# 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 should the wire be cut so that the total area enclosed by both is a maximum?

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...

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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.

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