Problem 17 - Thomas Calculus

Question

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

Step-by-Step Solution

Answer

The volume formula is \( V(x) = x(36-2x)(24-2x) \). Maximum volume occurs at \( x \approx 4.5 \) with \( V \approx 1296 \, \mathrm{in}^3 \).

Step 1: Define the Volume Formula

To find the volume of the box, notice that when you cut squares of side length \( x \) from each corner, the length and width of the box become \( 36 - 2x \) and \( 24 - 2x \) respectively. The height of the box is \( x \). Therefore, the volume \( V(x) \) is given by \( V(x) = x(36 - 2x)(24 - 2x) \).

Step 2: Determine the Domain of the Function

The value of \( x \) must be such that the length and width are positive after cutting the squares. Therefore, \( x \) must satisfy the inequalities \( 36 - 2x > 0 \) and \( 24 - 2x > 0 \). Solving these gives \( x < 18 \) and \( x < 12 \), respectively. Thus, the domain of \( V(x) \) is \( 0 < x < 12 \).

Step 3: Graph the Volume Function

Plot the function \( V(x) = x(36 - 2x)(24 - 2x) \) over the domain \( 0 < x < 12 \). The graph provides a visual representation and helps identify approximate locations of maximum volume.

Step 4: Find Maximum Volume Graphically

Observing the graph of \( V(x) \), identify the point where the volume reaches its maximum. Use graphing technology to estimate the maximum value of \( V(x) \) and the corresponding \( x \) value.

Step 5: Confirm Maximum Volume Analytically

To confirm the maximum volume analytically, take the derivative \( V'(x) \) and find critical points by setting \( V'(x) = 0 \). Analyze these critical points using the second derivative test or first derivative test within the defined domain to confirm the maximum.

Step 6: Find \( x \) for a Specific Volume

To find \( x \) that gives a volume of \( 1120 \, \mathrm{in}^3 \), solve the equation \( V(x) = 1120 \, \mathrm{in}^3 \). This may require numerical methods or graphing techniques to find an appropriate \( x \) that satisfies this equation.

Step 7: Discuss Domain Issues

The domain \( 0 < x < 12 \) ensures the box's dimensions remain positive, but also reflects practical constraints, such as the constructability and usability of the box. As \( x \) approaches either limit of the domain, the shape of the box becomes impractical or impossible.

Key Concepts

Volume Formula

To calculate the volume of the box formed from the cardboard sheet, we use a combination of its geometric properties. Initially, when four squares of side length \( x \) are removed from each corner, the dimensions of the base of the box change. The length thus becomes \( 36 - 2x \), since both corners on each end reduce the length by \( x \) each. Similarly, the width becomes \( 24 - 2x \). The height of the box equals the side of the square removed, which is \( x \). Hence, the volume \( V(x) \) is obtained by multiplying these dimensions:

• Base length: \( 36 - 2x \)
• Base width: \( 24 - 2x \)
• Height: \( x \)

Putting all these together, the volume is described by the equation \( V(x) = x(36 - 2x)(24 - 2x) \). This formula captures how different values of \( x \) impact the overall volume of the box.

Domain of a Function

The domain of a function in an optimization problem refers to all possible values that the variable, \( x \), can take. In this context, it's crucial to ensure that the dimensions of the box remain positive after cutting the squares. For the box, the length \( 36 - 2x \) and width \( 24 - 2x \) must both be greater than zero.

• Inequality for length: \( 36 - 2x > 0 \).
• Inequality for width: \( 24 - 2x > 0 \).

Solving these inequalities, we find:

• \( x < 18 \) from the length condition.
• \( x < 12 \) from the width condition.

The stricter condition is \( x < 12 \), since it overlaps the other. Thus, the domain of the function is \( 0 < x < 12 \), ensuring realistic, positive dimensions throughout its range.

Graphical Analysis

Graphical representation helps visualize the relationships in a problem like this one. By plotting the volume function \( V(x) = x(36 - 2x)(24 - 2x) \) over its domain, \( 0 < x < 12 \), we can observe how the volume changes as \( x \). It's insightful because the graph displays the rise and fall of the volume, helping locate its maximum.The maximum volume point usually appears as a peak or crest on the graph. Using graphing technology or manual sketching, you can identify this maximum visually. The value of \( x \) at this peak provides information about the optimal size of the cut squares to achieve the maximum possible volume.Analyzing the graph allows for an immediate understanding of how changes in \( x \) affect the volume, capturing complexities that might be hard to discern through equations alone.

Critical Points

Critical points are key in finding the maximum or minimum values in an optimization problem. They occur where the derivative of a function is zero or undefined. For the volume function \( V(x) = x(36 - 2x)(24 - 2x) \), these are found by solving \( V'(x) = 0 \).Taking the first derivative helps identify where changes in slope occur, which correspond to potential maximum or minimum volumes:

• Calculate the derivative \( V'(x) \).
• Set \( V'(x) = 0 \) to find critical points.

Further analysis with the second derivative test, or by evaluating the first derivative's sign changes around these points, confirms whether each critical point is a maximum or minimum. In the context of the problem, confirming the maximum analytically verifies what was observed graphically, ensuring that the found \( x \) indeed results in the largest box volume possible within the given domain.