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klimenkov
Group Title
Find the rectangle of maximal area inscribed in a circle of radius R.
 2 years ago
 2 years ago
klimenkov Group Title
Find the rectangle of maximal area inscribed in a circle of radius R.
 2 years ago
 2 years ago

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klimenkov Group TitleBest ResponseYou've already chosen the best response.0
dw:1342945269745:dw Something like this.
 2 years ago

huneya Group TitleBest ResponseYou've already chosen the best response.0
the maximum area the function is xy (if x is length and y is breadth and radius is r )
 2 years ago

waterineyes Group TitleBest ResponseYou've already chosen the best response.1
Sorry I am somewhat late than huneya.. Ha ha ha.. dw:1342945437729:dw
 2 years ago

klimenkov Group TitleBest ResponseYou've already chosen the best response.0
Use the Shiftbutton to draw circles instead of the ellipses.
 2 years ago

waterineyes Group TitleBest ResponseYou've already chosen the best response.1
Are you thinking it as ellipse???
 2 years ago

klimenkov Group TitleBest ResponseYou've already chosen the best response.0
Looks like a bit flattened.
 2 years ago

waterineyes Group TitleBest ResponseYou've already chosen the best response.1
So I supposed to draw a perfect circle for you?? You can't understand it other wise? Right??
 2 years ago

huneya Group TitleBest ResponseYou've already chosen the best response.0
http://www.themathpage.com/acalc/applied.htm
 2 years ago

klimenkov Group TitleBest ResponseYou've already chosen the best response.0
Don't be angry, please. It was just a joke.
 2 years ago

mukushla Group TitleBest ResponseYou've already chosen the best response.1
dw:1342945810260:dw
 2 years ago

waterineyes Group TitleBest ResponseYou've already chosen the best response.1
I am also joking.. Ha ha ha.
 2 years ago

mukushla Group TitleBest ResponseYou've already chosen the best response.1
dw:1342945859462:dw
 2 years ago

FoolAroundMath Group TitleBest ResponseYou've already chosen the best response.1
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 \(1\)
 2 years ago

waterineyes Group TitleBest ResponseYou've already chosen the best response.1
Yes @mukushla It has maximum area when the rectangle is a square, I guess..
 2 years ago

mukushla Group TitleBest ResponseYou've already chosen the best response.1
dw:1342945935930:dw
 2 years ago

mukushla Group TitleBest ResponseYou've already chosen the best response.1
so the maximum area is for a square with \(a=b=r \sqrt{2} \)
 2 years ago

waterineyes Group TitleBest ResponseYou've already chosen the best response.1
dw:1342946128295:dw
 2 years ago

waterineyes Group TitleBest ResponseYou've already chosen the best response.1
Using pythogoras theorem: \[r^2 = x^2 + y^2\] \[y = \sqrt{r^2  x^2}\] Area = (2x)(2y) = 4xy = \(\huge {4 (\frac{r^2  2x^2}{\sqrt{r^2  x^2}})}\)
 2 years ago

waterineyes Group TitleBest ResponseYou've already chosen the best response.1
For maximum and minimum put Area' = 0
 2 years ago

waterineyes Group TitleBest ResponseYou've already chosen the best response.1
Sorry the above one that I found is Area' and not Area..
 2 years ago

waterineyes Group TitleBest ResponseYou've already chosen the best response.1
So: \[ {4 (\frac{r^2  2x^2}{\sqrt{r^2  x^2}})} = 0 \implies r^2  2x^2 = 0\] \[x = \frac{r}{\sqrt{2}}\]
 2 years ago

mukushla Group TitleBest ResponseYou've already chosen the best response.1
always calculus is explicit way to solving problems :D
 2 years ago

waterineyes Group TitleBest ResponseYou've already chosen the best response.1
Now you can check this this is will be maximum by taking the second derivative.. So, \[x = y = \frac{r}{\sqrt{2}}\] Or: \[2x = 2y \implies Square\]
 2 years ago

klimenkov Group TitleBest ResponseYou've already chosen the best response.0
How can I give the "Best Response" not only one @mukushla? I like all the solutions.
 2 years ago

waterineyes Group TitleBest ResponseYou've already chosen the best response.1
Go with mukushla...
 2 years ago

mukushla Group TitleBest ResponseYou've already chosen the best response.1
@FoolAroundMath your approach is nicer than the others i gave a medal to foolaround math on behalf of u klimenkov
 2 years ago

klimenkov Group TitleBest ResponseYou've already chosen the best response.0
@FoolAroundMath and @waterineyes solutions are actually the same. One can get it using the substitution \(x=r\cosθ, \ y=r\sinθ\).
 2 years ago
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