To find the lengths that minimize the total area, we need to consider the formulas for the area of a circle and the area of a square.
1. Let's consider the piece of wire made into a circle first. We will use the formula for the circumference of a circle:
C = 2Ï€r, where C is the circumference and r is the radius.
In this case, if x ft of wire is used for the circle, we have:
C = x, so 2Ï€r = x, and solving for r, we get r = x / (2Ï€).
Now, the area of a circle is given by the formula:
A = πr^2, where A is the area of the circle.
Substituting the value of r we found earlier, we get:
A = π(x / (2π))^2 = πx^2 / (4π^2) = x^2 / (4π).
2. Now, let's consider the piece of wire made into a square. The perimeter of a square is equal to four times the length of one side. So, if (12 - x) ft of wire is used for the square, we have:
4s = (12 - x), where s is the length of one side of the square.
Solving for s, we get s = (12 - x) / 4 = (3 - x/4).
The area of a square is simply the side length squared:
A = s^2 = [(3 - x/4)]^2 = (9 - 3x/2 + x^2/16).
3. To minimize the total area, we need to find the values of x that minimize the expression A_total = A_circle + A_square.
A_total = x^2 / (4Ï€) + (9 - 3x/2 + x^2/16).
Taking the derivative of A_total with respect to x and setting it equal to zero, we can find the potential minimum:
d(A_total)/dx = 0.
Simplifying, we get: d(A_total)/dx = (8Ï€x - 6 + x/8) / (4Ï€) = 0.
Solving for x, we find x = 24 / (8Ï€ + 1).
4. To find the radius of the circle, we can substitute the value of x we found back into the equation for r:
r = x / (2Ï€) = (24 / (8Ï€ + 1)) / (2Ï€) = 12 / (4Ï€ + 1).
5. To find the length of each side of the square, we can substitute the value of x into our expression for s:
s = (3 - x/4) = (3 - (24 / (4Ï€ + 1))/4) = 3 - (6 / (4Ï€ + 1)).
Therefore, by using x = 24 / (8Ï€ + 1) ft of wire for the circle, we will minimize the total area. The radius of the circle will be 12 / (4Ï€ + 1) ft, and the length of each side of the square will be 3 - (6 / (4Ï€ + 1)) ft.