Problem 26 - Applied Calculus

Question

The perimeter of a rectangle is \(64 \mathrm{cm} .\) Find the lengths of the sides of the rectangle giving the maximum area.

Step-by-Step Solution

Answer

The rectangle is a square with sides of 16 cm.

Step 1: Understand the Problem

We need to find the dimensions of a rectangle with a fixed perimeter that gives the maximum area. In this case, the perimeter is given as 64 cm.

Step 2: Define Variables

Let the length of the rectangle be \( l \) and the width be \( w \). The perimeter formula is given by \( 2l + 2w = 64 \). We want to maximize the area, which is given by \( A = lw \).

Step 3: Simplify the Constraints

Since \( 2l + 2w = 64 \), divide the whole equation by 2 to get \( l + w = 32 \). This is our constraint for the dimensions of the rectangle.

Step 4: Express One Variable in Terms of the Other

From the constraint \( l + w = 32 \), express \( w \) in terms of \( l \): \( w = 32 - l \).

Step 5: Write the Area in Terms of One Variable

Substitute \( w = 32 - l \) into the area formula \( A = lw \) to get \( A = l(32 - l) = 32l - l^2 \).

Step 6: Find the Maximum Area Using Calculus

To find the maximum area, take the derivative of \( A \) with respect to \( l \): \( A'(l) = 32 - 2l \). Set the derivative equal to zero and solve: \( 32 - 2l = 0 \) gives \( l = 16 \).

Step 7: Verify the Maximum Area

Plug \( l = 16 \) back into the expression for \( w \): \( w = 32 - 16 = 16 \). Thus, both the length and the width of the rectangle are 16 cm. To verify, the area is \( A = 16 \times 16 = 256 \) cm², and these dimensions correspond to a square, which maximizes the area for a given perimeter.

Key Concepts

Rectangle Perimeter

The perimeter of a rectangle can be thought of as the distance all the way around the outside of the rectangle. It is commonly calculated using the formula:

• Perimeter (\(P\)) = \(2l + 2w\)

where \(l\) is the length and \(w\) is the width of the rectangle. In our exercise, the perimeter is fixed at 64 cm. This constraint helps us find the possible dimensions of the rectangle.

Once you know the perimeter, figuring out how length and width relate is crucial. By simplifying: \(l + w = 32\). This represents the total length you have to distribute between length and width to form the rectangle.

Each different set of \(l\) and \(w\) pair that satisfies \(l + w = 32\) will form a rectangle with a perimeter of 64 cm. But, to maximize area, it's important to choose the optimal values of \(l\) and \(w\).

Area Maximization

The goal is to maximize the area of a rectangle, which can be an exciting challenge in calculus. The area, \(A\), is given by the product of its length and width:

• Area (\(A\)) = \(lw\)

Area maximization in this context involves finding dimensions that yield the largest possible area while keeping the perimeter constant.

With the perimeter equation \(l + w = 32\) solved for \(w\), we get \(w = 32 - l\). Thus, the area equation becomes: \(A = l(32 - l)\). Expanding this gives a quadratic:

• \(A = 32l - l^2\)

The aim now is to find the peak of this quadratic, which tells us the maximum area the rectangle can enclose.

Interestingly, when \(l = w\), meaning the rectangle becomes a square, the area is maximized under a fixed perimeter. For our given problem, \(l = w = 16\) cm results in an area of \(256\) cm².

Derivative in Calculus

The derivative in calculus plays a vital role in optimizing functions, such as the area of a rectangle. It is a mathematical tool that helps us find where a function's slope is zero, indicating a potential maximum or minimum.

In our area maximization problem, we express the area \(A\) in terms of length \(l\). The expression is \(A = 32l - l^2\). To optimize, we need to find where its derivative equals zero:

• \(A'(l) = 32 - 2l\)

By setting \(A'(l) = 0\), we solve for \(l\) and find \(l = 16\), which indicates the point of maximum area.

This technique is commonly used in calculus to locate features of interest in graphs and solve many types of real-world problems, where optimizing measurable quantities is necessary. In our case, it confirms that setting \(l = w\) yields the maximum possible area for a rectangle under a fixed perimeter.