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Math Help - maximizing largest area of rectangle within a circle

  1. #1
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    maximizing largest area of rectangle within a circle

    find the dimensions of the rectangle of largest area that can be inscribed in a circle of radius r
    circle:x^2+y^2=r^2
    area of the rectangle = 2x2y
    y = root(r^2-x^2)
    area=(4x)[root(r^2-x^2)]
    Area'=4r^2+4xr-8x^2/root(r^2-x^2)
    4r^2+4xr-8x^2=(r+2x)(r-x)=0
    x=r or x=-r/2
    plugging x back into area formula
    area = 4(-r/2)[root(r^2-r^2/4)]
    area = -r(root3r^2)
    the answer is that the sides are [root(2)]r, and that it is actually a square.
    Not sure what I'm doing wrong..
    also, what happened to the Latex tutorial...?
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  2. #2
    MHF Contributor Unknown008's Avatar
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    Can you show us how you got that line?

    Area'=\dfrac{4r^2+4xr-8x^2}{\sqrt{r^2-x^2}}

    From this:

    Area = 4x\sqrt{r^2 - x^2}

    I'm getting (taking r as a constant):

    \begin{array}{cl}<br />
Area' &= 4x\cdot \dfrac12 (r^2 - x^2)^{-\frac12}\cdot -2x + (r^2 - x^2)^{\frac12}\cdot 4 \\<br /> <br />
&  \\<br /> <br />
& = \dfrac{-4x^2}{\sqrt{ r^2-x^2}} + 4(r^2 - x^2)^{\frac12} \\<br />
&  \\<br /> <br />
& = \dfrac{1}{\sqrt{r^2 - x^2}} \left(-4x^2 + 4(r^2 - x^2)\right) \\<br />
&  \\<br /> <br />
& = \dfrac{-4x^2 + 4r^2 - 4x^2}{\sqrt{r^2 - x^2}} \\<br />
&  \\<br /> <br />
& = \dfrac{ 4r^2 - 8x^2}{\sqrt{r^2 - x^2}}\end{array}
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  3. #3
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    ohh got it! I forgot that r is actually a constant because it is a radius, I treated it like a variable.
    Using your derviative I got Area=r(root2)r(root2)
    so each side would r(root2) since area = L x W
    thanks!
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  4. #4
    MHF Contributor Unknown008's Avatar
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    You're welcome
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