A piece of wire 40 m long is cut into two pieces. One piece is bent into a square and the other is bent into a circle. How much of the wire should go to the square to minimize the total area enclosed by both figures?

Let \( x \) meters be the length of the wire used to form the square, and \( 40 - x \) meters be the length of the wire used to form the circle.

### Step-by-Step Solution:

1. **Square Perimeter and Area:**
   - Since \( x \) meters of wire forms a square, each side of the square will be \( \frac{x}{4} \) meters.
   - The area of the square, \( A_{\text{square}} \), is:
     \[
     A_{\text{square}} = \left(\frac{x}{4}\right)^2 = \frac{x^2}{16}
     \]

2. **Circle Circumference and Area:**
   - The length of the wire for the circle is \( 40 - x \), so this is the circumference of the circle.
   - The radius \( r \) of the circle is:
     \[
     r = \frac{40 - x}{2\pi}
     \]
   - The area of the circle, \( A_{\text{circle}} \), is:
     \[
     A_{\text{circle}} = \pi r^2 = \pi \left(\frac{40 - x}{2\pi}\right)^2 = \frac{(40 - x)^2}{4\pi}
     \]

3. **Total Area:**
   - The total area, \( A_{\text{total}} \), is the sum of the areas of the square and the circle:
     \[
     A_{\text{total}} = A_{\text{square}} + A_{\text{circle}} = \frac{x^2}{16} + \frac{(40 - x)^2}{4\pi}
     \]

4. **Minimizing the Total Area:**
   - To find the value of \( x \) that minimizes \( A_{\text{total}} \), we take the derivative of \( A_{\text{total}} \) with respect to \( x \), set it to zero, and solve for \( x \):
     \[
     \frac{dA_{\text{total}}}{dx} = \frac{x}{8} - \frac{40 - x}{2\pi} = 0
     \]
   - Rearranging terms gives:
     \[
     \frac{x}{8} = \frac{40 - x}{2\pi}
     \]
     \[
     x \left(1 + \frac{4}{\pi}\right) = 40
     \]
     \[
     x = \frac{40}{1 + \frac{4}{\pi}}
     \]

5. **Calculating \( x \):**
   - Using the approximation \( \pi \approx 3.1416 \):
     \[
     x \approx \frac{40}{1 + \frac{4}{3.1416}} \approx \frac{40}{1 + 1.2732} \approx \frac{40}{2.2732} \approx 17.6 \text{ meters}
     \]

Thus, to minimize the total area, approximately **17.6 meters** of the wire should go to the square.

Would you like more details, or have any questions about this solution?

Here are some related questions you might find useful:
1. How would the solution change if we wanted to maximize the area instead?
2. What if the wire length were different, say 50 meters? How would that affect the optimal cut?
3. How would using different shapes (e.g., a triangle instead of a square) impact the solution?
4. Can we generalize this solution for any perimeter and shape combination?
5. What’s the significance of using derivatives in optimization problems like this?

**Tip:** When optimizing, consider expressing all areas and dimensions in terms of one variable to simplify finding minima or maxima.

Solve an optimization problem involving a 40-meter wire divided to form a square and a circle. Learn how to allocate the wire to minimize the total enclosed area, using geometry and calculus concepts like derivatives and area formulas. Ideal for high school students (Grades 10-12).

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