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Find the rectangle of maximal area inscribed in a circle of radius R.

Mathematics
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|dw:1342945269745:dw| Something like this.
the maximum area the function is xy (if x is length and y is breadth and radius is r )
Sorry I am somewhat late than huneya.. Ha ha ha.. |dw:1342945437729:dw|

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Other answers:

Use the Shift-button to draw circles instead of the ellipses.
Are you thinking it as ellipse???
Looks like a bit flattened.
So I supposed to draw a perfect circle for you?? You can't understand it other wise? Right??
http://www.themathpage.com/acalc/applied.htm
Don't be angry, please. It was just a joke.
|dw:1342945810260:dw|
I am also joking.. Ha ha ha.
|dw:1342945859462:dw|
|dw:1342945796754:dw| Area of rectangle = \(2R\cos{\alpha}\times 2R\sin{\alpha}\) So basically, it is to max the functions \(\sin{2\alpha}\) which is \(1\)
Yes @mukushla It has maximum area when the rectangle is a square, I guess..
|dw:1342945935930:dw|
so the maximum area is for a square with \(a=b=r \sqrt{2} \)
|dw:1342946128295:dw|
was i wrong
Using pythogoras theorem: \[r^2 = x^2 + y^2\] \[y = \sqrt{r^2 - x^2}\] Area = (2x)(2y) = 4xy = \(\huge {4 (\frac{r^2 - 2x^2}{\sqrt{r^2 - x^2}})}\)
For maximum and minimum put Area' = 0
Sorry the above one that I found is Area' and not Area..
So: \[ {4 (\frac{r^2 - 2x^2}{\sqrt{r^2 - x^2}})} = 0 \implies r^2 - 2x^2 = 0\] \[x = \frac{r}{\sqrt{2}}\]
always calculus is explicit way to solving problems :D
Now you can check this this is will be maximum by taking the second derivative.. So, \[x = y = \frac{r}{\sqrt{2}}\] Or: \[2x = 2y \implies Square\]
How can I give the "Best Response" not only one @mukushla? I like all the solutions.
Go with mukushla...
@FoolAroundMath your approach is nicer than the others i gave a medal to foolaround math on behalf of u klimenkov
@FoolAroundMath and @waterineyes solutions are actually the same. One can get it using the substitution \(x=r\cosθ, \ y=r\sinθ\).

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