A piece of wire is 14m long is cut into 2 pieces. One is bent into a square. The other is bent into an equilat

Question

A piece of wire is 14m long is cut into 2 pieces. One is bent into a square. The other is bent into an equilateral triangle. How should the wire be cut for the square so that the total area enclosed is a minimum?

Answer 1

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Let x be the length for the square and 14-x for the triangle.

The area of the square is (x/4)^2 = (x^2)/16.

The height of the triangle is sqrt(3)(14-x)/6 and the base of the triangle is (14-x)/3, so the area of the triangle is

1/2 bh = [(14-x)/6] [sqrt(3)(14-x)/6] = (14-x)^2 sqrt(3)/36

Total area A = (x^2)/16 + (14-x)^2 sqrt(3)/36

A = [9x^2 + 4(x^2 - 28x + 196) sqrt(3)] / 144

A = [(9 + 4 sqrt 3) x^2 - (112 sqrt 3) x + (784 sqrt 3)] / 144

dA/dx = [2(9 + 4 sqrt 3) x - (112 sqrt 3)] / 144 = 0

(9 + 4 sqrt 3) x = 56 sqrt 3

x = (56 sqrt 3) / (9 + 4 sqrt 3) = 6.0895 meters (Answer)

The area of the square is 2.32 m^2. The area of the triangle is 3.01 m^2, and the total area is 5.33 m^2.

Answer 2

for square 6.089 m

and for triangle 7.92