klimenkov
Find the rectangle of maximal area inscribed in a circle of radius R.
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klimenkov
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Something like this.
huneya
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the maximum area the function is xy (if x is length and y is breadth and radius is r )
waterineyes
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Sorry I am somewhat late than huneya..
Ha ha ha..
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klimenkov
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Use the Shift-button to draw circles instead of the ellipses.
waterineyes
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Are you thinking it as ellipse???
klimenkov
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Looks like a bit flattened.
waterineyes
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So I supposed to draw a perfect circle for you??
You can't understand it other wise?
Right??
klimenkov
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Don't be angry, please. It was just a joke.
mukushla
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waterineyes
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I am also joking..
Ha ha ha.
mukushla
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FoolAroundMath
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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
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Yes @mukushla It has maximum area when the rectangle is a square, I guess..
mukushla
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mukushla
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so the maximum area is for a square with \(a=b=r \sqrt{2} \)
waterineyes
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huneya
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was i wrong
waterineyes
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Using pythogoras theorem:
\[r^2 = x^2 + y^2\]
\[y = \sqrt{r^2 - x^2}\]
Area = (2x)(2y) = 4xy = \(\huge {4 (\frac{r^2 - 2x^2}{\sqrt{r^2 - x^2}})}\)
waterineyes
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For maximum and minimum put Area' = 0
waterineyes
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Sorry the above one that I found is Area' and not Area..
waterineyes
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So:
\[ {4 (\frac{r^2 - 2x^2}{\sqrt{r^2 - x^2}})} = 0 \implies r^2 - 2x^2 = 0\]
\[x = \frac{r}{\sqrt{2}}\]
mukushla
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always calculus is explicit way to solving problems :D
waterineyes
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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:
\[2x = 2y \implies Square\]
klimenkov
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How can I give the "Best Response" not only one @mukushla? I like all the solutions.
waterineyes
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Go with mukushla...
mukushla
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@FoolAroundMath
your approach is nicer than the others
i gave a medal to foolaround math on behalf of u klimenkov
klimenkov
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@FoolAroundMath and @waterineyes solutions are actually the same. One can get it using the substitution \(x=r\cosθ, \ y=r\sinθ\).