Problem 54 - Calculus Early Transcendentals

Question

All rectangles with an area of 64 have a perimeter given by \(P(x)=2 x+128 / x,\) where \(x\) is the length of one side of the rectangle. Find the absolute minimum value of the perimeter function on the interval \((0, \infty) .\) What are the dimensions of the rectangle with minimum perimeter?

Step-by-Step Solution

Answer

Answer: The dimensions of the rectangle with the minimum perimeter are 8 by 8, and the minimum perimeter is 32.

Step 1: Find the first derivative of P(x)

First, find the derivative of P(x) with respect to x to find the critical points: \[\frac{dP(x)}{dx} = \frac{d}{dx}(2x+\frac{128}{x}).\] Using the power rule for derivatives, we have: \[\frac{dP(x)}{dx} = 2 - \frac{128}{x^2}.\]

Step 2: Find the critical points of P(x)

Now, find the critical points by setting the derivative to 0 and solving for x: \[2 - \frac{128}{x^2}=0.\] Move the second term to the right side: \[\frac{128}{x^2}=2.\] Now, multiply both sides by \(x^2\) and divide by 2 to solve for x: \[x^2 = \frac{128}{2} = 64.\] Taking the square root, we obtain our critical point: \[x = 8.\]

Step 3: Find the second derivative of P(x)

To ensure that this critical point corresponds to a minimum perimeter, we need to find the second derivative of P(x) and check if it is positive or negative for the discovered critical point: \[\frac{d^2P(x)}{dx^2} = \frac{d}{dx}(2 - \frac{128}{x^2}).\] Using the power rule for derivatives again, we have: \[\frac{d^2P(x)}{dx^2} = \frac{256}{x^3}.\]

Step 4: Verify the critical point corresponds to a minimum

Evaluate the second derivative at the critical point (x = 8): \[\frac{d^2P(x)}{dx^2}\Big|_{x=8} = \frac{256}{8^3} = \frac{256}{512} = \frac{1}{2}.\] Since the second derivative is greater than 0 at x=8, we can conclude that the critical point corresponds to a minimum.

Step 5: Find the minimum perimeter and dimensions

Now, we have confirmed a critical point at x=8 which corresponds to a minimum. We can now find the minimum value of the perimeter function and the dimensions of the rectangle: The minimum perimeter is given by: \[P(8) = 2(8) + \frac{128}{8} = 16 + 16 = 32.\] Therefore, the absolute minimum value of the perimeter is 32. Since x represents the length of one side of the rectangle, we have a length of 8. To find the other side, we can use the fact that the area of the rectangle is 64. Therefore, \[8y = 64\] or \[y = 8.\] So, the dimensions of the rectangle with the minimum perimeter are 8 by 8.

Key Concepts

Critical Points

Finding critical points is an essential step in optimization problems. Critical points are where the first derivative of a function equals zero or does not exist. These points are significant as they can indicate potential maximum or minimum values of the function.
To find critical points, we do the following:

• First, compute the first derivative of the function.
• Set the derivative equal to zero to find the critical points.
• Solve the resulting equation for the variable.

In the given problem, we have the perimeter function of a rectangle with constant area, which gives us an equation based on one side's length. Critical points help us determine the extremities of these function values, crucial in finding the minimum perimeter.

Derivative

The derivative of a function gives us information about the function's rate of change. In calculus, the derivative is represented as \(\frac{d}{dx}\) and allows us to understand the behavior of a function.
For example, the derivative tells us whether the function is increasing or decreasing at a particular point, which is crucial in optimization tasks. In this problem, we derived the perimeter function: \(P(x) = 2x + \frac{128}{x}\).

• The derivative \(\frac{dP(x)}{dx} = 2 - \frac{128}{x^2}\) helps us pinpoint where changes in the function's values might lead to a minimum or maximum perimeter value.
• The slope of the line (given by the derivative) reaches zero at critical points, potentially indicating these extreme values.

Understanding derivatives is pivotal for finding optimal solutions in similar geometrical problems.

Second Derivative

After finding the critical points, the second derivative helps us determine the nature of these points. It tells us whether a critical point is a maximum, minimum, or a point of inflection.
A positive second derivative indicates a minimum (the graph is concave up at that point), whereas a negative second derivative shows a maximum (the graph is concave down). In this exercise:

• The second derivative is given by \(\frac{d^2P(x)}{dx^2} = \frac{256}{x^3}\).
• Evaluating it at the critical point \(x=8\) results in a positive value \(\frac{1}{2}\).

This confirms that at \(x=8\), the perimeter function indeed reaches a minimum. Thus, using the second derivative test is a robust method to ensure accurate identification of those minimum or maximum points.

Perimeter of Rectangle

The perimeter of a rectangle is a linear measure that represents the total distance around the edges of the shape. It is calculated by adding together twice the rectangle's length and twice its width. So, the general formula is \(P = 2l + 2w\).
In the context of this problem, we have a unique scenario where the rectangle's area is fixed at 64. Here, depending on one variable (the length \(x\)), the perimeter \(P(x)\) can be represented as \(2x + \frac{128}{x}\).

• This representation reflects the influence of the rectangle's area constraint, simplifying the problem to a one-variable dependence for optimization purposes.

By carefully analyzing and manipulating these geometric constraints, we found the dimensions leading to the smallest possible perimeter, balancing both length and width equally at 8 units.