# A piece of wire 25 m long is cut into two pieces. One piece is bent into a square and the other is bent into an equilateral triangle. How much wire should be used for the square in order to maximize the total area? The piece of wire is 25 m long and is cut into two pieces one of which is bent into a square and the other is bent into an equilateral triangle.

Let the length of the wire bent into a square be x, the length of the wire bent into...

The piece of wire is 25 m long and is cut into two pieces one of which is bent into a square and the other is bent into an equilateral triangle.

Let the length of the wire bent into a square be x, the length of the wire bent into an equilateral triangle is 25-x. The area of the square is `(x/4)^2` and the area of the equilateral triangle is `(sqrt 3/4)*((25-x)/3)^2`

The total area is A = `(x/4)^2 + (sqrt 3/4)*((25-x)/3)^2`

To maximize the area solve `(dA)/(dx) = 0`

`(dA)/(dx) = ((3^(3/2)+4)*x-100)/(8*3^(3/2))`

`((3^(3/2)+4)*x-100)/(8*3^(3/2))= 0`

=> `x = 100/(3^(3/2)+4) `

=> `x ~~ 10.874`

The length of the wire to be bent into a square is approximately 10.874 m

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