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Calculus

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A piece of wire 40 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 of the wire should go to the square to minimize the total area enclosed by both figures?

  • Calculus - ,

    let the side of the square be x
    let the side of the triangle be y
    4x + 3y = 40
    x = (40-3y)/4

    height of triangle from the 30-60-90 triangle ratio = √3/2 y
    area of triangle = (1/2)(y)(√3/2 y)
    = (√3/4)y^2

    area of square = [(40-3y)/4]^2
    = (1600 - 240y + 9y^2)/16
    = 100 - 15y + (9/16)y^2

    A = 100 - 15y + (9/16)y^2 + (√3/4)y^2
    dA/dy =-15 + 9/8 y + 2√3/4 y
    = 0 for a max/min

    √3/2 y + 9/8 y = 15
    4√3 y + 9y = 135
    y = 135/(4√3+9) = appr 8.476

    3y = 25.43 m
    then 4x = 14.573 m

    appr 14.57 m should go for the square

    check my arithmetic and algebra

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