A piece of wire 100 cm long is cut into two pieces. One piece is bent to form a square and the other to form a circle. Let x equal the length of the wire used to form the square.

A. write the function that represents the area of the two figures

B. determine the domain of the function

C. find the value(s) of x that yield a maximum and minimum area. Explain your reasoning.

the circumference of the circle is 100-x. So, the area is

a = (x/4)^2 + π((100-x)/2π)^2

This is just a parabola, so its vertex (minimum area) is easy to find.

For the maximum area, compare a at the ends of the domain.

A. The function that represents the area of the two figures can be written as follows:

For the square:
Area of the square = side^2
Since the perimeter of the square is equal to 4 times the side, we have:
4x = 100 - x (since the other piece of wire is used to form the circle)
x = (100 / 5) = 20 cm (length of each side of the square)
Area of the square = (side)^2 = (20)^2 = 400 cm^2

For the circle:
Circumference of the circle = 2πr
Since the perimeter of the circle is equal to the length of the wire used for the circle, we have:
2πr = 100 - x (since the other piece of wire is used for the square)
r = (100 - x) / (2π) (radius of the circle)

Area of the circle = πr^2
= π((100 - x) / (2π))^2
= (100 - x)^2 / (4π) cm^2

B. The domain of the function depends on the range of possible values for x. Since the length of the wire used to form the square cannot be negative or greater than the total length of the wire (100 cm), the domain can be defined as 0 ≤ x ≤ 100.

C. To find the values of x that yield a maximum and minimum area, we can analyze the behavior of the function.

Taking the derivative of the area function with respect to x, we have:

dA/dx = d[(100 - x)^2 / (4π)] / dx
= -2(100 - x) / (4π)

Setting this derivative equal to zero to find critical points:

-2(100 - x) / (4π) = 0
100 - x = 0
x = 100

Since the value of x lies within the domain, it is a critical point.

Evaluating the second derivative of the area function at the critical point:

d^2A/dx^2 = d[(-2(100 - x) / (4π))] / dx
= 2 / (4π)
= 1 / (2π)

Since the second derivative is positive, the critical point corresponds to the minimum area.

Thus, the value of x that yields the minimum area is x = 100 cm.

To find the maximum area, we need to evaluate the area function at the endpoints of the domain (0 and 100 cm):

For x = 0 cm:
Area of the circle = (100 - 0)^2 / (4π) = 2500 / (4π) ≈ 198.94 cm^2

For x = 100 cm:
Area of the square = (100)^2 = 10000 cm^2

Comparing the areas, we can see that the maximum area is obtained when x = 100 cm, which yields the square with an area of 10000 cm^2.

To solve this problem, we need to break it down into steps:

A. Writing the function that represents the area of the two figures:
Let's first consider the Square:
The wire is cut into two pieces of lengths x and (100 - x) cm. The length of each side of the square would be x/4 (since the perimeter of a square is 4 times one of its sides). Therefore, the area of the square is (x/4)^2 = x^2/16.

Now, let's consider the Circle:
The remaining piece of wire (100 - x) cm is used to form a circle. The circumference of the circle would be equal to the remaining wire, so it is (100 - x). To find the radius (r) of the circle, we can use the formula for circumference: C = 2πr. Solving for r, we have r = (100 - x)/(2π). The area of the circle is given by A = πr^2 = π((100 - x)/(2π))^2 = (100 - x)^2/(4π).

Thus, the function that represents the area of the square and the circle is:
A(x) = x^2/16 + (100 - x)^2/(4π).

B. Determining the domain of the function:
The domain of the function represents the possible values of x. In this case, since x represents the length of the wire used to form the square, it must be a positive value. Additionally, the total wire length is 100 cm, so x must be less than or equal to 100. Therefore, the domain of the function is 0 ≤ x ≤ 100.

C. Finding the value(s) of x that yield a maximum and minimum area:
To find the value(s) of x that yield a maximum or minimum area, we can take the derivative of the function A(x) and set it equal to zero.

A'(x) = (2x)/16 - 2(100 - x)/(4π) = 0

Simplifying the equation gives us:

x/8 - (100 - x)/(2π) = 0

Multiplying through by 8π gives us:

πx - 4(100 - x) = 0

πx - 400 + 4x = 0

(π + 4)x = 400

x = 400/(π + 4)

Since the domain is 0 ≤ x ≤ 100, we need to check if this value is within the specified range.

By substituting π ≈ 3.14, we find:

x ≈ 400/(3.14 + 4) = 400/7.14 ≈ 55.92

Thus, the value of x that yields a maximum and minimum area is approximately 55.92 cm.

To determine which value yields the maximum and minimum area, we can evaluate the second derivative of the function A(x). If A''(x) > 0, then x = 55.92 corresponds to a minimum. If A''(x) < 0, then x = 55.92 corresponds to a maximum.