Problem 50 - College Algebra and Calculus: An Applied Approach

Question

Dimensions of a Box An open box is to be made from a rectangular piece of material, 16 inches by 12 inches, by cutting equal squares from the corners and tuming up the sides (see figure). (a) Write the volume \(V\) of the box as a function of \(x\). Determine the domain of the function. (b) Sketch the graph of the function and approximate the dimensions of the box that yield a maximum volume. (c) Find values of \(x\) such that \(V=120\). Which of these values is a physical impossibility in the construction of the box? Explain. (d) What value of \(x\) should you use to make the tallest possible box with a volume of 120 cubic inches?

Step-by-Step Solution

Answer

The volume function is \(V(x) = x(16 - 2x)(12 - 2x)\) with a domain of \([0, 6]\). The maximum volume appears to occur at approximately \(x = 2\). For a volume of 120 cubic inches, multiple values of \(x\) will satisfy, with any negative values being physically impractical. The tallest box with a volume of 120 cubic inches is obtained at the valid \(x\) which is greatest.

Step 1: Formulate the Volume Function and its Domain

After the squares are cut off and the sides are turned up, the new length and width are \(16 - 2x\) and \(12 - 2x\) respectively, and the height is \(x\). Thus, the volume of the box is given by the function \(V(x) = x(16 - 2x)(12 - 2x)\). The squares cut off cannot be larger than half the smaller side of the rectangle since we need at least two squares. Thus, the domain of \(x\) is \([0, 6]\).

Step 2: Sketch Function Graph and Approximate Maximum Volume

Sketch the function \(V(x)\) within the domain. This is a cubic function with turning points. By analyzing the graph, the maximum volume occurs approximately at \(x = 2\) inches.

Step 3: Solve for \(x\) with \(V = 120\)

To find the values of \(x\) such that \(V = 120\), solve the equation \(x(16 - 2x)(12 - 2x) = 120\). There will be multiple solutions, but remember only those in the interval \([0, 6]\) are physically practical. The negative solution should be discarded, as it is not possible to cut off a negative square from the rectangle.

Step 4: Find Maximum Height Value for Defined Volume

To find the value of \(x\) that will give the maximum height for a box of volume 120 cubic inches, check the \(x\) values obtained in the last step. The value that gives the maximum height (remember \(x\) is also the height) while keeping within the 120 cubic inch volume will be picked.

Key Concepts

Volume Function

Understanding the volume function is crucial in solving optimization problems in geometry. In our exercise, the volume function for the open box, denoted as \(V(x)\), is derived from the dimensions created after cutting squares from each corner of a rectangular piece of material. The original material has dimensions of 16 inches by 12 inches, and the squares cut out have sides of length \(x\). Therefore, the new length and width of the material are \(16 - 2x\) and \(12 - 2x\) respectively, with \(x\) also representing the height of the box.

When you multiply these dimensions together, you get the volume function \(V(x) = x(16 - 2x)(12 - 2x)\), which represents the volume of the box as a function of the size of the squares that are cut out. This equation is the backbone of our optimization problem, as it will be used to determine the domain of possible values for \(x\) and ultimately find the dimensions that maximize the volume of the box.

Domain of Function

The domain of a function is the set of all possible input values (in this case, \(x\)) for which the function is defined. For the volume function \(V(x)\), the domain is restricted by the physical limitations of our box. Specifically, since the squares being cut cannot have sides larger than half the length of the shorter side of the original rectangle, the maximum value for \(x\) is 6 inches. Hence, the domain of the volume function is \([0, 6]\), where 0 represents the scenario where no material is removed, and 6 inches is the maximum size of the square cuts possible without negating the structure of the box. It's important for students to remember that the domain must be considered to keep solutions realistic and physical.

Graph of Function

Graphing the volume function \(V(x)\) provides a visual representation that can help identify the box dimensions that yield the maximum volume. The graph of a cubic function like ours will generally have a characteristic 'S' shape with turning points. By plotting \(V(x)\) within the domain \([0, 6]\), we observe that the volume increases to a certain point as \(x\) increases, but then begins to decrease past a specific value of \(x\).

This point where the volume is at its maximum before decline is the turning point we are interested in for this optimization problem. The sketch of the graph allows us to approximate this maximum volume point without exact calculations, but it is the calculus—in the form of finding the derivative and setting it to zero—that could give us the exact value of \(x\) for the maximum volume.

Maximum Volume Calculation

Calculating the maximum volume of the box is the crux of our optimization problem. To find the value of \(x\) that maximizes \(V(x)\), we can either complete a thorough analysis of the graph or apply calculus to derive the exact value mathematically. However, using the graph, the maximum volume occurs approximately at \(x = 2\) inches.

In addition to understanding the graph, solving the derivative of the volume function and setting it to zero provides the precise value of \(x\) for which the volume is maximum. Once we have this value, we cross-check it with our domain to ensure it's a feasible solution. Understanding this process enables students to solve complex volume optimization problems systematically and accurately, reinforcing their grasp of the overarching concept.