# Maximum area for minimum perimeter of a triangle

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 $2(\phi - 1).$

• Will do, thank you. – John Singac Jul 11 '18 at 16:34
• There's a problem of units when taking a ratio of area to perimeter. Depending on your choice of units the same triangle can have such a ratio that is arbitrarily large or arbitrarily small, and if you fix a unit of length, the ratio can be made arbitrarily large or arbitrarily small by changing the size of the triangle, – hardmath Jul 11 '18 at 16:37
• Welcome to stackexchange. Are you sure your calculus/algebra is right? They symmetry of Heron's formula implies symmetry in the triangle for which the area is maximum. Perhaps the problem is in the units, as @hardmath comments. – Ethan Bolker Jul 11 '18 at 16:37

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.

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.

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.