A piece of wire 20 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 maximize the total area?

Question

Luke Alvarez · Accepted Answer

Let #x# metres be the side of the square, so #4x# metres are the total amount of wire used for the square (with #0<=4x<=20rArr0<=x<=5#). For the equilater triangle #20-4x# metres of wire remain, and the side will be #(20-4x)/3#.

Given the side #l# of an equilatrer triangle, the height is, using the Pitagora's theorem, #h=l/2sqrt3#, so the area of the triangle is:

#A=l*l/2sqrt3*1/2=l^2sqrt3/4#

(#A=(b*h)/2#)

So the total area is:

#A=x^2+((20-4x)/3)^2sqrt3/4#.

#A'=2x+sqrt3/4*1/9 * 2(20-4x) * (-4)=#

#=2x-2sqrt3/9(20-4x)#.

#A'>0# if

#2x-2sqrt3/9(20-4x)>0rArr18x-40sqrt3+8sqrt3x>0rArr#

#2x(9+4sqrt3)>40sqrt3rArrx>(40sqrt3)/(2(9+4sqrt3))rArr#

#x>(20sqrt3)/(9+4sqrt3)*(9-4sqrt3)/(9-4sqrt3)rArr#

#x>(20*3(3sqrt3-4))/(81-48)rArrx>20/11(3sqrt3-4)=barx#.

So our function Area is decreasing in #[0,barx)# and growing in #(barx,5]# and so the point #(barx,A(barx))# is a minimum of the function. The conclusion is that maximum (absolute maximum) is in one of the two sides of the interval of definition of #x#, #0 or 5#.

If #x=0# the value of area will be: #A=100sqrt3/9~=19,2#.

If #x=5# the value of the area will be: #A=25#, that is greater than the other value.

In conclusion the maximum of the area will be in the case in which all the wire will be used for the square!

Charles Anctil · Answer

undefined To maximize the total area, half of the wire (10 meters) should be used for the square.