Deriving the Area Function

Oct 13, 2025 by wbrum.org 96 views

Introduction

In mathematics, optimization problems involve finding the maximum or minimum value of a function subject to certain constraints. In this article, we will explore the problem of maximizing the area of a rectangle formed by bending a piece of wire 48 cm in length. We will first show that the area of the rectangle can be expressed as a function of its width, and then calculate the maximum area that can be formed.

The Problem

A piece of wire, 48 cm in length, is bent in the shape of a rectangle. If the width of the rectangle is $x$ cm, we need to show that the area $A$ of the rectangle can be expressed as:

A = x(24 - x)

We also need to calculate the maximum area $A$ that can be formed.

Deriving the Area Function

To derive the area function, we need to consider the relationship between the length and width of the rectangle. Since the wire is bent in the shape of a rectangle, the perimeter of the rectangle is equal to the length of the wire, which is 48 cm. The perimeter of a rectangle is given by the formula:

P = 2l + 2w

where $l$ is the length and $w$ is the width of the rectangle. Since the perimeter is equal to 48 cm, we can write:

48 = 2l + 2w

We are given that the width of the rectangle is $x$ cm, so we can substitute $x$ for $w$ :

48 = 2l + 2x

Now, we need to express the length $l$ in terms of $x$ . Since the perimeter is equal to 48 cm, we can rearrange the equation to get:

2l = 48 - 2x

Dividing both sides by 2, we get:

l = 24 - x

Now, we can substitute this expression for $l$ into the formula for the area of a rectangle:

A = lw

Substituting $l = 24 - x$ and $w = x$ , we get:

A = x(24 - x)

Calculating the Maximum Area

To calculate the maximum area, we need to find the value of $x$ that maximizes the area function. We can do this by taking the derivative of the area function with respect to $x$ and setting it equal to zero:

\frac{dA}{dx} = 24 - 2x = 0

Solving for $x$ , we get:

x = 12

Now, we can substitute this value of $x$ into the area function to get:

A = x(24 - x)

A = 12(24 - 12)

A = 12 \times 12

A = 144

Therefore, the maximum area that can be formed is 144 square cm.

Conclusion

In this article, we have shown that the area of a rectangle formed by bending a piece of wire 48 cm in length can be expressed as a function of its width. We have also calculated the maximum area that can be formed, which is 144 square cm. This problem is a classic example of an optimization problem, where we need to find the maximum or minimum value of a function subject to certain constraints.

Mathematical Derivations

Derivation of the Area Function

P = 2l + 2w

where $l$ is the length and $w$ is the width of the rectangle. Since the perimeter is equal to 48 cm, we can write:

48 = 2l + 2w

We are given that the width of the rectangle is $x$ cm, so we can substitute $x$ for $w$ :

48 = 2l + 2x

Now, we need to express the length $l$ in terms of $x$ . Since the perimeter is equal to 48 cm, we can rearrange the equation to get:

2l = 48 - 2x

Dividing both sides by 2, we get:

l = 24 - x

Now, we can substitute this expression for $l$ into the formula for the area of a rectangle:

A = lw

Substituting $l = 24 - x$ and $w = x$ , we get:

A = x(24 - x)

Derivation of the Maximum Area

\frac{dA}{dx} = 24 - 2x = 0

Solving for $x$ , we get:

x = 12

Now, we can substitute this value of $x$ into the area function to get:

A = x(24 - x)

A = 12(24 - 12)

A = 12 \times 12

A = 144

Therefore, the maximum area that can be formed is 144 square cm.

Mathematical Formulas

Perimeter of a Rectangle

The perimeter of a rectangle is given by the formula:

P = 2l + 2w

where $l$ is the length and $w$ is the width of the rectangle.

Area of a Rectangle

The area of a rectangle is given by the formula:

A = lw

where $l$ is the length and $w$ is the width of the rectangle.

Derivative of the Area Function

The derivative of the area function with respect to $x$ is given by:

\frac{dA}{dx} = 24 - 2x

Mathematical Theorems

Theorem 1: Optimization Problem

If a function $f(x)$ is continuous and differentiable on an interval $[a, b]$ , and if $f'(x) = 0$ at some point $x = c$ in the interval, then $f(x)$ has a maximum or minimum value at $x = c$ .

Theorem 2: Derivative of a Product

If $f(x)$ and $g(x)$ are two functions that are differentiable at a point $x = a$ , then the derivative of their product is given by:

\frac{d(f(x)g(x))}{dx} = f'(x)g(x) + f(x)g'(x)

Mathematical Proofs

Proof of Theorem 1

Let $f(x)$ be a function that is continuous and differentiable on an interval $[a, b]$ . Suppose that $f'(x) = 0$ at some point $x = c$ in the interval. We need to show that $f(x)$ has a maximum or minimum value at $x = c$ .

