Maximum area for minimum perimeter of a triangle

Question

Does the fact that a triangle has the maximum area for a given perimeter when it is an equilateral triangle (Isoperimetric Property of Equilateral Triangles) imply that the ratio of a triangle's area to its perimeter will have a maximum when all side lengths are equal? I found the derivative of Heron's formula for area divided be perimeter for an isosceles triangle and this is not the case. The ratio is at a maximum when the third side is $\sqrt5 - 1$ times the length of the two equal sides, which interestingly enough comes out to be $2(\phi - 1).$

Answer 1

Well, even for a square, your question is not very interesting. The maximum of area-perimeter ratio is $+\infty$. Just magnify your triangle or square, area increases quadratically, while perimeter only linearly increases.

Answer 2

You should take a ratio of area to the perimeter squared to get meaningful (independent from the scale/units) results.

And yes, then the ratio is maximum for equilateral triangle.

Answer 3

Area and perimeter have different unites so the ratio changes with the size of triangle.

The ratio of the area of an equilateral triangle of size $a$ to its perimeter is $$R=\frac {a\sqrt 3}{12}$$

As you see, the ration grows with the size of triangle.

For $a=4\sqrt 3$ we have $R=1$ and the ratio grows to $\infty$ as the triangle enlarges.

Answer 4

This is a demo to show that the interpretation of Heron is wrong. I have not had good luck with him either so perhaps misreading him is just a common thing. Note: D is used for area because it is the fourth term of a cubic equation that often results in area computations and it avoids confusion with sides or vertices A,B,C of any triangles used.

1. Given a triangle with $P=240$ we will let $(\sqrt{5}-1)x$ be the base.

$$\big(\sqrt5 - 1\big)x+2x=240\implies x=\frac{240}{\sqrt{5}+1}\approx 74.16\\ \implies (\sqrt{5}-1)x\approx 91.67 \implies h=\sqrt{74.16^2-\bigg(\frac{91.67}{2}\bigg)^2}\approx 58.2997...\\ \implies D=area=\bigg(\frac{91.67}{2}\bigg)\cdot 58.2997\approx 2672.17$$

2. In comparison, an equilateral triangle where $P=240$ has sides of $80$. $$x=80\implies h=\sqrt{80^2-40^2}\approx 69.28\\ \implies D=area= 40\cdot 69.28\approx 2771.2$$

The equilateral triangle has the larger area for the same perimeter and the larger D/P ratio. $$ \frac{D_1}{P}=\frac{2672.17}{240}\approx 11.134\\ \frac{D_2}{P}=\frac{2771.2}{240}\approx 11.546$$