max area of isosceles triangle

[ace_rimmer]

Im looking for a formmul;ar i could use in a computer program to find the max area of an isosceles triangle where the sides adsd up to a given length.

Hope im describing this right all help would be great thanks you


-Ace

[krtxmrtz]

Let b each of the 2 equal sides and c the other one. Then, the perimeter is:

P = c + 2b

and the area:

A = ch/2 where h is the height:

h = Sqr[b² - (c/2)²]

so that,

A = (c / 2)*Sqr[b² - (c/2)²]

From the first equation above,

b = (P - c) / 2

and substituting this into the previous equation,

A = (c / 2)*Sqr{[(P - c) / 2]² - (c/2)²} = (c / 4) Sqr(P² - 2PC)

Now, to find the maximum area as a function of c, the derivative must be zero. Using R = Sqr(P² - 2PC):

0 = dA / dc = -(c / 4)*P / R + R / 4 = (P² - 2Pc) / 4R

so that

0 = P² - 2Pc

from which,

P = 3c
c = P / 3 and, of course, b = P / 3

and therefore it's a equilateral triangle as we already knew from intuition.

[Rassis]

Nice solution and a phrase of wisdom in perfect portuguese "Aquele que não pode perdoar destroi a ponte sobre a qual ele mesmo deve passar".

Rui

[krtxmrtz]

Nice solution and a phrase of wisdom in perfect portuguese "Aquele que não pode perdoar destroi a ponte sobre a qual ele mesmo deve passar".

Rui

Thanks, I found the phrase in a Brazilian web and I liked it very much. I thought it was fair to keep it in the original, although I might have set it to Catalan: "Aquell que no pot perdonar destrueix el pont sobre el qual ell mateix ha de passar".