Question 976989
Let L = the length of the wire
Let s = the perimeter of the square, then s/4 = the length of one side
Then the circumference of the circle, C = L - s
Area of the square = (s/4)^2
Area of the circle = C^2/(4*pi)
A = the sum of the areas = (s/4)^2 + C^2/(4*pi)
In terms of C, since s = L - C, we can write
A = (L-C)^2/16 + C^2/(4*pi)
For the sum of the areas to be a minimum, dA/dC = 0:
0 = 2(L-C)(-1)/16 + 2C/(4*pi)
0 = (C-L)/8 + C/(2*pi)
Solve for C in terms of L:
C(1/8 + 1/(2*pi)) = L/8
C(1 + 4/pi) = L -> C = L/(1 + 4/pi)
Substituting L = 16 gives C = 7.04 inches
Therefore, the perimeter of the square is s = 16 - 7.04 = 8.96 inches