Question

A piece of wire 40 m long is cut into two pieces. One piece is bent into a square and the other is bent into an equilateral triangle.
How much of the wire should go to the square to minimize the total area enclosed by both figures?

Answer 1

let the side of the square be x
let the side of the triangle be y
4x + 3y = 40
x = (40-3y)/4

height of triangle from the 30-60-90 triangle ratio = √3/2 y
area of triangle = (1/2)(y)(√3/2 y)
= (√3/4)y^2

area of square = [(40-3y)/4]^2
= (1600 - 240y + 9y^2)/16
= 100 - 15y + (9/16)y^2

A = 100 - 15y + (9/16)y^2 + (√3/4)y^2
dA/dy =-15 + 9/8 y + 2√3/4 y
= 0 for a max/min

√3/2 y + 9/8 y = 15
4√3 y + 9y = 135
y = 135/(4√3+9) = appr 8.476

3y = 25.43 m
then 4x = 14.573 m

appr 14.57 m should go for the square

check my arithmetic and algebra