A piece of wire 15 m long is cut into two pieces. One piece is bent into a square and the other is bent into an equilateral triangle. (Give your answers correct to two decimal places.)

Question

(a) How much wire should be used for the square in order to maximize the total area?

  m

(b) How much wire should be used for the square in order to minimize the total area?

 m

Answer 1

Let x be the length used for the square and y be the length used for the triangle

x+y=15

Each side of the square is x/4 and each side of the equilateral triangle is y/3.

Square Area, A_s=x²/16

Triangle Area, A_t=√3/4y²/9 = √3/36y²

Total area, A= x²/16 + √3/36y² = x²/16 + √3/36(15-x)²

A'(x) = x/8 - √3/18(15-x)

Set A'(x) to zero and solve for x

Find A(15), A(0), and A(x) to determine the what values of x give the maximum and minimum total areas.

Answer 2

Let we cut x m for square, then (15 - x) m left for triangle. Then total area A = (x/4)² + (15 - x)²/9·√3/4 =

x²/16 + (15 - x)²·√3/36; dA/dx = x/8 - (15 - x)√3/18 = 0; 9x - 4(15 - x)√3 = 0; 9x + 4x√3 = 60√3;

x = 60√3/(9 + 4√3) = 6.524468 m

(b) So minimum total area min A will be when x = 60√3/(9 + 4√3) m = 6.524468 m

(a) Ma ximum then will be on one of the ends interval [0, 15]

x = 15; A = 225/16 = 14.0625

x = 0 A = 225√3/36 = 10.82532

So, maximum area if all wire we use for square