# 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.

## Ok, I did that, no what ?

## 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.

## 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.

## glad you got rid of all those pesky fractions!

:-)

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

((15-x)/4)^2 + (x/3)^2*√3/4

## Well, that wire must be quite the contortionist! I hope it stretched properly before being bent. Let's dive deeper into this amusing wire conundrum.

Now, if we take one piece of the wire and bend it into a square, we know that a square has four sides of equal length. So, each side of the square would measure (15 m / 4 = 3.75 m).

As for the other piece of wire, we'll bend it into an equilateral triangle – a shape that has three sides of equal length. Since we used up one piece of wire for the square, we have 15 m - 3.75 m = 11.25 m left for the triangle.

Since all three sides of an equilateral triangle must be equal, each side would measure (11.25 m / 3 = 3.75 m).

As you can see, our wire chose a magical length of 3.75 m! That's enough to make both a square and an equilateral triangle, leaving no wire leftover.

Now, whether this wire wanted to be a square or a triangle, we may never know. But either way, it got all bent out of shape into some interesting geometric goodness!

## To find the lengths of the sides of the square and the equilateral triangle, we need to determine how much wire is used for each shape.

Let's denote the length of the wire used for the square as "s" and the length of the wire used for the equilateral triangle as "t".

We know that the total length of the wire is 15 meters, so we can write the equation:

s + t = 15

The perimeter of a square is given by P = 4s, so the length of the wire used for the square is equal to the perimeter:

s = P = 4s

Likewise, the perimeter of an equilateral triangle is given by P = 3t, so the length of the wire used for the triangle is equal to the perimeter:

t = P = 3t

From the two equations above, we can solve for s and t.

For the square:

s = 4s

Divide both sides of the equation by 4:

1 = s

Thus, the length of the sides of the square is 1 meter.

For the equilateral triangle:

t = 3t

Divide both sides of the equation by 3:

1 = t

Thus, the length of the sides of the equilateral triangle is also 1 meter.

Therefore, the square and the equilateral triangle both have sides with a length of 1 meter.