Problem 538 - Precalculus

Question

For the following exercises, construct a rational function that will help solve the problem. Then, use a calculator to answer the question. An open box with a square base is to have a volume of 108 cubic inches. Find the dimensions of the box that will have minimum surface area. Let \(x=\) length of the side of the base.

Step-by-Step Solution

Answer

The box dimensions that minimize surface area are 6 inches x 6 inches x 3 inches.

Step 1: Define Variables and Constraints

Let \( x \) be the length of the side of the square base, and \( h \) be the height of the box. The volume of the box is given by \( x^2 \cdot h = 108 \) cubic inches. We want to minimize the surface area.

Step 2: Express Height in Terms of Base Side

From the volume equation, \( x^2 \cdot h = 108 \), solve for \( h \). Thus, \( h = \frac{108}{x^2} \).

Step 3: Write the Surface Area Formula

The surface area of the box is the sum of the area of the base and four sides: \( A = x^2 + 4(xh) \). Substitute \( h = \frac{108}{x^2} \) to get \( A = x^2 + 4x \left( \frac{108}{x^2} \right) = x^2 + \frac{432}{x} \).

Step 4: Differentiate Surface Area Function

Differentiate the surface area function with respect to \( x \): \( A'(x) = 2x - \frac{432}{x^2} \).

Step 5: Find Critical Points

Set the derivative equal to zero to find critical points: \( 2x - \frac{432}{x^2} = 0 \). Solve for \( x \): \( 2x^3 = 432 \) or \( x^3 = 216 \). Thus, \( x = \sqrt[3]{216} = 6 \).

Step 6: Verify Minimum Surface Area

Use second derivative test or analyze change in \( A'(x) \) around \( x = 6 \) to confirm it's a minimum. The second derivative, \( A''(x) = 2 + \frac{864}{x^3} \), is positive for \( x > 0 \), indicating a local minimum.

Step 7: Calculate Dimensions

With \( x = 6 \), calculate \( h \) using \( h = \frac{108}{x^2} \): \( h = \frac{108}{36} = 3 \). The dimensions of the box are 6 inches by 6 inches by 3 inches.

Key Concepts

Volume Optimization

In mathematical optimization problems, volume optimization often plays a crucial role. When tasked with creating a container of a given volume, the goal is usually to determine the most efficient shape for maximum utility. For instance, in this exercise, the volume of the open box is fixed at 108 cubic inches.
Knowing the volume helps us determine the relationships between the dimensions of the box, such as the side of the square base and the height.To express this mathematically, we use the equation for volume: The volume is defined as the product of the area of the base and the height, specifically for a square base box, it's represented by \( x^2 \cdot h = 108 \), where \( x \) is the length of the base, and \( h \) is the height. Now, with this relationship, we can explore other properties like surface area based on fixed volume constraints.

Surface Area Minimization

Minimizing surface area is essential for designing efficient packaging while using minimal material. In our problem, the task is to minimize the surface area of a box with a fixed volume. To find the minimum surface area, you must analyze and adjust the box's dimensions.Given the volume equation, you already know that \( x^2 \cdot h = 108 \).To express the surface area, consider both the base and side faces of the box:

• Base Area: \( x^2 \)
• Side Faces Area: \( 4(xh) \)

Substituting \( h = \frac{108}{x^2} \) to express the surface area entirely in terms of \( x \), you get:\( A = x^2 + \frac{432}{x} \). By formulating the surface area in terms of a single variable, you can apply calculus techniques to find where this function is minimized.

Calculus Application

Calculus reveals the trends and properties of functions, making it invaluable for optimization problems. In this scenario, we use calculus to determine how the surface area function behaves as \( x \) changes.First, develop the equation for the surface area from the problem constraints and express it by substituting volume relations, as mentioned previously. Next, to find the minimum surface area, differentiate the surface area function with respect to \( x \):\[ A'(x) = 2x - \frac{432}{x^2} \].By setting this derivative to zero, you find critical points, which indicate potential minima or maxima in context of this problem. This function represents a typical case where calculus allows the finding of optimal values by observing the function's turning points.

Critical Points Analysis

Critical points in a function occur where its derivative is zero or undefined, representing potential maxima, minima, or points of inflection. These points are crucial in determining the optimal dimensions for the surface area minimization problem. In this exercise, after differentiating the surface area function, it's set equal to zero:\[ 2x - \frac{432}{x^2} = 0 \]. Solving this gives the critical point \( x = \sqrt[3]{216} = 6 \). This critical point represents a candidate for the box's side length.To confirm it's a minimum, analyze the second derivative, \( A''(x) = 2 + \frac{864}{x^3} \). Since the second derivative is positive for all \( x > 0 \), this indicates a local minimum at \( x = 6 \). Hence, using critical points analysis shows that the box's dimensions are optimally \( 6 \times 6 \times 3 \) inches, providing minimal surface area for a fixed volume.