Problem 1 - Thomas Calculus

Question

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

Step-by-Step Solution

Answer

The smallest perimeter is 16 inches, with dimensions 4 x 4 inches.

Step 1: Understand the Problem

You need to find the dimensions of a rectangle that minimize the perimeter, given its area is 16 square inches.

Step 2: Area of Rectangle

The area of a rectangle is given by the formula: \( Area = length \times width \). For this problem, \( length \times width = 16 \).

Step 3: Perimeter of Rectangle

The perimeter of a rectangle is given by \( P = 2(length + width) \). We want to minimize this perimeter.

Step 4: Express Width in Terms of Length

Using the area equation \( width = \frac{16}{length} \). Substitute into the perimeter equation to get \( P = 2(length + \frac{16}{length}) \).

Step 5: Differentiate Perimeter Function

Find the derivative of the perimeter function with respect to length: \( \frac{dP}{dl} = 2(1 - \frac{16}{l^2}) \).

Step 6: Find Critical Points

Set the derivative equal to zero and solve for \( l \): \( 1 - \frac{16}{l^2} = 0 \), which implies \( l^2 = 16 \), so \( l = 4 \).

Step 7: Determine Width

Substitute \( l = 4 \) back into the width equation: \( width = \frac{16}{4} = 4 \).

Step 8: Calculate Perimeter

Substitute the values of length and width into the perimeter formula to find minimum perimeter: \( P = 2(4 + 4) = 16 \).

Step 9: Verify Minimality

Since changing the dimensions alters the area without minimizing the perimeter further, a square with length and width as 4 in. each is appropriate and optimal.

Key Concepts

Perimeter Minimization

Minimizing the perimeter of a geometric figure like a rectangle while maintaining a constant area is a classic optimization problem in calculus. In essence, you want to keep the boundary as short as possible while the area inside remains the same.

For a rectangle with a given area, the perimeter is minimized when the rectangle is as close to being a square as possible. This happens because a square configuration maximizes the amount of area enclosed for a given perimeter.

To solve perimeter minimization problems:

• Start with the area equation, ensuring that the product of length and width equals the desired area.
• Next, express the perimeter in terms of one variable, making it ripe for differentiation.
• Find the minimum perimeter by deriving and analyzing a critical point.

In our problem, the optimal dimensions for a rectangle with an area of 16 square inches emerge when our shape becomes a square with a perimeter of 16 inches.

Derivative Application

Applying derivatives is a powerful tool in calculus, especially useful for finding maximum and minimum values, such as minimizing a perimeter. Differentiation gives us the rate of change of a function, allowing us to identify points where the function reaches its critical values — these are the maxima or minima.

Follow these steps to apply derivatives in optimization problems:

• Formulate the problem in terms of a function and express it with a single variable.
• Differentiating this function with respect to that variable provides the necessary information about its behavior.
• Set the derivative equal to zero to determine critical points, which can indicate maxima or minima.

In the context of the rectangle exercise, we used the formula for the perimeter and differentiated it to find the critical point where the perimeter is minimized. Setting the derivative to zero pinpointed this critical state, revealing that when both the length and width are 4 inches, the perimeter is minimized.

Rectangle Dimensions

Understanding the dimensions of a rectangle is essential in solving both geometric and optimization problems. The basic understanding of a rectangle involves its length and width, which together define its area and perimeter.

For an optimization problem like perimeter minimization, knowing how dimensions interact is crucial. A rectangle can continually shift dimensions while keeping its area constant — this flexibility requires us to strategically select the optimal dimensions.

Key points to consider about rectangle dimensions:

• The formula for the area, where length times width gives a constant value, is pivotal in finding compatible dimensions.
• The perimeter equation, which relies on the sum of length and width, guides us to minimize the perimeter.
• Reformulating either length or width based on the other using known constraints (like area) helps streamline problem-solving.

In our problem, by fixing the area at 16 square inches and minimizing the perimeter, we determine that a square, where both length and width equal 4 inches, is the optimal shape. This configuration perfectly highlights the interaction between dimensions when area and perimeter are mathematically balanced.