A fence must enclose an area that borders a straight river. We have a fixed amount of fence, say 100 meters.

I want to make the shape a semi-ellipse (half an ellipse).

What a/b ratio should I choose to maximize the area?

I cannot do this because the circumference of an ellipse is an insane equation. It involves an integral that I think is not solvable in indefinite form.

The area of an ellipse is much easier: ab*pi. (But it will be 1/2 of that.)

Optional Extra Info:

Technically a circle is a special case of an ellipse where a = b. In other words the ratio is 1:1. Circles are way easier so I did this example case.

P = $\frac{1}{2}2\pi r = 100$

r = $\frac{100}{\pi}$

A = $\frac{1}{2}\pi r^2$

A = $\frac{1}{2}\pi\frac{10,000}{\pi^2}$

A = $\frac{5,000}{\pi}$ ~= 1,591.549 sq m

Now I remember the classic rectangle version, where the optimal area occurs when x = 2y. Therefore I want to do a test case where a = 2b and see if it's bigger than the semi-circle area.

But I can't plug that in because the circumference of an ellipse involves an integral that I can't evaluate.

Even that would not be the final answer. The optimization asks what is the a/b ratio that maximizes area? I did 1 test case and wanted to do a 2nd, but to find the optimal I would have to take the derivative, set it to zero, find critical points, and figure it out.

Can anyone do this?

P.S., that area of about 1,600 sq m is about 30 percent bigger than the rectangle version where x = 2y. In that case I get A = 1250 sq m

  • 1
    $\begingroup$ as it happens, the semicircle is the best. The version of "constant mean curvature" for a curve in the plane is "arc of a circle." Also, in a "free boundary" problem, the (curve) is orthogonal to the boundary. Try putting some soap solution on a smooth surface and then blow some bubbles on it; they spread out after contact, you get half bubbles. $\endgroup$ – Will Jagy Feb 1 '17 at 0:51
  • $\begingroup$ @WillJagy I would really like to at least plug in the a = 2b example just to see what it is. Maybe someone else can do the integral. $\endgroup$ – DrZ214 Feb 1 '17 at 2:18
  • $\begingroup$ No, well known, arc length for an ellipse does not have closed form. It can be numerically estimated with a computer, but lots of work. $\endgroup$ – Will Jagy Feb 1 '17 at 2:33

You can get the optimum a/b ratio without regard to the perimeter. You only need to consider the perimeter to determine the respective values of a and b. The area of an ellipse is $πa^2\sqrt{1 - e^2}$, where e is the eccentricity. For given a, it's clear that the area is maximized by e = 0, so the ellipse is actually a circle with radius r = a = b. The same goes for a semi-ellipse and a semi-circle. To get a 100 meter fence, the radius is 100/π, as you said.


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