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!