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?

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Answer 1 · 2018-04-01T15:34:53

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

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.