A wire 10m long is cut into 2 pieces. one piece will be cut into a square the other piece will be shaped into a circle. Graph A= A(x) for what value of x is A smallest?

if the side of the square is x,

square has perimeter 4x
circle has circumference 10-4x

A(x) = x^2 + pi * ((10-4x)/(2pi))^2
= 1/pi ((4+pi)x^2 - 20x + 25)

dA/dx = 2/pi ((4+pi)x - 10)
dA/dx = 0 when x = 10/(4+pi)
since the parabola opens up, A is smallest there.

i don't understand how to graph what is x and y

As I said, let x be the side of the square. The perimeter is thus 4x.

The available length of wire is 10. So, the wire available for the circle is only 10-4x

area of square is x^2
area of circle is pi r^2
if the circumference of the circle is (10-4x), the radius is (10-4x)/2pi

so, the area of the combined figures is as shown above:

A(x) = 1/pi ((4+pi)x^2 - 20x + 25)
now, this is pre-cal, so you know about parabolas. You know that the vertex of a parabola ax^2+bc+c is at x = -b/2a.

Here, that would be x = 20/(2(4+pi)) = 10/(4+pi)

Since the coefficient of x^2 is positive, the parabola opens up, so the value of A(x) at the vertex is a minimum.

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:
2πr = 10 - x
r = (10 - x) / (2π)

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 / (4π)

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))/(4π)
Simplifying further:
A'(x) = x/8 - (20 - 2x) / (4π)

7. Setting A'(x) equal to zero:
x/8 - (20 - 2x) / (4π) = 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).