Question 450607
Suppose the side lengths are x, x, and 64-2x:


{{{drawing(300,300,0,10,0,10,


triangle(2,2, 8,2, 5,8)
locate(4.2,2,(64-2x)),
locate(2.9,5,x),
locate(6.9,5,x)



)
}}}

(64-2x won't show up so assume the base length is 64-2x)


By Heron's formula, we can find the area A of the triangle in terms of x (semiperimeter = 32)


*[tex \LARGE A = \sqrt{32(32-x)(32-x)(32-(64-2x))} = \sqrt{32(32-x)^2(2x-32)} = (32-x)\sqrt{64x-1024}]


Maximizing the area requires a little calculus. We can say that A is a function in terms of x, and find dA/dx. However, we can also note that the value of x that maximizes A will also be the value of x that maximizes A^2. To make things simpler, we can find A^2:


*[tex \LARGE A^2 = (32-x)^2(64x-1024)]


Here, we take the derivative with respect to x:


*[tex \LARGE \frac{d(A^2)}{dx} = -2(32-x)(64x-1024) + (32-x)^2(64)]


*[tex \LARGE \frac{d(A^2)}{dx} = (32-x)(-128x + 2048 + 2048 - 64x) = (32-x)(-192x + 4096)]


Here, the derivative is equal to zero when x = 32 or x = 64/3. Clearly, x = 64/3 maximizes the area of the triangle. This would also imply that the maximal area occurs when the triangle is equilateral.