Problem 1 - Thomas Calculus

Question

Minimizing perimeter for a rectangle whose area is \(16 \mathrm{in}^{2},\) and what are its dimensions?

Step-by-Step Solution

Answer

The rectangle has dimensions 4 inches by 4 inches.

Step 1: Define Variables

Let the length and width of the rectangle be \( l \) and \( w \), respectively. Since the area \( A \) of the rectangle is given by \( A = l \times w \), we have the equation \( l \times w = 16 \).

Step 2: Express Perimeter

The perimeter \( P \) of a rectangle is given by the formula \( P = 2l + 2w \). Our goal is to minimize this perimeter while maintaining the area constraint.

Step 3: Substitute and Simplify

Using the area constraint \( l \times w = 16 \), express \( w \) in terms of \( l \) as \( w = \frac{16}{l} \). Substitute this into the perimeter equation: \( P = 2l + 2\left(\frac{16}{l}\right) = 2l + \frac{32}{l} \).

Step 4: Differentiate Perimeter

Find the derivative of the perimeter function with respect to \( l \): \( \frac{dP}{dl} = 2 - \frac{32}{l^2} \).

Step 5: Find Critical Points

Set the derivative equal to zero to find critical points: \( 2 - \frac{32}{l^2} = 0 \). Solving for \( l \), we get \( l^2 = 16 \), which gives \( l = 4 \).

Step 6: Verify Minimum

Check the second derivative \( \frac{d^2P}{dl^2} = \frac{64}{l^3} \). For \( l = 4 \), the second derivative is positive, indicating a local minimum for the perimeter.

Step 7: Calculate Dimensions

With \( l = 4 \), find \( w \) using \( w = \frac{16}{4} = 4 \). The rectangle dimensions that minimize the perimeter while having the given area are 4 inches by 4 inches.

Key Concepts

Rectangle Perimeter

The perimeter of a rectangle is the total distance around its boundary. Imagine walking along the edges of a rectangular field; the total distance you travel is the perimeter. In mathematical terms, if you know the length (\( l \)) and width (\( w \)) of a rectangle, you can easily calculate its perimeter using the formula:
\[P = 2l + 2w\]This formula adds up the lengths of all four sides (two lengths and two widths) to give the perimeter.

For example, let's consider a rectangle with lengths of 4 inches. The perimeter would be calculated as:

• \( 2 imes 4 + 2 imes 4 = 16 \)

So, the perimeter is 16 inches. Understanding this concept is crucial when you aim to minimize or optimize a rectangle's dimensions with a given constraint like a fixed area.

Critical Points

In calculus, critical points are values of a variable where its function's derivative is zero or undefined. These points are significant because they can indicate where a function might be at a minimum or a maximum. In optimization problems, like minimizing the perimeter of a rectangle while keeping the area constant, finding critical points helps identify optimal solutions.

To find the critical points when minimizing a rectangle's perimeter, we first express the perimeter as a function of one variable. In our case, using the area constraint (\( l \times w = 16 \)), we substitute to express perimeter (\( P \)) solely in terms of one variable:
\[P = 2l + \frac{32}{l}\]Then, we find the derivative of \( P \) with respect to \( l \). The critical point occurs when:

• \( \frac{dP}{dl} = 0 \)

Solving \( 2 - \frac{32}{l^2} = 0 \) leads us to the critical point at \( l = 4 \), indicating a potential minimum or maximum perimeter at this dimension.

Second Derivative Test

The second derivative test is a tool used in calculus to determine the nature of critical points. Once we have identified a critical point, using the second derivative helps confirm if that point is a local minimum, maximum, or neither for the function under consideration.

For our problem, after finding the critical point where\( l = 4 \), we use the second derivative test to verify its nature. The second derivative of the perimeter function \( P \) with respect to \( l \) is:

• \( \frac{d^2P}{dl^2} = \frac{64}{l^3} \)

At \( l = 4 \), compute:

• \( \frac{d^2P}{dl^2} = \frac{64}{4^3} > 0 \)

A positive value of the second derivative indicates a local minimum. Therefore, the dimensions of \( l = 4 \) and \( w = 4 \) not only satisfy the area requirement but also ensure that the perimeter is minimized, thanks to applying the second derivative test successfully.