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calculus

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A five feet piece of wire is cut into two pieces. One Piece is bent into a square and the other is bent into an equilateral triangle. Where should the wire be cut so that the total area enclosed by both is minimum.

  • calculus -

    let each side of the equilateral triangle be 2x
    let each side of the square be y
    then 6x + 4y = 5
    y = (5-6x)/4

    Area of triangle:
    draw in the altiude, so we have a right angled triangle with hypotenuse 2x and base 1x, by Pythagorus the height is √3x (now do you see why I defined the side as 2x and not x ? )
    Area of triangle = (1/2)(2x)√3x = √3x^2
    area of square = y^2

    Area = √3x^2 + y^2
    = √3x^2 + ((5-6x)/4 )^2
    = √3x^2 + (1/16) (25 - 60x + 36x^2)
    = √3x^2 + 25/16 - (15/4)x + (9/4)x^2

    d(Area)/dx = 2√3x - 15/4 + (9/2)x
    = 0 for a min of Area

    x(2√3 + 4.5) = 3.75
    x = 3.75/(2√3+4.5) = appr .4709

    so each side of the triangle is 2x = .9417 ft
    and each side of the square = (5-6x)/4 = .5437 ft

    check my arithmetic, I should have written it out on paper first.

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