Question 379922
Suppose the two wire lengths are L and 10-L. We need to derive the formulas for the areas of a square and equilateral triangle given the perimeter.

Without loss of generality, let the perimeter of the square be L. Then, the side length is L/4 and the area is {{{L^2/16}}}.

If the perimeter of the triangle is 10-L, then each side length is {{{(10-L)/3}}}, and the height is {{{(10-L)(sqrt(3))/6 }}}. It follows that the area is {{{((10-L)^2*sqrt(3))/36 = (sqrt(3)/36)L^2 - (5sqrt(3)/9) L + 20sqrt(3)/9}}}.

Therefore, the total area is {{{((4sqrt(3) + 9)/144)L^2 - (5sqrt(3)/9)L + 20sqrt(3)/9}}}. As the function is a second degree function, the vertex occurs at L = -b/2a = 4.3496. This represents the minimum total area, as the function points upward. As this function has no absolute maximum, the maximum total area must occur at an endpoint (either L = 0 or L = 10). Checking, we find that L=10 provides the optimal area.