Problem 84 - College Algebra with Corequisite Support

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 with minimum surface area are 6 inches by 6 inches by 3 inches.

Step 1: Define the Variables

Let the side of the square base be denoted by \( x \) and the height of the box by \( h \). We need to find these values specifically for the box to have minimum surface area while maintaining a volume of 108 cubic inches.

Step 2: Volume Equation

Given the volume \( V \) of the box is 108 cubic inches, express it in terms of \( x \) and \( h \): \[V = x^2 h = 108.\] Solve for \( h \): \[ h = \frac{108}{x^2}. \]

Step 3: Surface Area Equation

The surface area \( S \) of the open-top box is the sum of the area of the base and the lateral sides. It's given by: \[S = x^2 + 4xh.\] Substitute \( h \) from the volume equation: \[ S = x^2 + 4x\left(\frac{108}{x^2}\right) = x^2 + \frac{432}{x}.\]

Step 4: Finding the Minimum Surface Area

To minimize the surface area \( S \), take the derivative of \( S \) with respect to \( x \) and set it to zero: \[ \frac{dS}{dx} = 2x - \frac{432}{x^2}.\]Solve \( \frac{dS}{dx} = 0 \):\[ 2x = \frac{432}{x^2}. \]Multiply both sides by \( x^2 \):\[ 2x^3 = 432. \]Solve for \( x \):\[ x^3 = 216 \quad \Rightarrow \quad x = 6.\]

Step 5: Calculating the Height

Use the value of \( x \) in the volume equation to find \( h \): \[ h = \frac{108}{x^2} = \frac{108}{6^2} = \frac{108}{36} = 3.\]

Step 6: Verify the Minimum

Double-check the minimum by evaluating the second derivative of \( S \). The second derivative is: \[ \frac{d^2S}{dx^2} = 2 + \frac{864}{x^3}. \]Substitute \( x = 6 \) and observe that \[ \frac{d^2S}{dx^2} = 2 + \frac{864}{216} = 6 > 0, \]confirming a local minimum.

Key Concepts

Rational Functions

Rational functions are mathematical expressions representing the ratio of two polynomials. In the context of calculus and optimization, rational functions play a crucial role because they allow us to define relationships between variables that are not linear.
In the problem of finding the dimensions of a box with minimal surface area given a fixed volume, we set up a rational function to express the surface area in terms of the side length of the base. Here, the rational function was derived by expressing the height of the box as a function of the base area and the volume constraint. This kind of function allows us to analyze how changes in one dimension can affect others, facilitating a systematic approach to optimization.

• A rational function generally takes the form \( f(x) = \frac{P(x)}{Q(x)} \) where \( P(x) \) and \( Q(x) \) are polynomials.
• These functions can illustrate behaviors like asymptotes, intercepts, and limits, all essential in understanding complex mathematical models.

Understanding rational functions is crucial for tackling problems involving multiple varying dimensions and constraints efficiently.

Surface Area Minimization

Surface area minimization involves finding the least possible surface area that meets certain required conditions, such as volume. This concept is widely applied in various fields, from engineering design to manufacturing, where reducing material use without compromising on required properties is crucial.
In our exercise, we considered an open box with a square base, and the goal was to find its dimensions that would result in the smallest surface area while keeping the volume at 108 cubic inches. We derived the surface area expression as a rational function in terms of the side length \( x \): \( S = x^2 + \frac{432}{x} \). By differentiating this function, we could determine the point where the surface area is minimized.

• Minimizing surface area involves using calculus to find critical points where the rate of change is zero.
• This is often solved by finding the first derivative of the surface area function, then setting it to zero to find potential candidates for minimum values.
• Surface area minimization ensures material efficiency and cost-saving in practical applications.

Mastering surface area minimization through calculus allows us to reliably determine optimal shapes and sizes for three-dimensional objects.

Volume Constraints

Volume constraints dictate that the total volume of a space or object must equal a specific value, no matter how the other dimensions adjust. In the exercise, our volume constraint was 108 cubic inches for an open box with a square base. This constraint was essential to find the height of the box in terms of its base side length because it helps prioritize solutions that respect the fixed volume requirement.
By setting up the equation \( V = x^2h = 108 \), we solved for the height: \( h = \frac{108}{x^2} \). This expression was then used to plug back into the surface area equation. Working with volume constraints ensures that the solution satisfies basic physical restrictions or design specifications.

• Volume constraints lead the derivation of relationships between different dimensions, ensuring the desired volume is achieved.
• They are critical in problems involving physical limitations and space requirements.
• Handling volume constraints requires precise attention to how each dimension affects the overall volume directly.

Successfully incorporating volume constraints into an optimization problem is crucial for achieving realistic and practical results.

Derivatives in Calculus

Derivatives are fundamental tools in calculus that measure how a function's output value changes as its input changes. In optimization problems like the one discussed, derivatives help identify points where functions such as surface area reach their maximum or minimum values.
The step-by-step solution involved taking the derivative of the surface area function \( S = x^2 + \frac{432}{x} \) with respect to \( x \). This resulted in the expression for the derivative: \( \frac{dS}{dx} = 2x - \frac{432}{x^2} \). By setting this expression to zero, we found the critical point for minimum surface area.

• Derivatives indicate the slope or rate of change of functions, essential for finding critical points.
• Setting a derivative to zero helps locate local minima or maxima, which are vital to optimization.
• The second derivative test confirms whether these points are indeed minimums, based on the slope's change.

Understanding how to apply derivatives allows you to not only solve for minimums and maximums in functions but also to interpret their significance through the lens of calculus.