Optimizing Area A Guide To Cutting Wire For Circles And Squares

Jul 18, 2025 by ADMIN 64 views

Hey everyone! Let's dive into an interesting math problem that combines geometry and optimization. We're going to take a 60-inch wire, cut it into two pieces, and use those pieces to form a circle and a square. Our goal is to figure out how to cut the wire to either maximize or minimize the total area enclosed by the two shapes. Sounds fun, right? Let's get started!

Problem Setup: Wire, Circle, and Square

First, let’s define our terms. We have a wire that is 60 inches long. We're cutting this wire into two pieces. One piece, which we'll call x, will be used to form a square. The remaining piece, which will be 60 - x, will be used to form a circle. Our ultimate goal is to find a formula for the total area enclosed by both the square and the circle, and then figure out how to optimize this area.

Setting Up the Equations: Square

Let's start with the square. We know the perimeter of the square is x. Since a square has four equal sides, we can find the length of one side by dividing the perimeter by 4. So, the side length of the square is x / 4. The area of a square is the side length squared, which means the area of our square, which we'll call A_square, is:

A_square = (x / 4)^2 = x^2 / 16

This equation tells us how the area of the square changes based on the length of the wire we use for it. The more wire we use (larger x), the bigger the square and the larger its area.

Setting Up the Equations: Circle

Now, let's move on to the circle. The circumference of the circle is the length of the remaining piece of wire, which is 60 - x. The formula for the circumference of a circle is C = 2πr, where r is the radius. So, we have:

2πr = 60 - x

We need to find the radius (r) in terms of x so we can calculate the area of the circle. Dividing both sides of the equation by 2π, we get:

r = (60 - x) / (2π)

Now that we have the radius, we can find the area of the circle. The formula for the area of a circle is A = πr^2. Plugging in our expression for r, we get the area of our circle, A_circle:

A_circle = π * [(60 - x) / (2π)]^2

Let's simplify this a bit:

A_circle = π * (60 - x)^2 / (4π^2)

A_circle = (60 - x)^2 / (4π)

This equation shows how the area of the circle changes with the length of the wire used for it. If we use more wire for the square (larger x), we'll have less wire for the circle, and its area will be smaller.

Combining the Areas: Total Area Formula

Now that we have the areas of both the square and the circle, we can find the total area, A_total, by simply adding them together:

A_total = A_square + A_circle

Substituting the formulas we derived earlier:

A_total = (x^2 / 16) + ((60 - x)^2 / (4π))

This is the formula we've been working towards! It tells us the total area enclosed by the square and the circle, depending on the length x of the wire used for the square. Now, we can use this formula to explore how to optimize the area. This total area formula is the key to solving our problem.

To reiterate the formula, the total area A_total enclosed by the square and circle is a function of x, the length of the wire used for the square, and is given by:

A_total(x) = (x^2 / 16) + ((60 - x)^2 / (4π))

Where:

x is the length of the wire used for the square (in inches).
60 - x is the length of the wire used for the circle (in inches).

Optimizing the Area: Minimizing and Maximizing

Now comes the fun part: optimization! We have a formula for the total area, and we want to find out how to cut the wire (x) to either minimize or maximize this area. This is a classic calculus problem, but we can also use some logical reasoning to get a sense of what's happening.

Minimizing the Area

To minimize the area, we want to find the value of x that makes A_total as small as possible. This involves finding the critical points of the function A_total(x) and determining whether they correspond to a minimum or maximum. We'll also need to check the endpoints of our interval (0 and 60) since the minimum or maximum could occur there.

