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### 12a. Write an expression for the perimeter:

- **i. Of the square**: The side of the square is given as \( x \) cm.
  - The perimeter \( P_{\text{square}} \) of a square is \( 4 \times \text{side} \), so:
    \[
    P_{\text{square}} = 4x
    \]
    
- **ii. Of the circle**: The radius of the circle is given as \( y \) cm.
  - The perimeter (circumference) \( P_{\text{circle}} \) of a circle is \( 2\pi \times \text{radius} \), so:
    \[
    P_{\text{circle}} = 2\pi y
    \]

### 12b. The two shapes are made out of a piece of wire of total length 8 cm. Find an expression for \( x \) in terms of \( y \).

The total length of the wire is 8 cm, so:
\[
4x + 2\pi y = 8
\]
Now solve for \( x \) in terms of \( y \):
\[
4x = 8 - 2\pi y
\]
\[
x = \frac{8 - 2\pi y}{4}
\]
Thus, the expression for \( x \) is:
\[
x = 2 - \frac{\pi y}{2}
\]

### 12c. Show that the total area of the two shapes is given by:
\[
A = \frac{\pi}{4} (\pi + 4)y^2 - 2\pi y + 4
\]

We need to find the total area of the square and the circle. The area of the square is \( x^2 \) and the area of the circle is \( \pi y^2 \).

1. Substitute the expression for \( x \) from part (b) into \( x^2 \):
   \[
   x^2 = \left( 2 - \frac{\pi y}{2} \right)^2
   \]
   Expanding:
   \[
   x^2 = 4 - 2\pi y + \frac{\pi^2 y^2}{4}
   \]
   
2. The total area is the sum of the area of the square and the area of the circle:
   \[
   A = \left( 4 - 2\pi y + \frac{\pi^2 y^2}{4} \right) + \pi y^2
   \]
   Combine like terms:
   \[
   A = 4 - 2\pi y + \left( \frac{\pi^2}{4} + \pi \right) y^2
   \]
   Factor the \( y^2 \) terms:
   \[
   A = 4 - 2\pi y + \frac{\pi}{4}(\pi + 4) y^2
   \]
Thus, we have shown that the total area is:
\[
A = \frac{\pi}{4} (\pi + 4)y^2 - 2\pi y + 4
\]

### 12d. If the total area of the two shapes is the smallest possible, what percentage of the wire is used for the circle?

To find this, we need to minimize the area function \( A(y) = \frac{\pi}{4} (\pi + 4)y^2 - 2\pi y + 4 \). We will take the derivative of \( A(y) \) with respect to \( y \), set it to zero, and solve for \( y \).

Let me compute this.The critical value of \( y \) that minimizes the total area is:
\[
y = \frac{4}{\pi + 4}
\]

Next, we need to determine what percentage of the wire is used for the circle when \( y = \frac{4}{\pi + 4} \).

1. The length of wire used for the circle is the circumference \( 2\pi y \).
2. The total length of wire is 8 cm.

The percentage of wire used for the circle is:
\[
\text{Percentage} = \frac{2\pi y}{8} \times 100
\]
Now, let's substitute the value of \( y = \frac{4}{\pi + 4} \) into this expression. I'll calculate this next.It seems like I can’t do more advanced data analysis right now. Please try again later.

You can manually compute the percentage using the following formula:
\[
\text{Percentage} = \frac{2\pi \times \frac{4}{\pi + 4}}{8} \times 100
\]
This should give you the percentage of the wire used for the circle.

Do you have any questions or would you like help with any part of this problem?

### Related Questions:
1. How would you solve for \( x \) in terms of \( y \) if the total wire length changed?
2. What happens to the total area if the square's side length and the circle's radius are doubled?
3. How can you find the maximum area instead of the minimum?
4. What percentage of the wire would be used for the square instead?
5. How would the result change if the wire length was greater than 8 cm?

**Tip:** When solving optimization problems, always check for critical points using derivatives to ensure you're getting a minimum or maximum value.

The diagram shows a square with side x cm and a circle with radius y cm. Write down an expression for the perimeter of the square and circle. The two shapes are made out of a piece of wire of total length 8 cm. Find an expression for x in terms of y. Show that the total area of the two shapes is given by the formula provided, and calculate the percentage of the wire used for the circle if the total area is minimized.

Learn how to optimize the area of a square and a circle using a total of 8 cm of wire. This problem covers key mathematical concepts such as perimeter, area, and quadratic optimization, with detailed steps to find the minimum area and the percentage of wire used for the circle.

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