## klimenkov 3 years ago Find the rectangle of maximal area inscribed in a circle of radius R.

1. klimenkov

|dw:1342945269745:dw| Something like this.

2. huneya

the maximum area the function is xy (if x is length and y is breadth and radius is r )

3. waterineyes

Sorry I am somewhat late than huneya.. Ha ha ha.. |dw:1342945437729:dw|

4. klimenkov

Use the Shift-button to draw circles instead of the ellipses.

5. waterineyes

Are you thinking it as ellipse???

6. klimenkov

Looks like a bit flattened.

7. waterineyes

So I supposed to draw a perfect circle for you?? You can't understand it other wise? Right??

8. huneya
9. klimenkov

Don't be angry, please. It was just a joke.

10. mukushla

|dw:1342945810260:dw|

11. waterineyes

I am also joking.. Ha ha ha.

12. mukushla

|dw:1342945859462:dw|

13. 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 $$1$$

14. waterineyes

Yes @mukushla It has maximum area when the rectangle is a square, I guess..

15. mukushla

|dw:1342945935930:dw|

16. mukushla

so the maximum area is for a square with $$a=b=r \sqrt{2}$$

17. waterineyes

|dw:1342946128295:dw|

18. huneya

was i wrong

19. waterineyes

Using pythogoras theorem: $r^2 = x^2 + y^2$ $y = \sqrt{r^2 - x^2}$ Area = (2x)(2y) = 4xy = $$\huge {4 (\frac{r^2 - 2x^2}{\sqrt{r^2 - x^2}})}$$

20. waterineyes

For maximum and minimum put Area' = 0

21. waterineyes

Sorry the above one that I found is Area' and not Area..

22. waterineyes

So: ${4 (\frac{r^2 - 2x^2}{\sqrt{r^2 - x^2}})} = 0 \implies r^2 - 2x^2 = 0$ $x = \frac{r}{\sqrt{2}}$

23. mukushla

always calculus is explicit way to solving problems :D

24. waterineyes

Now you can check this this is will be maximum by taking the second derivative.. So, $x = y = \frac{r}{\sqrt{2}}$ Or: $2x = 2y \implies Square$

25. klimenkov

How can I give the "Best Response" not only one @mukushla? I like all the solutions.

26. waterineyes

Go with mukushla...

27. mukushla

@FoolAroundMath your approach is nicer than the others i gave a medal to foolaround math on behalf of u klimenkov

28. klimenkov

@FoolAroundMath and @waterineyes solutions are actually the same. One can get it using the substitution $$x=r\cosθ, \ y=r\sinθ$$.