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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

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  1. klimenkov Group Title
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    |dw:1342945269745:dw| Something like this.

    • 2 years ago
  2. huneya Group Title
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    the maximum area the function is xy (if x is length and y is breadth and radius is r )

    • 2 years ago
  3. waterineyes Group Title
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    Sorry I am somewhat late than huneya.. Ha ha ha.. |dw:1342945437729:dw|

    • 2 years ago
  4. klimenkov Group Title
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    Use the Shift-button to draw circles instead of the ellipses.

    • 2 years ago
  5. waterineyes Group Title
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    Are you thinking it as ellipse???

    • 2 years ago
  6. klimenkov Group Title
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    Looks like a bit flattened.

    • 2 years ago
  7. waterineyes Group Title
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    So I supposed to draw a perfect circle for you?? You can't understand it other wise? Right??

    • 2 years ago
  8. huneya Group Title
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    http://www.themathpage.com/acalc/applied.htm

    • 2 years ago
  9. klimenkov Group Title
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    Don't be angry, please. It was just a joke.

    • 2 years ago
  10. mukushla Group Title
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    |dw:1342945810260:dw|

    • 2 years ago
  11. waterineyes Group Title
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    I am also joking.. Ha ha ha.

    • 2 years ago
  12. mukushla Group Title
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    |dw:1342945859462:dw|

    • 2 years ago
  13. FoolAroundMath Group Title
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    |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
  14. waterineyes Group Title
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    Yes @mukushla It has maximum area when the rectangle is a square, I guess..

    • 2 years ago
  15. mukushla Group Title
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    |dw:1342945935930:dw|

    • 2 years ago
  16. mukushla Group Title
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    so the maximum area is for a square with \(a=b=r \sqrt{2} \)

    • 2 years ago
  17. waterineyes Group Title
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    |dw:1342946128295:dw|

    • 2 years ago
  18. huneya Group Title
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    was i wrong

    • 2 years ago
  19. waterineyes Group Title
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    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
  20. waterineyes Group Title
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    For maximum and minimum put Area' = 0

    • 2 years ago
  21. waterineyes Group Title
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    Sorry the above one that I found is Area' and not Area..

    • 2 years ago
  22. waterineyes Group Title
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    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
  23. mukushla Group Title
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    always calculus is explicit way to solving problems :D

    • 2 years ago
  24. waterineyes Group Title
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    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
  25. klimenkov Group Title
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    How can I give the "Best Response" not only one @mukushla? I like all the solutions.

    • 2 years ago
  26. waterineyes Group Title
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    Go with mukushla...

    • 2 years ago
  27. mukushla Group Title
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    @FoolAroundMath your approach is nicer than the others i gave a medal to foolaround math on behalf of u klimenkov

    • 2 years ago
  28. klimenkov Group Title
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    @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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