A community for students.
Here's the question you clicked on:
 0 viewing
klimenkov
 3 years ago
Find the rectangle of maximal area inscribed in a circle of radius R.
klimenkov
 3 years ago
Find the rectangle of maximal area inscribed in a circle of radius R.

This Question is Closed

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

huneya
 3 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
 3 years ago
Best ResponseYou've already chosen the best response.1Sorry I am somewhat late than huneya.. Ha ha ha.. dw:1342945437729:dw

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

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

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

waterineyes
 3 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
 3 years ago
Best ResponseYou've already chosen the best response.0Don't be angry, please. It was just a joke.

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

FoolAroundMath
 3 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
 3 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
 3 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
 3 years ago
Best ResponseYou've already chosen the best response.1dw:1342946128295:dw

waterineyes
 3 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
 3 years ago
Best ResponseYou've already chosen the best response.1For maximum and minimum put Area' = 0

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

waterineyes
 3 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
 3 years ago
Best ResponseYou've already chosen the best response.1always calculus is explicit way to solving problems :D

waterineyes
 3 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
 3 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
 3 years ago
Best ResponseYou've already chosen the best response.1Go with mukushla...

mukushla
 3 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
 3 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 \(x=r\cosθ, \ y=r\sinθ\).
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.