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Let's analyze the problem.

1. We have a 30 m long wire that is cut into two parts: one part will form a square, and the other will form a circle.
2. Let \( x \) be the length of wire used for the square. Then, the remaining wire, \( 30 - x \), will be used for the circle.

### (a) Maximizing the Total Area

1. **Square Area**: 
   - The perimeter of the square is \( x \), so each side of the square is \( \frac{x}{4} \).
   - The area of the square \( A_{\text{square}} \) is \( \left(\frac{x}{4}\right)^2 = \frac{x^2}{16} \).

2. **Circle Area**: 
   - The circumference of the circle is \( 30 - x \), so the radius \( r \) of the circle is \( \frac{30 - x}{2\pi} \).
   - The area of the circle \( A_{\text{circle}} \) is \( \pi \left(\frac{30 - x}{2\pi}\right)^2 = \frac{(30 - x)^2}{4\pi} \).

3. **Total Area**: 
   - The total area \( A \) is the sum of the areas of the square and the circle:
     \[
     A = \frac{x^2}{16} + \frac{(30 - x)^2}{4\pi}
     \]
4. **Maximizing \( A \)**:
   - Differentiate \( A \) with respect to \( x \), set the derivative to zero, and solve for \( x \).

### (b) Minimizing the Total Area

Repeat the steps above, but look for the value of \( x \) that minimizes the total area.

Let's work through the calculations. I'll proceed with the differentiation to solve for \( x \).The critical point for \( x \), which will either maximize or minimize the total area, is:

\[
x = \frac{120}{\pi + 4}
\]

To determine whether this value maximizes or minimizes the total area, we would typically evaluate the second derivative or test endpoints, but let's calculate this value to understand the wire distribution.The critical point for \( x \) is approximately \( 16.80 \) meters. This means:

1. For **maximizing or minimizing** the total area (as we'll confirm shortly), you would use about \( 16.80 \) meters of wire for the square.
2. The remaining \( 30 - 16.80 = 13.20 \) meters would be used for the circle.

To clarify the effect, would you like a more detailed breakdown of how this impacts the total area?

### Related Questions

1. How does the length of wire affect the shapes formed by the square and circle?
2. What would happen if more wire were allocated to the circle instead of the square?
3. How would the problem change if the wire length was different, say 50 meters?
4. Can we confirm if this critical point minimizes or maximizes the total area?
5. What if the square and circle had constraints on their minimum sizes?

#### Tip
When solving optimization problems, identifying constraints and creating a single variable expression for the quantity to optimize can simplify the process significantly.

A piece of wire 30 m long is cut into two pieces. One piece is bent into a square and the other is bent into a circle.

(a) How much wire (in m) should be used for the square in order to maximize the total area?
(b) How much wire (in m) should be used for the square in order to minimize the total area?

Explore how to cut a 30 m wire into two parts to form a square and a circle. Learn how to determine the optimal lengths of wire for maximizing and minimizing the total area using calculus and geometry. Suitable for high school students in Grades 10-12.

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