To find the value of x for which the area, A, is smallest, we need to express the area of each piece in terms of x and then determine how it varies as x changes.
Let's break down the problem step by step:
1. Let's assume the wire is cut into two pieces: one for a square and the other for a circle.
Let's denote the length of the wire used for the square as "x" and the remaining wire for the circle as "10 - x".
2. The wire used for the square will form a square with each side measuring x/4, since a square has four equal sides.
Consequently, the area of the square, A_s, can be expressed as:
A_s = (x/4)^2 = x^2/16
3. The remaining wire, 10 - x, is used to form a circle. The circumference of the circle is equal to the length of the remaining wire, i.e., 10 - x.
Therefore, we can find the radius, r, of the circle using the formula for circumference:
2Ï€r = 10 - x
r = (10 - x) / (2Ï€)
4. The area of the circle, A_c, can be calculated using the formula:
A_c = πr^2 = π((10 - x) / (2π))^2 = (10 - x)^2 / (4π)
Now that we have the expressions for the area of both the square and the circle, we can compare them to find the value of x for which A is smallest.
5. Graph A=A(x) represents the total area, which is the sum of the area of the square and the area of the circle:
A(x) = A_s + A_c = x^2/16 + (10 - x)^2 / (4Ï€)
To find the value of x that minimizes the area A, we can differentiate A(x) with respect to x and set the derivative equal to zero.
6. Differentiating A(x) with respect to x:
A'(x) = (1/16)(2x) - (2(10 - x))/(4Ï€)
Simplifying further:
A'(x) = x/8 - (20 - 2x) / (4Ï€)
7. Setting A'(x) equal to zero:
x/8 - (20 - 2x) / (4Ï€) = 0
Now, solving for x will give us the value for which A is smallest. However, this equation cannot be solved analytically. A numerical approach, such as using a calculator or a computer, would be required to find the value of x that minimizes the area A based on the graph A=A(x).