math Help - Optimizing area within a parabola

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    Optimizing area within a parabola

    Find the largest rectangle that is inside the graph of the parabola $y = x^2$ below the line $y = a$ (a is an unspecified constant value), with the top side of the rectangle on the horizontal line $y = a$.
    The explanation says we want $x$ to be in $[0, \sqrt{a}]$. What does $ \sqrt{a} $ mean?
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    Re: Optimizing area within a parabola

    Is it the point where $y=a$ intersects the parabola?

    We set $x^2 = a$ and then solve for $x$?
    Last edited by maxpancho; August 21st 2014 at 03:34 PM.
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    Re: Optimizing area within a parabola

    Quote Originally Posted by maxpancho View Post
    The explanation says we want $x$ to be in $[0, \sqrt{a}]$. What does $ \sqrt{a} $ mean?
    ??? The square root of a, of course.

    Is it the point where y=a intersects the parabola?

    We set x 2 =a and then solve for x ?
    Yes, the line y= a intersects y= x^2 where x= \sqrt{a}. Since the given region is to lie inside (above) the graph of y= x^2 and below y= a, any point in the region must have -\sqrt{a}\le x\le \sqrt{a}.

    They are probably assuming a few things not explicitly stated: The largest rectangle will have sides parallel to the x and y axes. The top edge will be on the line y= a. The largest rectangle will be symmetric about y= 0. With a little thought you should see that those things follow from "symmetry conditions". For example, if a rectangle has vertical sides from -a to b with b< a, the we can get a larger rectangle by extending the right side from b to a.

    Given that, such a rectangle must have vertices at (-x, a), (x, a), on the line y= a, and, going straight down to the parabola, (-x, x^2), (x, x^2). The horizontal sides of that rectangle, from x to -x, have length 2x and the vertical sides. from a to x^2, have length [tex]a- x^2[tex] The area to be maximized is (2x)(a- x^2)= 2ax- 2x^3.
    Last edited by HallsofIvy; August 21st 2014 at 05:01 PM.
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