Question

A piece of wire `26 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 wire should be used for the square in order to minimize the total area?

Answer 1

The length of wire used for the square should be `color(red)(11.309` meters

Explanation:

Let's start by expressing what we know mathematically.

Let `s` be the length of wire used for the square.

Let `t` be the length of wire used for the triangle.

Let `A_S` be the area of the square.

Let `A_T` be the area of the triangle.

One side of the square is `s/4`, so we know that

`A_S=(s/4)^2`

The formula for the area of an equilateral triangle is

`A=sqrt3/4a^2` where `a` is the length of one side

and one side of our triangle is `t/3`, so we know that

`A_T=sqrt3/4(t/3)^2`

The problem states that we want to find a value for `s` such that

`s+t=26`

and

`A_S+A_T=A_(S+T)` is at a minimum

Let's find an expression for `A_(S+T)` as a function of `s`

It will be necessary to express `t` in terms of `s`, so let's write

`t=26-s`

`A_(S+T)=(s/4)^2+sqrt3/4((26-s)/3)^2`

`A_(S+T)=1/16s^2+sqrt3/4((676-52s+s^2)/9)`

`A_(S+T)=1/16s^2+sqrt3/36s^2-(13sqrt3)/9s+(169sqrt3)/9`

`A_(S+T)=(9+4sqrt3)/144s^2-(13sqrt3)/9s+(169sqrt3)/9`

Graphing this, we get a parabola:

graph{(9+4sqrt3)/144x^2-(13sqrt3)/9x+(169sqrt3)/9 [-64.2, 105.36, -7.5, 77.3]}

This function has a global minimum at

`s=(312sqrt3-416)/11~~11.309`

Therefore the answer is to use 11.309 meters of wire for the square, and the remaining 14.691 meters of wire for the triangle.