Problem 1 - Calculus: Early Transcendentals

Question

Find the dimensions of the rectangle of largest area having fixed perimeter \(100 .\)

Step-by-Step Solution

Answer

The dimensions of the rectangle are 25 by 25, forming a square.

Step 1: Understand the Problem

We need to find the dimensions of a rectangle that will maximize the area while keeping the perimeter fixed at 100.

Step 2: Define Variables

Let's denote the length of the rectangle as \( l \) and the width as \( w \). The perimeter of the rectangle is given by the formula \( P = 2l + 2w \).

Step 3: Apply Perimeter Condition

Since the perimeter is fixed at 100, we have the equation \( 2l + 2w = 100 \). Simplifying, we get \( l + w = 50 \).

Step 4: Express Width in Terms of Length

From the equation \( l + w = 50 \), solve for \( w \), giving \( w = 50 - l \).

Step 5: Write the Area Function

The area \( A \) of the rectangle is given by \( A = l \times w \). Substitute \( w = 50 - l \) into the area function: \( A = l(50 - l) = 50l - l^2 \).

Step 6: Find the Maximum Area

To maximize the area, we need to find the critical points of the function \( A = 50l - l^2 \). This quadratic function can be rewritten as \( A = -l^2 + 50l \).

Step 7: Compute the Derivative and Critical Points

Find the derivative of \( A \), \( A' = -2l + 50 \). Set \( A' = 0 \) to find critical points: \( -2l + 50 = 0 \). Solving, we get \( l = 25 \).

Step 8: Determine the Corresponding Width

Substitute \( l = 25 \) back into the equation \( w = 50 - l \), to find \( w = 25 \).

Step 9: Verify the Maximum

Since the function \( A = -l^2 + 50l \) opens downward (coefficient of \( l^2 \) is negative), the critical point \( l = 25 \) is indeed a maximum. The corresponding width is also 25.

Key Concepts

Rectangle area maximization

Rectangle area maximization is a classic optimization problem in calculus where you aim to find dimensions that will give the largest possible area. Imagine you have a rectangle and you can adjust its length and width, but the total distance around the rectangle—called the perimeter—must stay the same.
In this exercise, the perimeter of our rectangle is fixed at 100. When maximizing the area, you want to explore different combinations of length and width that still comply with this perimeter constraint.
Ultimately, the goal is to rearrange the rectangle's sides in such a way that the area is at its maximum possible value, given that perimeter limit.

Quadratic functions

Quadratic functions are a type of polynomial function where the highest degree of the variable is squared. They are often written in the form:

• \[f(x) = ax^2 + bx + c\]

In our scenario, after setting and rearranging our area function for the rectangle, it becomes a quadratic function:

• \[A = -l^2 + 50l\]

Quadratic functions usually graph as a parabola. The key characteristic here is the 'a' value, which determines whether the parabola opens upward or downward. For this problem, the 'a' value is negative, indicating the parabola opens downwards.
This is important because in such scenarios, the vertex of the parabola represents the maximum point. Here, finding this vertex helps in determining the maximum area.

Perimeter constraint

The perimeter constraint is essential in optimization problems. It sets a boundary or a limit to what can be adjusted within the problem. For a rectangle, the perimeter is calculated by summing up all the sides:

• \[P = 2l + 2w\]

In this exercise, the perimeter is fixed at 100, leading to the equation:

• \[l + w = 50\]

This tells us that for every possible change in length, there has to be a compensating change in width to keep the perimeter constant. This constraint is the key that allows us to transform the problem into a solvable equation, making it easier to explore changes in dimensions.

Critical points and derivatives

Critical points are crucial in identifying where a function reaches its highest or lowest values. These are found by taking the derivative of the function and setting it to zero. In the context of our rectangle problem, the area function's critical points will tell us when the area is maximized.
The derivative of the area function

• \[A' = -2l + 50\]

is calculated by differentiating the function

• \[A = -l^2 + 50l\]

Setting

• \[A' = 0\]

and solving for \( l \), we find the critical point where \( l = 25 \). This critical point is where the area is maximized because the parabola from our quadratic function opens downward. Ensuring this maximum involves confirming that our critical point leads to the maximum possible area, aligning with the vertex of the parabola.