Closed curve

Dec 2015
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Earth
Find maximum possible area of a closed curve with perimeter L.
My approach : \(\displaystyle \max A = \frac{1}{2} \int_{0}^{\alpha } \max y^2 (\alpha ) d\alpha=\frac{1}{2}\int_{0}^{\alpha} c^2 d\alpha =\frac{1}{2}c^2 \alpha\) which is the circle's sector area and \(\displaystyle \frac{dy^2 (\alpha ) }{d\alpha}=0\).
So \(\displaystyle r=\frac{L}{2\pi}\) and \(\displaystyle A=r^2 \pi = \frac{L^2 \pi }{4\pi^2 }=\frac{L^2 }{4\pi }\).
 
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romsek

Math Team
Sep 2015
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now find the max possible length of a closed curve enclosing area $A$!
 

mathman

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May 2007
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Max length is "infinite." Rectangle with pair of opposite sides is x and other pair is L. Total length is 2(L+x) and area is xL=A. Fix A and let x-> 0, then L -> $\infty$..
 

romsek

Math Team
Sep 2015
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I was thinking more along the lines of taking the perimeter of any shape with area A and fractalizing it somehow, there are a variety of recursive methods.
The area enclosed essentially stays the same while the length of the perimeter goes to infinity.

But yours is much simpler and works just as well!
 

SDK

Sep 2016
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541
USA
Find maximum possible area of a closed curve with perimeter L.
My approach : \(\displaystyle \max A = \frac{1}{2} \int_{0}^{\alpha } \max y^2 (\alpha ) d\alpha=\frac{1}{2}\int_{0}^{\alpha} c^2 d\alpha =\frac{1}{2}c^2 \alpha\) which is the circle's sector area and \(\displaystyle \frac{dy^2 (\alpha ) }{d\alpha}=0\).
So \(\displaystyle r=\frac{L}{2\pi}\) and \(\displaystyle A=r^2 \pi = \frac{L^2 \pi }{4\pi^2 }=\frac{L^2 }{4\pi }\).
I have no idea what you are trying to compute, but the expression, $\int_{0}^{\alpha } \max y^2 (\alpha ) d\alpha$ is not well defined at all so anything coming after it is nonsense.
 
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Dec 2015
1,078
166
Earth
It must be \(\displaystyle \frac{1}{2} \int_{0}^{\phi_0 } f^2 (\phi) d\phi\) , the area in polar coordinates.
\(\displaystyle \frac{1}{2} \int_{0 }^{2\pi} c^2 d\phi = c^2 \pi \) , where c-radius of circle. \(\displaystyle c=L/2\pi\).
 
Last edited:

mathman

Forum Staff
May 2007
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I was thinking more along the lines of taking the perimeter of any shape with area A and fractalizing it somehow, there are a variety of recursive methods.
The area enclosed essentially stays the same while the length of the perimeter goes to infinity.

But yours is much simpler and works just as well!
My idea was based on what Gerrymandered districts look like.