# Math Help - maximizing largest area of rectangle within a circle

1. ## 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...?

2. 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}
Area' &= 4x\cdot \dfrac12 (r^2 - x^2)^{-\frac12}\cdot -2x + (r^2 - x^2)^{\frac12}\cdot 4 \\

& \\

& = \dfrac{-4x^2}{\sqrt{ r^2-x^2}} + 4(r^2 - x^2)^{\frac12} \\
& \\

& = \dfrac{1}{\sqrt{r^2 - x^2}} \left(-4x^2 + 4(r^2 - x^2)\right) \\
& \\

& = \dfrac{-4x^2 + 4r^2 - 4x^2}{\sqrt{r^2 - x^2}} \\
& \\

& = \dfrac{ 4r^2 - 8x^2}{\sqrt{r^2 - x^2}}\end{array}$

3. 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!

4. You're welcome