Problem 43 - Calculus Single Variable

Question

A wire of length \(L\) can be shaped into a circle or a square, or it can be cut into two pieces, one of which is formed into a circle and the other a square. How is the minimum total area obtained?

Step-by-Step Solution

Answer

Cutting the wire and allocating part of it to each shape minimizes the total area.

Step 1: Define the Problem

We are tasked with finding the configuration of a wire when shaped into either a circle, a square, or both, that minimizes the total area of the resulting shapes. The wire has a fixed length of \(L\).

Step 2: Express Areas in Terms of Length

For a circle with circumference \(C\) equal to the length used, the radius \(r\) is \(\frac{C}{2 \pi}\) and the area \(A_c\) is \(\pi \left(\frac{C}{2\pi}\right)^2\). For the square with side \(a\) equal to \(\frac{L}{4}\), the area \(A_s\) is \(\left(\frac{L}{4}\right)^2\).

Step 3: Consider Combined Configurations

If the wire is cut into two parts, one forming a circle and the other a square, assign lengths \(x\) and \(L-x\) to the circle and square respectively. The area of the circle \(A_c\) is \(\pi \left(\frac{x}{2\pi}\right)^2\) and of the square \(A_s\) is \(\left(\frac{L-x}{4}\right)^2\).

Step 4: Write Total Area Expression

The total area \(A\) combining both shapes is given by the equation: \[A = \pi \left(\frac{x}{2\pi}\right)^2 + \left(\frac{L-x}{4}\right)^2\].

Step 5: Find Critical Points

To minimize the total area, take the derivative \(\frac{dA}{dx}\), set it to zero, and solve for \(x\). Then, compute \(\frac{d^2A}{dx^2}\) to confirm a minimum at the critical points.

Step 6: Evaluate Special Cases

Check the areas when the wire is entirely a circle (\(x=L\)), entirely a square (\(x=0\)), and when it is cut for both shapes to find the configuration that yields the minimum area.

Key Concepts

Area Minimization

Minimizing the total area formed by geometric shapes is a classic problem in calculus optimization. Consider a wire of fixed length, and imagine shaping it into various forms like a circle, square, or both. The goal is to find the configuration that results in the smallest total area.
This problem is not just theoretical; it has practical applications. For instance, designing structures or objects with minimal material usage can save costs and resources. In our problem, we set up mathematical expressions for the area of each shape formed by parts of the wire.

• Circle: The area depends on the radius, which is linked to the length of wire allocated to form it.
• Square: The area is determined by the side length, which is one-fourth of the wire's length if entirely used.

To solve the problem effectively, we must express the areas in terms of the wire's total length. The next step involves calculus to find configurations that minimize this combined area.

Derivatives

Derivatives are pivotal in calculus for analyzing how a function changes. They help identify points where a function might reach its minimum or maximum. In our area minimization problem, derivatives guide us in finding the optimal point for splitting the wire into a circle and square.
When we set up the expression for total area, denoted by \(A\), as a function of the wire divided into two portions forming a circle and a square, derivatives become essential. The first derivative \(\frac{dA}{dx}\) tells us how the area changes as the wire distribution changes between the two shapes.

• Finding where \(\frac{dA}{dx} = 0\) helps us locate potential points of area minimization, referred to as critical points.
• The second derivative \(\frac{d^2A}{dx^2}\) confirms whether these critical points yield a minimum or maximum area.

Taking derivatives and solving equations allows us to significantly cut down on possibilities and focus on configurations that truly minimize the area.

Critical Points

Critical points are vital in determining where a function, like our area expression, reaches extreme values. These points occur where the first derivative of the function equals zero or is undefined.
In the context of our problem, critical points guide us to configurations where the wire's distribution into a circle and a square either minimizes or maximizes the total area.

• First, compute \(\frac{dA}{dx}\) and find values of \(x\) that solve \(\frac{dA}{dx} = 0\). These values are the critical points.
• Second, apply the second derivative test using \(\frac{d^2A}{dx^2}\) to determine if these points yield a minimum area. If the second derivative is positive at a critical point, it is a minimum.

By analyzing critical points, we identify optimal wire divisions that result in the smallest possible total area.

Geometric Shapes

Geometric shapes play a central role in calculus optimization through their unique properties and formulas. In this scenario, we use circles and squares. Each shape’s area calculation depends directly on the configuration of the wire.
For a circle formed by the wire, the key elements are:

• The radius is derived from the wire length dedicated to the circle.
• The area is computed using the radius in the formula \(A = \pi r^2\).

For a square made from the wire:

• The side of the square is formed by dividing the available wire length by four if all is used for a square.
• The area is then \(A = a^2\), where \(a\) is the side length.

Given this, we can manipulate the formulas to find how different wire lengths impact the total combined area, leading us to solutions that minimize the space covered by these shapes.