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?

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.