# Maximising the area of a triangle of known perimeter

How would I find the maximum area of a triangle given a fixed perimeter?

I assume that it would be an equilateral triangle, but I'm finding it very hard proving it.

I started by drawing a triangle of side lengths a, b and P -(a+b) with angles of alpha, beta and gamma.

I then used the A=(ab*sine(gamma))/2 formula for the area. But have been hitting a definite brick wall. I'm guessing calculus is necessary, but I'm struggling to see how.

Can anyone help?

Thanks,
Unredeemed.

tiny-tim
Homework Helper
Hi Unredeemed! I started by drawing a triangle of side lengths a, b and P -(a+b) with angles of alpha, beta and gamma.

I then used the A=(ab*sine(gamma))/2 formula for the area.

But what is your formula for the area as a function of a b and P (only) ?

Well, I got from the cosine rule:

(P-(a+b))^2=a^2+b^2-2abcos($$\gamma$$)
So cos($$\gamma$$)=((P-(a+b))^2-a^2-b^2)/(-2ab)

Here, I got a bit confused, because I need sin($$\gamma$$), but only have cos($$\gamma$$).

Do I need to use the fact that cos($$\gamma$$-1/2$$\pi$$)=sin($$\gamma$$)?

How would I find the maximum area of a triangle given a fixed perimeter?

You can do it by finding the minimum perimeter of a triangle given a fixed area.

disregardthat
Use Heron's formula: The area A = $$\sqrt{s(s-a)(s-b)(s-c)}$$, where $$s = \frac{a+b+c}{2}$$, and a,b,c are the sides of the triangle. The perimeter is fixed, so you want to maximize the expression (s-a)(s-b)(s-c). If you have heard of the AM-GM inequality