Twenty feet of wire is used to form an equilateral triangle

Twenty feet of wire is used to form an equilateral triangle and a square. How much wire should be used so that the total enclosed area is maximized?

I thought that the function would be 3/4x^2 + (20-3x / 4)^2 = A but this doesn\'t work because there must be a maxiumum x that can be found when graphed (and there are only minimums with this function).
What am I doing wrong with that function?

Solution

sum of perimeter of equ. trinagle + square = 20 feet

Let x be lenght of one side of equilateral triangle and y be length of side of square:

3x+4y =20 ----(1)

Total Area = sqrt3x^2/4 + y^2

= sqrt3x^2/4 + {(20 -3x)/4}^2

= x^2(sqrt3/4) + 25 +9x^2/16 - 120/16x

= 0.995x^2 - 7.5x +100

Maximum area is found at vertex x = -(-7.5)/(2*0.995) = 3.768

So, wire length for triangle = 11.30 ft

wire length of sqaure = 8.69 ft