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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.

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  • $\begingroup$ Generally, the closer you are to a circle, the more area you get. Thus, for instance, a regular $n$-gon will have less area than a regular $(n+1)$-gon of the same perimeter. So a regular $3$-gon (also known as an equilateral triangle) has less area than a regular $4$-gon (also known as a square) of the same perimeter. It's that what you're looking for? $\endgroup$
    – Arthur
    Sep 25, 2017 at 7:09

2 Answers 2

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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.

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  • $\begingroup$ From the pointview of calculus, even without the equations I already knew it. My point is, let's say that every side of this polygon is equal or more than 1 (working with natural numbers). Will there be a polygon that has less area than every other polygon that could be shaped by this conditions? $\endgroup$
    – user484279
    Sep 25, 2017 at 7:15
  • $\begingroup$ @MarcoI.Domínguez Actually in this rhombus every side is greater or equal 1. $\endgroup$ Sep 25, 2017 at 7:18
  • $\begingroup$ Now I'm really lost. Could you (if it's not too much to ask for) explain it in more detail? $\endgroup$
    – user484279
    Sep 25, 2017 at 7:23
  • $\begingroup$ Each side of rombs have side length 16 $\endgroup$ Sep 25, 2017 at 7:43
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Look, here we are talking about two kinds of two dimensional figures.

$1$ Figures whose area depend upon just one variable like circle (radius), square (side), equilateral triangle (side) etc.

$2$ Figures whose area depend upon two variable like rectangle (length and breadth), ant triangle other than equilateral triangle.

Now for the $2-D$ figures of first kind, we know that area and perimeter both depend on just one variable. So we cannot try to minimize the area as we will have to touch perimeter for it. For instance you cannot think of minimizing $\pi\times r^2$ without varying r and you cannot expect circumference to be same after varying r.

But in $2-D$ figures of second kind, you have got that opportunity. Since area depends on two variables so you can select two variables such that one is very large while other one is small thus making perimeter adjustable to the range you want and area tending to zero.

Consider a rope of length $44m$. You can make a square from it with side length $11 m$ and get the area $121 m^2$ while you cab also make a rectangle with dimensions $21.5m,0.5m$ and get the area $10.75m^2$. Further you can adjust length and breadth to $21.999999m,0.000001m$ to get even smaller area. Conclusively as breadth (or length) tends to zero, the area will also tend to zero keeping perimeter same by manipulating length (or breadth).

You can try the same with triangles too.

So, sadly answer of your question is no!! There is no minimum area for a given perimeter.

I discussed about maximum area for given perimeter here ( How to relate areas of circle, square, rectangle and triangle if they have same perimeter??). Take a look.

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