I'm really new to the world of pure mathematics and proofs. I was recently watching a playlist made by Bill Shilito (in YouTube) called Introduction to Higher Mathematics. The first example given is: if you are given 64m of fence to make a pen, what is the largest area that you could get with a shape of that perimeter? I'm not sure yet but I was told that the most area that you can get from any perimeter x is the one you get with a circle.
Now, I started thinking then… is there any polygon that will always have less area than any other polygon? I know that, if you think from calculus perspective you can always get a shape that has less area than the shape before that, but it is not my point. My question is not about going on, and on forever with the same shape.
To make sure that my point is (kind of) clear I will use the same example from the video.
If I am given 64m of fence, I can have a circle with a circunference of 64. We know that circunference = pi * diameter, therefore, diameter = circunference/pi. We solve for diameter and we get (aprox.) 20.37m, and we know that radius = diameter/2, then, the radius is (aprox.) 10.18m. Now we know all we want to get the area which is pi * (10.18)^2 and we get an area of (aprox.) 325.94m^2.
Now, is there any polygon that given its perimeter of 64m, its area will always be less than every other polygon of same perimeter but different sides? For example, if I have a square with 16m of perimeter, will there be a different polygon with the same perimeter but less area? Considering that the different polygon already is builded as the maximum area.
I hope I made myself clear. If you think that the answer is too long, I'd appreciate if you gave bibliographic references about the topic so I can search more on my own (I'm not doing it right now because, to be honest, I have homework to do and I don't really remember very well anything about euclidean geometry).
Thank you very much.
Best Answer
Let's take a rhombus $ABCD$ with side length $a=16$ and distance between vertices $A$ and $C$ equal $2d\in(0,32)$.
Then we can compute it's area: $$P=2d\sqrt{256-d^2}$$
Now le'ts take a sequence of such rombs $(R_n)$ such that a distance $2d$ is in rombus $R_n$ is equal $\frac{2}{n}$.
$$\lim_{n\to \infty}P(R_n) = \lim_{n\to \infty}\frac{2}{n}\sqrt{256-\left(\frac{1}{n}\right)^2}=0$$
Thus for every area $X>0$ we can find a rhombus $R_n$ such that $P(R_n)<X$, so there is no polygon with minimal area.