# Simple closed curve area?

1. Aug 23, 2006

### waht

What steps would you have to take to prove that an area of a simple closed curve is maximized if it is a circle?

I believe you would have to do some calculus of variations but i'm not sure.

2. Aug 24, 2006

### HallsofIvy

Yes, you would have to use "calculus of variations" to find a function that maximizes or minimizes a functional. If the boundary of the area is given by parametric equations, x= f(t), y= g(t), then, from Green's theorem the area is given by
$$\int\int dxdy= \int (\frac{\partial f}{\partial y}- \frac{\partial f}{\partial x}) dt[/itex] That second functional can be used to derive the Euler equation. 3. Aug 24, 2006 ### tieu sory, but is this uni stuff? or highschool? cause ive never seen green's theorem :tongue2: 4. Aug 24, 2006 ### stefannm I agree Greens theorem is a good place to start; however, I am not sure you could maximize this without some other constraints, ie the perimeter is equal to a constant or something. Without the another constraint, the area of any shape could be arbitrarily large. For greens theorem, you can find it in an undergrad mulitivariable calculus book. 5. Aug 24, 2006 ### waht Thanks I forgot to mention the perimeter has to be minimized. How would you go exactly from Green's theorem to Euler's lagrange equation? Is the closed intergral the functional then? 6. Aug 24, 2006 ### StatusX One easier way to do it, though maybe not completely rigorous, is to assume the shape has mirror symmetry about some axis (this is reasonable, since if one shape satisfies the maximal area property, so will its mirror image, and so if there is a unique such shape, it must have this symmetry). Then you make the x-axis this axis of symmetry, and find a function y, corresponding to one half of the shape (this also assumes half of the shape passes the vertical line test and so corresponds to a function, which is maybe less defensible). The problem reduces to finding a function y(x) such that the perimeter: [tex] P=2 \int_0^L \sqrt{1+\left(\frac{dy}{dx}\right)^2 } dx$$

is minimal subject to the constraint:

$$A(y)=2 \int_0^L y dx=A_0$$

(note how this is equivalent to maximizing the area for a constant perimeter) The Euler Lagrange equations are modified to include a constraint in a manner similar to the method of Lagrange multipliers. Namely, if you want to extremize the integral of f(y,y',x) subject to the constraint that the integral of g(y,x) is some constant, y must satisfy:

$$\frac{\partial f}{\partial y} -\frac{d}{dx} \left( \frac{\partial f}{\partial y'}\right) - \lambda \frac{\partial g}{\partial y} =0$$

where $\lambda$ is a constant. I'm not sure how you would include a constraint that depended on y', which is why I rephrased the question to minimize perimeter. You should be able to show that y is the equation of a semi-circle, as expected.

Last edited: Aug 24, 2006
7. Aug 25, 2006

### HallsofIvy

Well, no, the problem can't be to maximize the area and minimize the perimeter with the same curve. It's very likely that the problem is to maximize the area with a given perimeter. The "dual problem" would be to minimize the perimeter for a given area.