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?

Question

Deegan Conner

Physics

28 May, 02:24

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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Answers (1)

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Shania King · Answer 1

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