Optimization Problem

A piece of wire 12m long is cut into two pieces. One piece is bent into in a square and the other is bent into an equilateral triangle. How should the wire be cut so that the total area enclosed is a minimum?

x+y=12
A = (x/4)^2 + 1/2(y/3)(sqrt(y/6)) <-- Is this correct? And how do I continue the problem?

Let's let 'x' be one of the pieces of the wire after it's cut. The other piece will be 12 - x. Using our first piece of wire for the square and the second piece for the triangle, respectively gives:

x/4 & (12 - x)/3.

Note that (12 - x) is the base of the equilateral triangle. Therefore, the height is sqrt(3)/2 * (12 - x)/3.

A = (1/16)x^2 + (1/2)(sqrt(3)/6)(12 - x) * (12 - x)/3) <--- This is in form (1/2 * h * b)
A = (1/16)x^2 + [sqrt(3)(12 - x)^2] / 36
A ' = x/8 - [sqrt(3)(12 - x)] / 18
A ' = x/8 - 12sqrt(3)/18 + sqrt(3)x/18
A ' = x/8 - 2sqrt(3)/3 + sqrt(3)x/18
A ' = (3x - 16sqrt(3)) / 24 + sqrt(3)x/18
A ' = (9x - 48sqrt(3) + 4xsqrt(3)) / 72

Now let's set the numerator equal to 0.

9x - 48sqrt(3) + 4xsqrt(3) = 0
9x + 4xsqrt(3) = 48sqrt(3)
x(9 + 4sqrt(3)) = 48sqrt(3)
x = 48sqrt(3)/(9 + 4sqrt(3))
x = 5.21 m

Check your endpoints to check which of the three values will give you the lowest x amount.

Hope this helped.