A piece of wire m long is cut into two pieces. One piece is bent into a square and 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? Round your answer to the nearest hundredth.

Question

A piece of wire  m long is cut into two pieces. One piece is bent into a square and 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? Round your answer to the nearest hundredth.

Quizplus · Accepted Answer

Let x be the length of wire used for the square, then the remaining length of wire (11 - x) is used for the equilateral triangle.

For the square, we need 4 equal sides of length x/4, so the perimeter of the square is x.

For the equilateral triangle, each side has length (11-x)/3, so the perimeter of the triangle is (11-x).

We want to minimize the total area enclosed, which is the sum of the area of the square and the area of the equilateral triangle.

Area of Square = (x/4)^2 = x^2/16
Area of Equilateral Triangle = sqrt(3)/4 * (11-x)/3)^2= (sqrt(3)/36) * (11-x)^2

Total Area = x^2/16 + (sqrt(3)/36) * (11-x)^2

To find the minimum value of the total area, we take the derivative of the equation and set it equal to zero:

d(Total Area)/dx = (x/8) - (sqrt(3)/18)(11-x) = 0
x/8 = (sqrt(3)/18)(11-x)
x = 8(sqrt(3)+1)/5

Therefore, the wire should be cut into a piece of length x = 8(sqrt(3)+1)/5 for the square, and 11-x = 3.2 for the equilateral triangle.

The minimum total area is:
x^2/16 + (sqrt(3)/36) * (11-x)^2 
= (8(sqrt(3)+1)/5)^2/16 + (sqrt(3)/36) * (11-8(sqrt(3)+1)/5)^2
≈ 2.605

The answer closest to 2.61 is (B) 11eaa3e5_55bb_7c6d_9f8f_ed128bac7c4b_TB5971_11.

A Piece of Wire M Long Is Cut into Two