Since $f(x)$ is continuous and differentiable on the interval $[a, b]$ , it is also continuous on the closed interval $[a, c]$ and $[c, b]$ . By the Extreme Value Theorem, $f(x)$ has a maximum and minimum value on each of these intervals.

Let $M$ be the maximum value of $f(x)$ on the interval $[a, c]$ , and let $m$ be the minimum value of $f(x)$ on the interval $[c, b]$ . Since $f(x)$ is continuous on the closed interval $[a, b]$ , it is also continuous on the closed interval $[a, c] \cup [c, b]$ .

By the Intermediate Value Theorem, there exists a point $x = d$ in the interval $[a, c] \cup [c, b]$ such that $f(d) = M$ . Since $f(x)$ is continuous on the closed interval $[a, c] \cup [c, b]$ , it is also continuous on the closed interval $[a, d] \cup [d, b]$ .

By the Extreme Value Theorem, $f(x)$ has a maximum and minimum value on each of these intervals. Let $M'$ be the maximum value of $f(x)$ on the interval $[a, d]$ , and let $m'$ be the minimum value of $f(x)$ on the interval $[d, b]$ .

Since $f(x)$ is continuous on the closed interval $[a, d] \cup [d, b]$ , it is also continuous on the closed interval $[a, c] \cup [c, b]$ . By the Intermediate Value Theorem, there exists a point $x = e$ in the interval $[a, c] \cup [c, b]$ such that $f(e) = M'$ .

Q: What is the problem of optimizing the area of a rectangle?

A: The problem of optimizing the area of a rectangle involves finding the maximum or minimum value of the area of a rectangle subject to certain constraints. In this case, we are given a piece of wire 48 cm in length, and we need to bend it into the shape of a rectangle with a width of $x$ cm.

Q: How do we derive the area function for the rectangle?

A: To derive the area function, we need to consider the relationship between the length and width of the rectangle. Since the wire is bent in the shape of a rectangle, the perimeter of the rectangle is equal to the length of the wire, which is 48 cm. We can use the formula for the perimeter of a rectangle to express the length in terms of the width.

Q: What is the formula for the perimeter of a rectangle?

A: The formula for the perimeter of a rectangle is:

P = 2l + 2w

where $l$ is the length and $w$ is the width of the rectangle.

Q: How do we express the length in terms of the width?

A: Since the perimeter is equal to 48 cm, we can write:

48 = 2l + 2x

We can then rearrange the equation to get:

2l = 48 - 2x

Dividing both sides by 2, we get:

l = 24 - x

Q: What is the area function for the rectangle?

A: Now that we have expressed the length in terms of the width, we can substitute this expression into the formula for the area of a rectangle:

A = lw

Substituting $l = 24 - x$ and $w = x$ , we get:

A = x(24 - x)

Q: How do we calculate the maximum area of the rectangle?

A: To calculate the maximum area, we need to find the value of $x$ that maximizes the area function. We can do this by taking the derivative of the area function with respect to $x$ and setting it equal to zero:

\frac{dA}{dx} = 24 - 2x = 0

Solving for $x$ , we get:

x = 12

Now, we can substitute this value of $x$ into the area function to get:

A = x(24 - x)

A = 12(24 - 12)

A = 12 \times 12

A = 144

Therefore, the maximum area that can be formed is 144 square cm.

Q: What is the significance of this problem?

A: This problem is a classic example of an optimization problem, where we need to find the maximum or minimum value of a function subject to certain constraints. It is a fundamental concept in mathematics and has many real-world applications, such as designing buildings, bridges, and other structures.

Q: How can we apply this concept to real-world problems?

A: We can apply this concept to real-world problems by using optimization techniques to find the maximum or minimum value of a function subject to certain constraints. For example, we can use this concept to design buildings, bridges, and other structures that are safe and efficient.

Q: What are some common applications of optimization techniques?

A: Some common applications of optimization techniques include:

Designing buildings, bridges, and other structures
Optimizing production processes in manufacturing
Optimizing supply chains and logistics
Optimizing financial portfolios and investments
Optimizing energy consumption and efficiency

Q: What are some common challenges in optimization problems?

A: Some common challenges in optimization problems include:

Finding the optimal solution
Dealing with constraints and limitations
Handling non-linear and non-convex problems
Dealing with uncertainty and risk
Balancing competing objectives and priorities

Q: How can we overcome these challenges?

A: We can overcome these challenges by using advanced optimization techniques, such as:

Linear and non-linear programming
Dynamic programming
Stochastic programming
Robust optimization
Multi-objective optimization

We can also use advanced software tools and algorithms, such as:

Optimization software packages
Machine learning and artificial intelligence algorithms
Simulation and modeling tools

By using these techniques and tools, we can overcome the challenges in optimization problems and find the optimal solution.