A wire 10 meters ong is to be cut into two pieces. One piece wil be shaped as a square, and the other piece will be shaped as a circle. (piece one is 4x, piece two is 10-4x)


a)Express the total area A enclosed by the pieces of wire as a function of the length x of the side of the square.

b)What is the domain of A?

c) Graph A=A(x). For what value of x is A smallest?

a) The length of the wire is 10 meters, so we can express the length of the square as 4x (each side of the square is x), and the length of the circle as 10-4x (since it's the remaining wire length after the square is formed).

The area of the square is A(square) = x^2, and the area of the circle is A(circle) = πr^2, where the radius r can be calculated as half of the circumference of the circle, which is (10-4x)/2.

Adding the areas together, we get the total area A enclosed by the pieces of wire:

A(x) = x^2 + π((10-4x)/2)^2

b) The domain of A(x) depends on the possible values of x. Since x represents the length of one side of the square, it cannot be negative. Also, the wire length cannot be used to make an infinitely small square or circle, so x must be less than or equal to 10/4 (2.5 meters).

Therefore, the domain of A(x) is 0 ≤ x ≤ 2.5.

c) To graph A(x), we need to plot the values of A for different values of x within the domain.

To find the value of x for which A is smallest, we can find the minimum point on the graph. This can be achieved by finding the critical points where the derivative of A(x) equals zero and checking for the minimum.

Differentiating A(x) with respect to x:

A'(x) = 2x + π(10-4x)(-2)/2 = 2x - 2π(10-4x)

Setting A'(x) = 0 and solving for x:

2x - 2π(10-4x) = 0
2x - 20π + 8πx = 0
10π = 10x
x = π

So, at x = π, the area A is expected to be at its minimum.

a) To express the total area A enclosed by the pieces of wire as a function of the length x of the side of the square, we need to find the dimensions of each piece.

Let's assume the length of the side of the square is x. Since the wire is 10 meters long, the remaining piece (the wire not used for the square) will have a length of 10 - x.

The area of the square is given by A₁ = side^2 = x^2.
The area of the circle can be calculated using the circumference formula C = 2πr, where r is the radius. In this case, since the remaining wire is used for the circle, the circumference of the circle is equal to 10 - x. So we have 10 - x = 2πr. Rearranging this equation, we get r = (10 - x) / (2π). The area of the circle is given by A₂ = πr^2 = π((10 - x) / (2π))^2 = (10 - x)^2 / (4π).

The total area A is the sum of the areas of the square and the circle:
A(x) = A₁ + A₂ = x^2 + (10 - x)^2 / (4π).

b) The domain of A is the set of all possible values for x. In this case, the domain is constrained by the length of the wire. The wire is 10 meters long, so x must be a value such that x + (10 - x) = 10, or simply x ≤ 10.

c) To graph A = A(x), we plot the values of A for different values of x from the domain. To find the value of x for which A is smallest, we need to find the minimum of the function A(x).

Let's differentiate A(x) with respect to x and set it equal to zero to find the critical point:
dA(x)/dx = 2x - 2(10 - x) / (4π) = 0.

Simplifying the equation, we get:
2x - (20 - 2x) / (4π) = 0
8x - 20 + 2x = 0
10x = 20
x = 2.

Therefore, the value of x for which A is smallest is x = 2.

a) To express the total area A enclosed by the pieces of wire as a function of the length x of the side of the square, we need to find the area of both the square and the circle.

The length of the wire used for the square will be the perimeter of the square, which is 4 times the length of one side. So, the length of the wire used for the square is 4x.

The length of the wire used for the circle is the remaining wire after subtracting the length used for the square, which is 10 - 4x.

The area of the square is calculated by squaring the length of its side, so the area of the square is x^2.

The area of the circle is given by the formula A = πr^2, where r is the radius of the circle. Since the wire used for the circle is a circumference, we can find the radius by dividing the circumference by 2π. So, the radius of the circle is (10 - 4x) / (2π). Substituting this value into the formula, the area of the circle is π((10 - 4x) / (2π))^2.

Therefore, the total area A is the sum of the areas of the square and the circle:

A = x^2 + π((10 - 4x) / (2π))^2
= x^2 + (10 - 4x)^2 / (4π)

b) The domain of A(x) is the set of possible values for x that make sense in the context of the problem.

In this case, x represents the length of the side of the square. The side length cannot be negative since it is a physical measurement. Also, the side length must be less than or equal to half of the wire length, since the other piece should have a positive length.

Therefore, the domain of A is 0 ≤ x ≤ 5.

c) To graph A = A(x), we can plot the function for different values of x within the domain and observe the shape of the graph.

The graph of A(x) will have x-values ranging from 0 to 5. For each x-value, calculate the corresponding value of A using the expression derived in part (a).

To find the value of x for which A is smallest, we can observe the graph and identify the minimum point.

However, without specific values for x or the constants involved, it is not possible to provide an exact graph or identify the precise x-value at which A is smallest.

area square = x^2

diameter circle = d = (10-4x)/pi
so area circle = pi d^2/4
= (1/4pi)(10-4x)^2

A = x^2 +(1/4pi)(100 -80x +16 x^2)
= (1/pi)pi x^2 + (1/pi) (25-20x+4x^2)
= (1/pi)( (pi+4)x^2 -20 x + 25)
that is a parabola, look for zeros and vertex
note that negative A is meaningless as is negative x.
negative 10-4x is also meaningless so right off the bat you know x>10/4 is outside the domain
so x can only be between 0 and 10/4