Question

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.

Answer 1

Ok, I did that, no what ?

Answer 2

The triangle's perimeter is x, and the square's is 15-x. So, the area is

((15-x)/4)^2 + x^2*√3/4

Now do with that what you will.

Answer 3

If typed fully , this is one of the standard optimization problems, and it probably asked for a minimum sum of the areas.
I will define my variables different from Steve's, mainly to avoid fractions
let each side of the square be x
let each side of the equilateral triangle be 2y
so we have 4x + 6y = 15
x = (15-6y)/4

A = x^2 + (1/2)(y)(√3y)
= (15-6y)^2 /16 + (√3/2) y^2

dA/dy = (2/16)(15-6y)(-6) + √3y
= (-3/4)(15-6y) + √3y
= -45/4 + 9y/2 + √3y = 0 for a max/min
-45/4 + 9y/2 + √3y = 0
-45 +18y + 4√3y = 0
y(18 + 4√3) = 45
y = 45/(18+4√3) = appr 1.8052

so we need 6y for the triangle or 10.8312 m ,
leaving 15-10.8312 or 4.1688 m for the square.

Answer 4

glad you got rid of all those pesky fractions!

:-)

Answer 5

oh, yeah. My equation was wrong anyway, as it should have been

((15-x)/4)^2 + (x/3)^2*√3/4