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Let's solve this problem step-by-step.

### Given Information:
A 7-meter long iron wire is cut into two pieces:
- One piece forms a square.
- The other piece forms a circle.

Let:
- \( x \) = length of the wire used for the square.
- \( 7 - x \) = length of the wire used for the circle.

### Part (a): Maximize the Sum of the Areas

1. **Formulas for the areas of the shapes:**
   - **Square:** If \( x \) meters of wire is used, each side of the square is \( \frac{x}{4} \).
     - Area of square \( = \left(\frac{x}{4}\right)^2 = \frac{x^2}{16} \).
   - **Circle:** If \( 7 - x \) meters of wire is used, the circumference of the circle is \( 7 - x \).
     - Radius of the circle \( = \frac{7 - x}{2\pi} \).
     - Area of circle \( = \pi \left(\frac{7 - x}{2\pi}\right)^2 = \frac{(7 - x)^2}{4\pi} \).

2. **Objective Function:**
   - We want to maximize the total area \( A \), given by:
     \[
     A(x) = \frac{x^2}{16} + \frac{(7 - x)^2}{4\pi}
     \]

3. **Optimization Process:**
   - To find the maximum, differentiate \( A(x) \) with respect to \( x \) and set the derivative equal to zero.

### Part (b): Minimize the Sum of the Areas

The approach will be similar, but this time we are looking to minimize \( A(x) \) instead of maximizing it.

Let's proceed with the calculations for both parts.The solutions for each part are as follows:

1. **For Maximum Area:**
   - To maximize the area, we check the area values at the endpoints \( x = 0 \) and \( x = 7 \), and at the critical point \( x = \frac{28}{\pi + 4} \approx 2.91 \) meters.
   - The maximum total area occurs at **\( x = 0 \)**, giving an area of approximately **3.90 square meters**.

   So, for part (a), the wire should not be cut for the square, and all 7 meters should be used to form the circle to maximize the area.

2. **For Minimum Area:**
   - To minimize the area, we compare the values at the endpoints and the critical point.
   - The minimum area occurs at **\( x \approx 2.91 \)** meters, which gives a minimized area of approximately **1.72 square meters**.

   Thus, for part (b), the wire should be cut such that approximately **2.91 meters** are used for the square.

### Answers:
- **(a)** Length of the piece of wire used to produce the square: **0.00 m**
- **(b)** Length of the piece of wire used to produce the square: **2.91 m**

Would you like more details on the steps, or do you have further questions?

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### Follow-up Questions:
1. What would happen to the solution if the wire length was different (e.g., 10 meters)?
2. How would the problem change if the square and circle required specific minimum areas?
3. Can you explain why the maximum area is achieved when all the wire is used for the circle?
4. How could calculus be used to verify the second derivative and confirm the maximum/minimum?
5. What real-life applications could a similar problem be related to?

### Tip:
In optimization problems involving area, it's essential to clearly define the variables and set up the equation correctly. This setup simplifies finding and interpreting critical points.

A 7-meter long iron wire is cut in two. One piece is to form a square and the other a circle.

a) Where should you cut the wire to maximize the sum of the areas from the two figures that have been produced?
Length of the piece of wire used to produce the square: ________ m

b) Where should you cut the wire to minimize the sum of the areas from the two figures that have been produced?
Length of the piece of wire used to produce the square: ________ m

Solve an optimization problem involving a 7-meter wire cut to form a square and a circle. Learn how to maximize and minimize the combined area using calculus and geometry principles.

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