Maximising the area of a triangle of known perimeter

  • Thread starter Unredeemed
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  • #1
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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.
 

Answers and Replies

  • #2
tiny-tim
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Hi Unredeemed! :smile:
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) ?
 
  • #3
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Well, I got from the cosine rule:

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

Here, I got a bit confused, because I need sin([tex]\gamma[/tex]), but only have cos([tex]\gamma[/tex]).

Do I need to use the fact that cos([tex]\gamma[/tex]-1/2[tex]\pi[/tex])=sin([tex]\gamma[/tex])?
 
  • #4
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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.
 
  • #5
disregardthat
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Use Heron's formula: The area A = [tex]\sqrt{s(s-a)(s-b)(s-c)}[/tex], where [tex]s = \frac{a+b+c}{2}[/tex], 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.
 

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