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

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

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

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

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

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

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

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

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

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

FoolAroundMath
 2 years ago
Best ResponseYou've already chosen the best response.1dw: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 \(1\)

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

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

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

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

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

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

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

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

waterineyes
 2 years ago
Best ResponseYou've already chosen the best response.1Now 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
 2 years ago
Best ResponseYou've already chosen the best response.0How can I give the "Best Response" not only one @mukushla? I like all the solutions.

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

mukushla
 2 years ago
Best 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

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