Maximizing Area with a Given Perimeter

In summary: The constraint actually has to be a curve, since you have to specify a shape, after all.In summary, to prove that an area of a simple closed curve is maximized if it is a circle, one would need to use calculus of variations and Green's theorem. The perimeter must also be minimized, which can be done by finding a function y(x) using the Euler-Lagrange equations. This function would be the equation of a semi-circle, as expected.
  • #1
waht
1,501
4
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.
 
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  • #2
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##
That second functional can be used to derive the Euler equation.
 
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  • #3
HallsofIvy said:
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
[tex]\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.

sory, but is this uni stuff? or high school? cause I've never seen green's theorem :-p
 
  • #4
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
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
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[/tex]

is minimal subject to the constraint:

[tex] A(y)=2 \int_0^L y dx=A_0[/tex]

(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:

[tex] \frac{\partial f}{\partial y} -\frac{d}{dx} \left( \frac{\partial f}{\partial y'}\right) - \lambda \frac{\partial g}{\partial y} =0[/tex]

where [itex]\lambda[/itex] 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.
 
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  • #7
waht said:
I forgot to mention the perimeter has to be minimized.

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.
 

FAQ: Maximizing Area with a Given Perimeter

1. What is a simple closed curve area?

A simple closed curve area is a geometric shape that is formed by a continuous line that starts and ends at the same point, without any self-intersections or crossings.

2. How is the area of a simple closed curve calculated?

The area of a simple closed curve can be calculated using various mathematical formulas, depending on the shape of the curve. For example, the area of a circle can be calculated using the formula A = πr^2, where r is the radius of the circle.

3. What are some examples of simple closed curve areas?

Examples of simple closed curve areas include circles, ellipses, squares, rectangles, triangles, and many other geometric shapes. These shapes can also be combined to form more complex simple closed curves.

4. What are some real-life applications of simple closed curve areas?

Simple closed curve areas have many practical applications in fields such as architecture, engineering, and physics. For example, the area of a simple closed curve can be used to determine the surface area of 3-dimensional objects, or to calculate the amount of material needed for construction projects.

5. How are simple closed curve areas different from open curves?

The main difference between simple closed curve areas and open curves is that open curves do not have a defined start or end point, and they may intersect or cross over themselves. Simple closed curve areas, on the other hand, are always continuous and do not intersect or cross over themselves.

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