The 60 cm long wire should be cut so that you have 2 lengths of 30 cm each.

Let 1 of the cut pieces total length be ##x## Then the other piece is length ##60-x##

Let the area for square 1 be ##A_1## Let the area for square 2 be ##A_2## Let the sum of the areas be ##A_s##

All sides of a square are of equal length so:

##A_1=(x/4)^2##

##A_2=((60-x)/4)^2##

##A_s=A_1+A_2=(x/4)^2+((60-x)/4)^2##

##A_s= x^2/16+(x^2-120x+3600)/16##

##A_s= (2x^2)/16- 120/16x+3600/16##

##A_s=1/8x^2-15/2x+225## ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ This is a quadratic and as the ##x^2## term is positive it is of general shape of ##uu##

Thus the minimum area (##A_s##) is at the vertex. Let me show you a trick to determining the vertex ##x## value.

Write as ##A_s=1/8(x^2-(8xx15)/2x)+225##

##A_s=1/8(x^2-color(red)((cancel(8)^4xx15)/(cancel(2)^1))x)+225##

##color(green)(“The above is the beginning of the process to ‘complete the square'”)##

##x_(“vertex”)= (-1/2)xx(color(red)(-4xx15)) = +30##

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ So the 60 cm long wire should be cut so that you have 2 lengths of 30 cm each.