Find the Derivative:

First, we need to find the derivative of A_total(x) with respect to x. This will give us the rate of change of the area as we change the length of wire used for the square. Let's differentiate our total area formula:
```
A_total(x) = (x^2 / 16) + ((60 - x)^2 / (4π))
```
```
A'_total(x) = d/dx [(x^2 / 16) + ((60 - x)^2 / (4π))]
```
```
A'_total(x) = (2x / 16) + (2(60 - x)(-1) / (4π))
```
```
A'_total(x) = (x / 8) - ((60 - x) / (2π))
```
This derivative tells us how the total area changes as we vary x. Now, we need to find where this rate of change is zero.
Set the Derivative to Zero and Solve:

To find the critical points, we set the derivative equal to zero and solve for x:
```
(x / 8) - ((60 - x) / (2π)) = 0
```
Let's solve this equation:
```
(x / 8) = ((60 - x) / (2π))
```
```
x / 8 = (60 - x) / (2π)
```
Multiply both sides by 8 and 2π to eliminate the fractions:
```
2πx = 8(60 - x)
```
```
2πx = 480 - 8x
```
Now, let's isolate x:
```
2πx + 8x = 480
```
```
x(2π + 8) = 480
```
```
x = 480 / (2π + 8)
```
```
x = 240 / (π + 4)
```
This is our critical point! Let's approximate this value:
```
x ≈ 240 / (3.14159 + 4) ≈ 240 / 7.14159 ≈ 33.605
```
So, x ≈ 33.605 inches is a critical point where the derivative is zero.
Determine if it's a Minimum:

To determine if this critical point is a minimum, we can use the second derivative test. We need to find the second derivative of A_total(x):
```
A'_total(x) = (x / 8) - ((60 - x) / (2π))
```
```
A"_total(x) = d/dx [(x / 8) - ((60 - x) / (2π))]
```
```
A"_total(x) = (1 / 8) - (-1 / (2π))
```
```
A"_total(x) = (1 / 8) + (1 / (2π))
```
Since A"_total(x) is positive, the critical point is indeed a minimum.

Check Endpoints:

We also need to check the endpoints of our interval, which are x = 0 (all wire for the circle) and x = 60 (all wire for the square).

If x = 0, then:

A_total(0) = (0^2 / 16) + ((60 - 0)^2 / (4π))

A_total(0) = 0 + (3600 / (4π))

A_total(0) ≈ 3600 / (4 * 3.14159) ≈ 286.479

If x = 60, then:

A_total(60) = (60^2 / 16) + ((60 - 60)^2 / (4π))

A_total(60) = (3600 / 16) + 0

A_total(60) = 225

Evaluate A_total at the Critical Point:

Now, let's evaluate A_total at our critical point x ≈ 33.605:

A_total(33.605) = ((33.605)^2 / 16) + ((60 - 33.605)^2 / (4π))

A_total(33.605) ≈ (1129.296 / 16) + ((26.395)^2 / (4π))

A_total(33.605) ≈ 70.581 + (696.706 / (4 * 3.14159))

A_total(33.605) ≈ 70.581 + 55.444

A_total(33.605) ≈ 126.025

Conclusion for Minimum Area:

Comparing the values, we find that the minimum total area occurs when x ≈ 33.605 inches, with a total area of approximately 126.025 square inches. This means to minimize the total area, you should use about 33.605 inches of the wire for the square and the remaining 26.395 inches for the circle.

Maximizing the Area

Now, let's consider maximizing the area. Looking at our endpoint values, we see that A_total(0) ≈ 286.479 square inches and A_total(60) = 225 square inches. The maximum area occurs when x = 0, meaning all the wire is used to form the circle.

In summary:

To minimize the total area, use approximately 33.605 inches for the square and 26.395 inches for the circle. The minimum area is approximately 126.025 square inches.
To maximize the total area, use all 60 inches of wire for the circle, resulting in an area of approximately 286.479 square inches.

Final Thoughts

This problem beautifully illustrates how math can be applied to real-world scenarios. By understanding the formulas for area and using calculus to optimize, we were able to determine how to cut a wire to achieve either the smallest or largest possible combined area of a square and a circle. It's all about finding the right balance and using the tools of mathematics to solve interesting problems. Keep exploring, guys, and have fun with math!