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...
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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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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