Problem 1 - The Calculus with Analytic Geometry

Question

Find the area of the largest rectangle having a perimeter of \(200 \mathrm{ft}\).

Step-by-Step Solution

Answer

The area of the largest rectangle is 2500 square feet.

Step 1 - Define Variables

Let the length of the rectangle be denoted by \(l\) and the width be denoted by \(w\).

Step 2 - Establish the Perimeter Formula

Recall that the perimeter \(P \) of a rectangle is given by: \[ P = 2l + 2w \. \] Given that the perimeter is \(200 \) ft, we can set up the equation: \[ 2l + 2w = 200. \]

Step 3 - Simplify the Perimeter Equation

Simplify the equation by dividing both sides by 2: \[ l + w = 100. \] This means the sum of the length and the width is \(100 \) ft.

Step 4 - Express Width in Terms of Length

Solve for \(w\) in terms of \(l\) by rearranging the equation: \[ w = 100 - l. \]

Step 5 - Write the Area Formula

The area \(A \) of a rectangle is found by multiplying the length and the width: \[ A = l \times w. \]

Step 6 - Substitute for Width

Substitute the expression for \(w\) from Step 4 into the area formula: \[ A = l \times \(100 - l \). \]

Step 7 - Formulate the Area Function

Write the area function \(A\) as a quadratic equation: \[ A = 100l - l^2. \]

Step 8 - Find the Maximum Area

To find the maximum area of a quadratic function \(A = -l^2 + 100l\), locate the vertex. The vertex of a parabola \(ax^2 + bx + c\) occurs at \[ l = -\frac{b}{2a}. \] Here, \(a = -1\) and \(b = 100\), so: \[ l = -\frac{100}{2(-1)} = 50. \]

Step 9 - Calculate Corresponding Width

Substitute \(l = 50 \) back into the equation for \(w\): \[ w = 100 - 50 = 50. \]

Step 10 - Calculate the Maximum Area

Finally, calculate the maximum area using length \(50 \) ft and width \(50 \) ft: \[ A = 50 \times 50 = 2500 \text{ square feet}. \]

Key Concepts

Perimeter

The perimeter of a shape is the total length around it. For a rectangle, the perimeter is the sum of all its sides.
Think of the perimeter as if you are putting a fence around a garden.
To find the perimeter of a rectangle, you use the formula:
\textbf{1. Perimeter Formula for Rectangle:}
\[ P = 2l + 2w \] where \(l\) is the length and \(w\) is the width.

Given a perimeter of 200 feet, the equation is:
\[ 2l + 2w = 200 \]
Simplifying this by dividing by 2 gives:
\[ l + w = 100 \]
So, the sum of the length and width of the rectangle is 100 feet.
This formula helps us relate the length and width to the perimeter of the rectangle.

Quadratic Functions

A quadratic function is an algebraic expression of the form:
\[ f(x) = ax^2 + bx + c \]
In this problem, we are dealing with the area of a rectangle, which can be represented as a quadratic function.
Using the relationships we've established:
1. Length plus width equals 100: \( l + w = 100 \)
2. Width in terms of length: \( w = 100 - l \)
3. Area formula of a rectangle: \(A = l \times w \)
Substituting for \(w\), the area becomes:
\[ A = l \times (100 - l) \]
Simplifying this gives us a quadratic equation:
\[ A = 100l - l^2 \]
In this form, the area \( A \) is a quadratic function of the length \( l \). It shows how the area changes as the length varies.

Vertex of a Parabola

The vertex of a parabola is the highest or lowest point of the graph of a quadratic function. For functions that open downwards, like our area function, the vertex represents the maximum value.
The quadratic equation for area is:
\[ A = 100l - l^2 \]
In general, the vertex of a quadratic function \( ax^2 + bx + c \) occurs at:
\[ x = -\frac{b}{2a} \]
Here, \( a = -1 \) and \( b = 100 \). Substituting these values in gives:
\[ l = -\frac{100}{2(-1)} = 50 \]
So, the length \( l \) at the vertex is 50 feet. This value of the length gives the maximum area of the rectangle.

Algebraic Expressions

An algebraic expression is a combination of variables, numbers, and operations. In this problem, we used algebraic expressions to solve for the maximum area.
Starting with the perimeter equation:
\[ 2l + 2w = 200 \]
We simplified and expressed width (\