Problem 17 - The Calculus with Analytic Geometry

Question

A piece of wire \(10 \mathrm{ft}\) long is cut into two pieces. One piece is bent into the shape of a circle and the other into the shape of a square. How should the wire be cut so that (a) the combined area of the two figures is as small as possible and (b) the combined area of the two figures is as large as possible?

Step-by-Step Solution

Answer

(a) Wire should be cut to maximize the length for the square. (b) Wire should be cut to maximize the length for the circle.

Step 1 - Define Variables

Let the length of the wire used for the circle be denoted as \(x\) feet. Thus, the length of the wire used for the square will be \(10 - x\) feet.

Step 2 - Express the Circumference

The circumference of the circle is equal to its wire length, so \(2\pi r = x\), where \(r\) is the radius of the circle.

Step 3 - Find the Radius

From \(2\pi r = x\), solve for \(r\): \(r = \frac{x}{2\pi}\).

Step 4 - Find the Area of the Circle

The area of the circle is given by \(A_{\text{circle}} = \pi r^2\). Substitute \(r = \frac{x}{2\pi}\) into the formula to get: \(A_{\text{circle}} = \pi \left( \frac{x}{2\pi} \right)^2 = \frac{x^2}{4\pi}\).

Step 5 - Express the Perimeter of the Square

The perimeter of the square equals the wire length used, which is \(10 - x\).

Step 6 - Find the Side Length of the Square

Each side of the square is \(\frac{10 - x}{4}\) feet.

Step 7 - Find the Area of the Square

The area of the square is given by \(A_{\text{square}} = \left( \frac{10 - x}{4} \right)^2 = \frac{(10 - x)^2}{16}\).

Step 8 - Combine the Areas

The combined area of the circle and the square is \(A_{\text{total}} = \frac{x^2}{4\pi} + \frac{(10 - x)^2}{16}\).

Step 9 - Differentiate the Combined Area

To find the extreme values, differentiate \(A_{\text{total}}\) with respect to \(x\): \(\frac{dA_{\text{total}}}{dx} = \frac{d}{dx} \left( \frac{x^2}{4\pi} + \frac{(10 - x)^2}{16} \right)\).

Step 10 - Set Derivative to Zero

Solve for critical points by setting \(\frac{dA_{\text{total}}}{dx} = 0\).

Step 11 - Solve for Minimum Area

To minimize \(A_{\text{total}}\), solve \(\frac{dA_{\text{total}}}{dx} = 0\), and confirm the nature using the second derivative test.

Step 12 - Solve for Maximum Area

To maximize \(A_{\text{total}}\), check relative values and boundary points (0 and 10).

Key Concepts

Circle Area

The area of a circle is one of the fundamental concepts in basic geometry and calculus optimization problems. To find it, you need to know the radius of the circle. When a piece of wire is bent into a circle, its total length forms the circumference of the circle, which can be mathematically expressed as:
\(2\pi r = x\)
Here, \(x\) represents the length of the wire. By solving for \(r\), we get: \(r = \frac{x}{2\pi}\).
Once you have the radius, the area of the circle can be calculated using the formula:
\[ A_{\text{circle}} = \pi r^2 \]
Substituting the expression for \(r\) into the formula for the area gives:
\[ A_{\text{circle}} = \pi \left( \frac{x}{2\pi} \right)^2 = \frac{x^2}{4\pi} \].
This process allows us to express the area of the circle solely in terms of the wire length \(x\).

Square Area

The area of a square is another essential concept and is slightly different from the area of a circle. When a piece of wire is bent into a square shape, its total length forms the perimeter of the square. This relationship is given by:
\(4s = 10 - x\).
Here, \(s\) represents the side length of the square, and \(10 - x\) is the wire length used for the square. By solving for \(s\), we get: \[ s = \frac{10 - x}{4} \]
Once you have the side length, calculating the area of the square is straightforward. The area can be expressed as:
\[ A_{\text{square}} = s^2 = \left( \frac{10 - x}{4} \right)^2 = \frac{(10 - x)^2}{16} \].
This formula expresses the area solely in terms of the wire length \(10 - x\), allowing us to evaluate how changes in \(x\) and \(10 - x\) affect the total area.

Differentiation

Differentiation is a core concept in calculus and is used to find the rates at which quantities change. In this optimization problem, differentiation helps us find the minimum and maximum combined areas of the circle and square. We start with the total area expression:
\[A_{\text{total}} = \frac{x^2}{4\pi} + \frac{(10 - x)^2}{16} \]
To find the extreme values of \(A_{\text{total}}\), we need to differentiate it with respect to \(x\):
\[ \frac{dA_{\text{total}}}{dx} = \frac{d}{dx} \left( \frac{x^2}{4\pi} + \frac{(10 - x)^2}{16} \right) \]
Setting the first derivative to zero gives the conditions for critical points:
\[ \frac{dA_{\text{total}}}{dx} = 0 \]
Solving this equation provides the value of \(x\) that either minimizes or maximizes the combined area. We use the second derivative test or boundary points to confirm the nature (minimum or maximum) of these critical points. In this problem, applying these techniques allows us to determine the optimal wire length to use for the circle and the square, ensuring the smallest or largest possible combined area.