calculus
posted by Kyle .
A five feet piece of wire is cut into two pieces. One Piece is bent into a square and the other is bent into an equilateral triangle. Where should the wire be cut so that the total area enclosed by both is minimum.

let each side of the equilateral triangle be 2x
let each side of the square be y
then 6x + 4y = 5
y = (56x)/4
Area of triangle:
draw in the altiude, so we have a right angled triangle with hypotenuse 2x and base 1x, by Pythagorus the height is √3x (now do you see why I defined the side as 2x and not x ? )
Area of triangle = (1/2)(2x)√3x = √3x^2
area of square = y^2
Area = √3x^2 + y^2
= √3x^2 + ((56x)/4 )^2
= √3x^2 + (1/16) (25  60x + 36x^2)
= √3x^2 + 25/16  (15/4)x + (9/4)x^2
d(Area)/dx = 2√3x  15/4 + (9/2)x
= 0 for a min of Area
x(2√3 + 4.5) = 3.75
x = 3.75/(2√3+4.5) = appr .4709
so each side of the triangle is 2x = .9417 ft
and each side of the square = (56x)/4 = .5437 ft
check my arithmetic, I should have written it out on paper first.
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