Calculus
posted by Sckricks .
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?

let the side of the square be x
let the side of the triangle be y
4x + 3y = 40
x = (403y)/4
height of triangle from the 306090 triangle ratio = √3/2 y
area of triangle = (1/2)(y)(√3/2 y)
= (√3/4)y^2
area of square = [(403y)/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
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