Problem 13 - Essential Calculus Early Transcendentals

Question

(a) Show that of all the rectangles with a given area, the one with smallest perimeter is a square. (b) Show that of all the rectangles with a given perimeter, the one with greatest area is a square.

Step-by-Step Solution

Answer

The rectangle with minimal perimeter for a given area and maximal area for a given perimeter is a square in both cases.

Step 1: Define the Problem for Part (a)

We need to find the rectangle with the smallest perimeter that has a given area, \( A \). For a rectangle, if \( x \) is the length and \( y \) is the width, then its area is \( xy = A \) and the perimeter is \( P = 2x + 2y \).

Step 2: Express Perimeter in terms of one variable for Part (a)

From \( xy = A \), express \( y \) as \( y = \frac{A}{x} \). Substitute this into the perimeter equation to get \( P(x) = 2x + 2\frac{A}{x} \).

Step 3: Find Critical Points for Part (a)

To find the minimum perimeter, take the derivative of \( P(x) \) with respect to \( x \): \( P'(x) = 2 - \frac{2A}{x^2} \). Set \( P'(x) = 0 \) to find the critical points: \( 2 = \frac{2A}{x^2} \), which simplifies to \( x^2 = A \), hence \( x = \sqrt{A} \).

Step 4: Verify the Minimum for Part (a)

To confirm \( x = \sqrt{A} \) is a minimum, use the second derivative test. Compute \( P''(x) = \frac{4A}{x^3} \), which is positive for \( x > 0 \), indicating a local minimum. Substituting \( x = \sqrt{A} \) gives \( y = \sqrt{A} \), proving the rectangle is a square.

Step 5: Define the Problem for Part (b)

We now find the rectangle with the largest area that has a given perimeter, \( P \). For a rectangle, \( x \) is the length and \( y \) is the width, making the perimeter \( P = 2x + 2y \) and the area \( A = xy \).

Step 6: Express Area in terms of one variable for Part (b)

From \( 2x + 2y = P \), express \( y \) as \( y = \frac{P}{2} - x \). Substitute into the area equation to get \( A(x) = x \left( \frac{P}{2} - x \right) = \frac{P}{2}x - x^2 \).

Step 7: Find Critical Points for Part (b)

Find the derivative of \( A(x) \) with respect to \( x \): \( A'(x) = \frac{P}{2} - 2x \). Set \( A'(x) = 0 \) to find critical points: \( \frac{P}{2} = 2x \), solving for \( x \) yields \( x = \frac{P}{4} \).

Step 8: Verify the Maximum for Part (b)

To confirm \( x = \frac{P}{4} \) is a maximum, use the second derivative test: \( A''(x) = -2 \), which is negative, indicating a local maximum. Substituting \( x = \frac{P}{4} \), get \( y = \frac{P}{4} \), showing the rectangle is a square.

Key Concepts

Critical Points

Critical points are essential in solving optimization problems in calculus. They help us find where a function might reach a minimum or maximum value. For this, we first need to take the first derivative of the function we are trying to optimize.
Critical points occur where the first derivative is zero or undefined. Let's break down the problem of minimizing the perimeter in part (a) of our exercise. We started with the perimeter function:

• Perimeter, expressed in one variable: \( P(x) = 2x + 2\frac{A}{x} \).
• Taking the derivative: \( P'(x) = 2 - \frac{2A}{x^2} \).

Setting \( P'(x) = 0 \), we find the critical points where the perimeter could be minimal. Solving \( 2 = \frac{2A}{x^2} \) gives us that \( x^2 = A \), hence at \( x = \sqrt{A} \), the critical point is found.
In part (b), to maximize the area with a given perimeter, we find the critical points similarly with the area function:

• Area, expressed in one variable: \( A(x) = \frac{P}{2}x - x^2 \).
• Taking the derivative: \( A'(x) = \frac{P}{2} - 2x \).

Setting \( A'(x) = 0 \) helps us solve for \( x = \frac{P}{4} \). These critical points guide us to analyze where the function has extremal values.

Second Derivative Test

The second derivative test is a powerful tool for verifying whether the critical points we found are indeed minima or maxima. In this exercise, for both parts (a) and (b), the second derivative test comes in handy after finding the critical points.
For part (a), we have verified our critical point for minimum perimeter by computing the second derivative of the perimeter function:

• Second derivative: \( P''(x) = \frac{4A}{x^3} \) which is positive for \( x > 0 \).

The positive sign of \( P''(x) \) implies a local minimum at \( x = \sqrt{A} \). Then, since the rectangle becomes a square, we have indeed found the rectangle with the smallest perimeter for a given area.
For part (b), to verify the maximum area, we analyze:

• Second derivative: \( A''(x) = -2 \) which is negative.

A negative \( A''(x) \) indicates a local maximum at \( x = \frac{P}{4} \). This confirms that the rectangle achieves maximal area as a square when both sides are equal.

Perimeter and Area of Rectangles

Rectangles have dimensions typically represented as length and width. Calculating perimeter and area are fundamental when solving optimization problems.
The perimeter of a rectangle is calculated by adding all the sides together:

• Formula: \( P = 2x + 2y \) where \( x \) and \( y \) are the rectangle's length and width.

In the exercise, we sought rectangles with a set area or perimeter:

• In part (a), the given area \( A = xy \) was a constraint. The task was minimizing \( P \) under this fixed area.
• In part (b), the given perimeter \( P \) was fixed. The task was maximizing \( A = xy \).

To make calculations involving such constraints, equations are often expressed with one variable by substituting for either \( x \) or \( y \), as seen in our step-by-step solution.
This approach simplifies the optimization process, allowing us to utilize calculus tools like derivatives to find critical points and verify solutions using concepts like symmetry, leading to the realization that a square shape offers optimal solutions for both problems.