# Closed curve

#### idontknow

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
now find the max possible length of a closed curve enclosing area $A$!

#### idontknow

now find the max possible length of a closed curve enclosing area $A$!
btw can you provide a hint ?

Math Team

#### mathman

Forum Staff
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 -> $\infty$..

• idontknow and romsek

#### romsek

Math Team
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

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.

• idontknow

#### idontknow

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. $$\displaystyle c=L/2\pi$$.

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#### mathman

Forum Staff
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.