Proving Circle Has Smallest Perimeter for Same Area

[pixel01]

Hi everybody,

In physics and mathematics, we often use the theorem that the circle alway has the smallest perimeter compared to all other geometric shapes of the same area. Anyone can tell me how to prove that?
Thanks for reading.

 

[arildno]

Well, assuming the perimeter is a smooth curve, you can show with calculus of variations that the circle minimizes the perimeter for a given area.

I haven't seen the proof valid for ALL perimeters, but I'm sure it exists somewhere.

 

[tacman]

Hi there,

To prove that the circle has the smallest perimeter for the same area, we can use the concept of optimization. This means that we want to find the minimum value of a function, in this case, the perimeter, while keeping the area constant.

Let's start by considering a circle with radius r and its area A = πr^2. We can express the perimeter of this circle as P = 2πr. Now, let's consider another shape with the same area A. We can express its perimeter as a function of its dimensions, let's say P(x,y). Since the area is constant, we can write x and y in terms of r, which gives us x = √(A/π) and y = √(A/π).

Now, we can substitute these values into the perimeter function of the other shape, giving us P(√(A/π), √(A/π)). We can then use the concept of partial derivatives to find the critical points of this function. The critical points are the values of x and y that make the derivative of P(x,y) with respect to x and y equal to 0.

After finding the critical points, we can use the second derivative test to determine whether these points are minimum, maximum, or saddle points. If the second derivative is positive, then the critical point is a minimum, which means that the perimeter is minimized when x = √(A/π) and y = √(A/π). This corresponds to a circle with radius r = √(A/π).

Therefore, we can conclude that the circle has the smallest perimeter for the same area, as it is the only shape that satisfies the condition of having a minimum perimeter while keeping the area constant. I hope this helps to prove the theorem you mentioned. Let me know if you have any further questions.