Question 404474
Since the perimeters of the two squares are x and L-x respectively, then the side lengths are {{{x/4}}} and {{{(L-x)/4}}}. The sums of the two areas are:


{{{A(x) = (x/4)^2 + ((L-x)/4)^2 = (x^2)/16 + (x^2 - 2Lx + L^2)/16 = (2x^2 - 2Lx + L^2)/16}}}


The minimum occurs when {{{A(x)}}} is minimized, which also happens when {{{2x^2 - 2Lx + L^2}}} is minimized. Either using the vertex of the parabola or by differentiating, we obtain {{{x = -(-2L)/4 = L/2}}}, so the minimum area occurs when the wire is cut into two pieces of 8 cm.