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22 February, 20:34

A 2 feet piece of wire is cut into two pieces and once a piece is bent into a square and the other is bent into an equilateral triangle. Where should the wire cut so that the total enclosed area by both shapes is at the maximum?

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  1. 22 February, 22:25
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    Let x ft be used to make square and 2-x ft be used to make equilateral triangle.

    each side of square = x/4

    area of square = (x / 4) ²

    Each side of triangle

    = (2-x) / 3

    Area of triangle = 1/2 (2-x) ²/9 sin 60

    = √3 / 36 x (2-x) ²

    Total area

    A = (x / 4) ² + √3 / 36 (2-x) ²

    For maximum area

    dA/dx = 0

    1/16 (2x) - √3 / 36 x2 (2-x) = 0

    x / 8 - √3 (2-x) / 18 = 0

    x / 8 - √3/9 + √3/18 x = 0

    x (1/8 + √3/18) = √3/9

    x (.125 +.096) =.192

    x =.868 ft
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