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Chapter 3: Problem 44

GENERAL: Area What is the area of the largest rectangle whose perimeter is 100 feet?

Short Answer

Expert verified
The maximum area is 625 square feet, occurring when the rectangle is a square with sides of 25 feet.

Step by step solution

01

Understanding the Problem

We need to find the dimensions of a rectangle with the largest possible area, given that its perimeter should be exactly 100 feet. The perimeter, \(P\), of a rectangle is calculated as \(P = 2l + 2w\), where \(l\) is the length and \(w\) is the width.
02

Calculate Perimeter in Terms of Variables

Given the perimeter \(P = 100\), we substitute into the formula: \[ 2l + 2w = 100 \]Simplifying, we find:\[ l + w = 50 \] This equation will be used to express one variable in terms of the other.
03

Express Width in Terms of Length

From the equation \(l + w = 50\), solve for width \(w\):\[ w = 50 - l \]We now have \(w\) in terms of \(l\).
04

Formulate the Area Equation

The area \(A\) of a rectangle is given by \(A = l imes w\). Substitute \(w = 50 - l\) into this equation:\[ A = l(50 - l) = 50l - l^2 \]This formula represents the area of the rectangle in terms of \(l\).
05

Find the Length for Maximum Area

To maximize \(A = 50l - l^2\), calculate the derivative \(\frac{dA}{dl} = 50 - 2l\). Set the derivative to zero to find critical points:\[ 50 - 2l = 0 \]\[ 2l = 50 \]\[ l = 25 \]
06

Determine the Width Using the Length

Substituting \(l = 25\) into the perimeter equation \(l + w = 50\), we find:\[ 25 + w = 50 \]\[ w = 25 \]This means our rectangle is a square with side lengths of 25 feet.
07

Confirm the Solution by Calculating the Area

Substitute \(l = 25\) and \(w = 25\) back into the area formula \(A = l imes w\):\[ A = 25 imes 25 = 625 \] square feet. This confirms the maximum area.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding the Area of a Rectangle
When talking about the **area** of a rectangle, we鈥檙e referring to the amount of space enclosed within its four sides. The formula for the area is simple:
  • Area \(A\) = Length \(l\) \( \times \) Width \(w\).
This means you multiply the rectangle鈥檚 length by its width to find out how much space it covers.
A key part of this exercise involves finding the largest possible area of a rectangle with a fixed perimeter.
  • The concept of area helps us understand how to best utilize limited boundaries to enclose a maximum space.
Exploring Perimeter Constraints
Perimeter constraints define the total length around the border of a rectangle. This is represented by:
  • Perimeter \(P = 2l + 2w\).
For the problem at hand, the perimeter is fixed at 100 feet. This constraint helps us relate the width and length of the rectangle.Simplified, we have:
  • \( l + w = 50 \)
This equation shows the total semi-perimeter, guiding us in rewriting one variable in terms of another, which simplifies our calculations.
Understanding perimeter constraints is crucial in setting up equations needed to derive maximum area from a fixed boundary.
Using Derivatives in Calculus
Derivatives are powerful tools in calculus used to find the rate of change of a quantity. In this exercise, we apply derivatives to identify where a rectangle's area reaches its maximum value.
First, you express the area equation based on the given perimeter:
  • \( A = 50l - l^2 \)
Then, compute the derivative of the area function with respect to \(l\):
  • \( \frac{dA}{dl} = 50 - 2l \)
Setting the derivative equal to zero helps find critical points:
  • \( 50 - 2l = 0 \)
  • This gives \( l = 25 \)
Identifying critical points through derivatives helps reveal where the function鈥檚 maximum or minimum values are.
The Optimization Problem
An optimization problem involves finding the best solution from a set of available options. In this scenario, it鈥檚 about maximizing the area of a rectangle given a perimeter constraint. By using calculus, specifically derivatives, we factored in all constraints to find optimal dimensions.
For the rectangle:
  • When \( l = 25 \), from the perimeter equation \( l + w = 50 \), we also get \( w = 25 \).
  • This configuration results in a square, as a square often provides the maximum area for a given perimeter.
Calculating the area confirms:
  • \( A = 25 \times 25 = 625 \) square feet.
Optimizing problems involve strategically using mathematical tools to derive the best possible outcome, as shown in finding the largest area effectively.

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