Maximising the area of a triangle of known perimeter

[Unredeemed]

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]

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) ?

 

[Unredeemed]

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)?

 

[lark]

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

http://en.wikipedia.org/wiki/AM-GM#The_inequality

You can find an upper bound, and by that the maximum value of the area. As you suspected, this is when a=b=c, that is when the triangle is equilateral.