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
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.