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

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klimenkovBest ResponseYou've already chosen the best response.0
dw:1342945269745:dw Something like this.
 one year ago

huneyaBest 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 )
 one year ago

waterineyesBest ResponseYou've already chosen the best response.1
Sorry I am somewhat late than huneya.. Ha ha ha.. dw:1342945437729:dw
 one year ago

klimenkovBest ResponseYou've already chosen the best response.0
Use the Shiftbutton to draw circles instead of the ellipses.
 one year ago

waterineyesBest ResponseYou've already chosen the best response.1
Are you thinking it as ellipse???
 one year ago

klimenkovBest ResponseYou've already chosen the best response.0
Looks like a bit flattened.
 one year ago

waterineyesBest 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??
 one year ago

huneyaBest ResponseYou've already chosen the best response.0
http://www.themathpage.com/acalc/applied.htm
 one year ago

klimenkovBest ResponseYou've already chosen the best response.0
Don't be angry, please. It was just a joke.
 one year ago

mukushlaBest ResponseYou've already chosen the best response.1
dw:1342945810260:dw
 one year ago

waterineyesBest ResponseYou've already chosen the best response.1
I am also joking.. Ha ha ha.
 one year ago

mukushlaBest ResponseYou've already chosen the best response.1
dw:1342945859462:dw
 one year ago

FoolAroundMathBest 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 \(1\)
 one year ago

waterineyesBest ResponseYou've already chosen the best response.1
Yes @mukushla It has maximum area when the rectangle is a square, I guess..
 one year ago

mukushlaBest ResponseYou've already chosen the best response.1
dw:1342945935930:dw
 one year ago

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

waterineyesBest ResponseYou've already chosen the best response.1
dw:1342946128295:dw
 one year ago

waterineyesBest 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 = \(\huge {4 (\frac{r^2  2x^2}{\sqrt{r^2  x^2}})}\)
 one year ago

waterineyesBest ResponseYou've already chosen the best response.1
For maximum and minimum put Area' = 0
 one year ago

waterineyesBest ResponseYou've already chosen the best response.1
Sorry the above one that I found is Area' and not Area..
 one year ago

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

mukushlaBest ResponseYou've already chosen the best response.1
always calculus is explicit way to solving problems :D
 one year ago

waterineyesBest 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: \[2x = 2y \implies Square\]
 one year ago

klimenkovBest ResponseYou've already chosen the best response.0
How can I give the "Best Response" not only one @mukushla? I like all the solutions.
 one year ago

waterineyesBest ResponseYou've already chosen the best response.1
Go with mukushla...
 one year ago

mukushlaBest 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
 one year ago

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