Maximum/triangle problem (1 Viewer)

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Show that the maximum area of a triangle corresponds to the triangle being equilateral.

I start by making y the height of the triangle and x a leg.
We have two formulas (for area)

A = xy/2
A = sqrt(s(s-a)(s-b)(s-x))

I'm thinking that in order to find the maximum, we must make dA/dx = 0 and show that x=a=b . Any suggestions on how to do this?

Forgive me if this is really easy, I've been out of school for some time and have forgotten alot--trying to learn again.
 
I think you can use the arithmetic mean-geometric means inequality: [tex] \frac{a+b}{2} \geq \sqrt{ab} [/tex]


[tex] (s-a)(s-b)(s-c) \leq (\frac{(s-a)+(s-b)+(s-c)}{3})^{3} [/tex]
 
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0rthodontist

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Are you trying to find the maximum-area triangle with a given perimeter? You have to specify the terms of the problem. There is no such thing as just a plain "maximum-area" triangle since you can always make a triangle larger.
 
Hmmm...yes. I was thinking that showing, given legs x a b, that the triangle has a maximum area when x=a=b, and that proving this first was a necessary step to the problem. "Assume a perimeter of 30 and find the largest area.", can be done in the head, but I wan't to figure out how to prove it.

So, the chapter I'm working on teaches maxima and minima, and it seems logical that the problem involves finding the derivative of the area with respect to x (one leg) and finding the maximum of that derivative. I'm just having a problem figuring-out what that formula is to derive.
 
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