A 2 m piece of wire is cut into 2 pieces. One piece is bent to form a square and the other piece is bent to form an equilateral triangle.Where...

The length of the wire is 2 m. It is cut into two pieces
one of which is used to create a square and the other to create an equilateral
triangle.

Let the length of the wire used to create the
square be L. The length of the wire used to  create the equilateral triangle is 2 -
L.

The area of the square formed is (L/4)^2. The area of
the equilateral triangle formed is (sqrt 3)(2 - L)^2/4.

The
total area enclosed is (L/4)^2 + (sqrt 3)(2 - L)^2/4.

We
have to maximize A = (L/4)^2 + (sqrt 3)(2 - L)^2/4

dA/dL =
2L/16 + (sqrt 3)/4 * 2*(2 - L)

2L/16 + (sqrt 3)/4 * 2*(2 -
L) = 0

=> L/8 + (sqrt 3)/2 * (2 - L) =
0

=> L/4 + sqrt 3 * (2 - L) =
0

=> L/4 - sqrt 3*L + 2*sqrt 3 =
0

=> L(sqrt 3 - 1/4) = 2*sqrt
3

=> L = 8*sqrt 3/(4*sqrt 3 -
1)

The wire should be cut to create two
pieces,one with a length of 8*sqrt 3/(4*sqrt 3 - 1) to make the square and the other
with a length 2 - 8*sqrt 3/(4*sqrt 3 - 1) to make the
triangle.