klimenkov
  • klimenkov
Find the rectangle of maximal area inscribed in a circle of radius R.
Mathematics
chestercat
  • chestercat
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klimenkov
  • klimenkov
|dw:1342945269745:dw| Something like this.
anonymous
  • anonymous
the maximum area the function is xy (if x is length and y is breadth and radius is r )
anonymous
  • anonymous
Sorry I am somewhat late than huneya.. Ha ha ha.. |dw:1342945437729:dw|

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