Question

A piece of wire 14 m long is cut into two pieces. One piece is bent into a square and the other is bent into an equilateral triangle.
(a) How much wire should be used for the square in order to maximize the total area?

= 14 m

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

= ????

Answer 1

let each side of the square be x m
let each side of the equilateral triangle be 2x
(that way, the height is √3y, from the ratio of the 30-60-90° triangle)

a) for a max area, you are right, all should be used for the square

b) 4x + 6y = 14
2x + 3y = 7
x = (7-3y)/4 OR y = (7-2x)/3

area = x^2 + (1/2)(2y)(√3y)
= x^2 + √3 y^2
= x^2 + √3 ((7-2x)/3)^2

= x^2 + (√3/9)(49 - 28x + 4x^2)
d(area)/dx = 2x + (√3/9)(-28 + 8x) = 0 for a max of area

2x = √3/9(28 - 8x)
18x = 28√3 - 8√3x
x(18 + 8√3) = 28√3
x = 28√3/(18+8√3) = appr 1.522

need 4 x's for the square, <b>so 6.09 m for the square</b>, leaving 7.9 m for the triangle for a minimum total area.

check my arithmetic, should have written it out first.