# 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...

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?

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### 1 Answer

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**