Problem 14 - Calculus Early Transcendentals

Question

A wire of length 12 in can be bent into a circle, bent into a square, or cut into two pieces to make both a circle and a square. How much wire should be used for the circle if the total area enclosed by the figure(s) is to be (a) a maximum \(\quad\) (b) a minimum?

Step-by-Step Solution

Answer

Use all wire for a circle for max area; all for a square for min area.

Step 1: Understand the Problem

We have a wire of length 12 inches which can be bent into a circle, a square, or divided to form both. Our goal is to maximize or minimize the total area enclosed.

Step 2: Define the Variables

Let \( x \) be the length of wire used for the circle, meaning \( 12 - x \) will be used for the square.

Step 3: Write the Formulas for Areas

For a circle, the wire used is equivalent to the circumference, so \( x = 2\pi r \), giving area \[ A_{circle} = \pi r^2 = \frac{x^2}{4\pi} \]. For a square with side \( s = \frac{12-x}{4} \), its area is \[ A_{square} = \left(\frac{12-x}{4}\right)^2 \].

Step 4: Establish the Total Area Function

The total area function \( A(x) \) combining both circle and square is: \[A(x) = \frac{x^2}{4\pi} + \left( \frac{12-x}{4} \right)^2.\]

Step 5: Differentiate the Total Area Function

Differentiate \( A(x) \) with respect to \( x \) to find the critical points: \[A'(x) = \frac{x}{2\pi} - \frac{12-x}{8}.\] Set \( A'(x) = 0 \) to find critical points.

Step 6: Solve for Critical Points

Setting \( A'(x) = 0 \) gives: \[\frac{x}{2\pi} = \frac{12-x}{8},\] solving this equation gives the critical point \( x = \frac{48\pi}{8 + 2\pi} \).

Step 7: Evaluate the Endpoints

Consider the endpoints: when \( x = 0 \), all wire for the square; when \( x = 12 \), all wire for the circle.

Step 8: Determine Maximum and Minimum Area

Calculate \( A(0) \), \( A(12) \), and \( A\left(\frac{48\pi}{8 + 2\pi}\right) \) to determine where the maximum and minimum total areas occur.

Key Concepts

Calculus

Calculus is a powerful mathematical tool that helps us understand changes and how different quantities relate to each other. It is typically divided into two main branches: differential calculus and integral calculus. Differential calculus focuses on the concept of the derivative, which measures how a function changes as its input changes. It is like finding the slope at a specific point on a curve. By understanding these rates of change, we can solve problems related to motion, growth, and optimization.

Integral calculus, on the other hand, deals with accumulation of quantities and the area under curves. This branch helps us determine quantities where there is a continuous flow or coverage, like calculating distance traveled or the volume under a surface. Both branches together give us the full picture of how variables interact in mathematical models.

In our optimization problem, we primarily use differential calculus to find the derivative of the area function. This helps us determine the critical points where the total enclosed area might reach its maximum or minimum.

Maximum and Minimum Problems

Maximum and minimum problems are all about finding the largest or smallest value that a function can take, given certain constraints. In calculus, solving these problems usually involves finding the critical points of a function using its derivative.

For our exercise, we need to maximize or minimize the total area enclosed by a circle and a square formed by wire. This involves:

• Finding an expression for the total area as a function of wire length used.
• Calculating the derivative to determine where the area function may peak or dip.

These problems have everyday applications, such as optimizing manufacturing efficiency or minimizing material costs. The basic steps always include creating a sensible mathematical model, differentiating the function, then using critical points and endpoints to ensure we have found the true maximum or minimum.

Critical Points

Critical points are specific values of a variable where the function's derivative is zero or undefined. This is important because critical points are potential locations for local maxima or minima of a function.

In the context of our problem, we derive the total area function and find critical points by setting its derivative equal to zero. It helps locate points where changes in the wire allocation can result in maximum or minimum areas. Generally:

• Take the derivative of your total area function, \( A(x) \).
• Set the derivative equal to zero, \( A'(x) = 0 \), to find critical points.
• Solve for the variable, in this case, wire length for the circle, \( x \).

After finding the critical points, we compare the function's value at these points and any endpoints, ensuring the solution aligns with the conditions we are exploring.

Geometry in Calculus

Geometry plays a crucial role in calculus, especially when dealing with optimization problems involving shapes and areas. Calculus provides tools to analyze geometrical figures, enabling us to determine properties like area, volume, and lengths effectively.

In our exercise, we use geometric formulas to express the areas of a circle and a square. These geometric concepts and their relations help frame the problem in a mathematical context, making use of:

• The formula for the circumference of a circle, used to resolve wire length into the radius.
• The area relation of a circle, \( A_{circle} = \pi r^2 \).
• The square's side length derived from remaining wire, utilized in calculating its area, \( A_{square}. \)

By integrating geometry with calculus, we efficiently set up a mathematical model that simplifies the process of finding optimal solutions for real-world problems.