Question

Designing a Suitcase A 24- by 36-in. sheet of cardboard is folded in half to form a 24- by 18-in. rectangle as shown in the figure. Then four congruent squares of side length x are cut from the corners of the folded rectangle. The sheet is unfolded, and the six tabs are folded up to form a box with sides and a lid. (a) Write a formula \(V(x)\) for the volume of the box. (b) Find the domain of \(V\) for the problem situation and graph \(V\) over this domain. (c) Use a graphical method to find the maximum volume and the value of \(x\) that gives it. (d) Confirm your result in part (c) analytically. (e) Find a value of \(x\) that yields a volume of 1120 \(\mathrm{in}^{3}\) . (f) Writing to Learn Write a paragraph describing the issues that arise in part (b).

Short answer

The function representing the volume of the box is \( V(x) = x(24-2x)(18-2x) \), valid for values of \( x \) that lies in the domain \( 0 \leq x \leq 9 \). For maximum volume, find the highest point in the graph of \( V(x) \) or by setting its derivative equal to zero and confirming a maximum. For a volume of 1120 in³, set \( V(x) = 1120 \) and solve for \( x \). Finally, the domain restrictions in part (b) arise due to the physical constraints of the problem.

Step-by-step solution

Formulate the volume function

The volume of a box can be expressed as \( V = l \times w \times h \), where \( l \), \( w \), and \( h \) represent the length, width, and height of the box, respectively. By cutting squares of side length \( x \) from each corner of our 24 by 18 in. rectangle, we create a box with dimensions \( l = 24-2x \), \( w = 18-2x \), and \( h = x \), since the box's height is equal to the side length of the cut squares. Substituting these into the volume formula gives \( V(x) = x(24-2x)(18-2x) \) cubic in.

Determine the domain of V

The domain of \( V \) is determined by the values of \( x \) that yield a real and meaningful (in this context, non-negative) volume. As \( x \) is the length of the cut squares, it must be less than half the smaller dimension of the rectangle, namely 18 in., to avoid overlap. So, the domain in this scenario is \( 0 \leq x \leq 9 \).

Plot the function \( V(x) \)

On graph paper, plot the function \( V(x) = x(24-2x)(18-2x) \) across the domain \( 0 \leq x \leq 9 \). The graph shape should reveal the volume's maximum value.

Find the maximum volume graphically

Inspect the graph and determine the maximum point of \( V(x) \) which corresponds to the optimal side length \( x \) for the cut squares and the maximum volume of the box.

Analytical confirmation

To confirm the result analytically, calculate the derivative of \( V(x) = x(24-2x)(18-2x) \), set it equal to zero, and solve for \( x \). Then, double-check that this value yields a maximum (versus a minimum or saddle point) by either analyzing the change in \( V'(x) \) around this value or employing the second derivative test.

Obtaining volume of 1120 cubic inches

To find a value of \( x \) that results in a volume of 1120 cubic inches, set \( V(x) = 1120 \) and solve for \( x \).

Discussion of issues in part (b)

In part (b) of this task, the challenge is understanding how the problem constraints dictate which values of \( x \) are feasible, resulting in the domain. In this situation, \( x \) must be non-negative, as it represents a length, but it also must be small enough to avoid overlap of the cut corners. These restrictions define the domain of the function.

Key concepts

Volume of a Box

When tasked with the problem of maximizing the storage space of a suitcase, understanding how to calculate the volume of a box becomes crucial. The volume is the amount of three-dimensional space enclosed by a box, which can be found by multiplying its length, width, and height. In our case, after cutting out squares of side length x, the new dimensions of the box are length = 24 - 2x, width = 18 - 2x, and height = x. Hence, the volume function is represented as:
\[V(x) = x(24 - 2x)(18 - 2x)\]
Clearly demonstrating the direct relationship between the side length of the cut squares and the resulting volume of the box. To ensure the best understanding, visual aids like illustrations can be particularly helpful alongside these equations.

Domain of a Function

Defining the domain of a function is vital as it illustrates all the possible input values the function can accept. For the suitcase problem, the domain represents the constraints within which the side length x of the cut squares must lie. Specifically, x must be positive but also less than half the narrowest dimension of the cardboard, to avoid producing negative or non-sensical dimensions for the box. Consequently, the domain for our function, V(x), is:
\[0 \leq x \leq 9\]
Understanding the domain of a function is not only about mathematical correctness but also about the practical aspects of the problem at hand. It's a good practice to always question whether a value within the domain makes sense in the real-world context of the problem.

Graphical Method

Using a graphical method to visualize the relationship between variables can be immensely beneficial. It allows one to see how changing the size of the cutout squares impacts the volume of the box. By plotting V(x) over its domain, a curve is revealed, usually reaching a maximum point. This highest point on the graph indicates both the maximum volume the box can hold and the optimal square size x. To facilitate comprehension, students should be encouraged to experiment with graphing software or simply use graph paper to sketch these types of functions. Observing the shape of the graph can provide immediate insights into the problem, emphasizing the power of visual learning.

Analytical Confirmation

While visual methods provide great intuition, analytical confirmation using calculus solidifies one's findings. This step entails taking the derivative of V(x), setting it to zero, and solving for x to find critical points. These critical points are where the function either reaches a maximum, minimum, or a point of inflection. By employing the derivative, we ensure that our results are not just approximations or subject to error from graph interpretation. Calculating the derivative and analyzing it strengthens the mathematical understanding and guarantees the accuracy of the solution, which is why this step is a cornerstone of calculus and its applications.

Derivatives

Derivatives are the cornerstone of optimization problems in calculus. They represent the rate at which a function is changing at any given point. By finding the derivative of V(x) and setting it equal to zero, we can determine where the function has a horizontal tangent line — which indicates a potential maximum or minimum value. The concept of derivatives extends beyond textbook problems, having real-life applications such as analyzing rates of change in physics, economics, and biology. Understanding derivatives is fundamental for students as it equips them with tools to solve a wide range of dynamic problems.