Problem 79 - Algebra and Trigonometry

Question

Volume of a Box An open box is to be constructed from a piece of cardboard 20 cm by 40 cm by cutting squares of side length x from each corner and folding up the sides, as shown in the figure. $$ \begin{array}{l}{\text { (a) Express the volume } V \text { of the box as a function of } x .} \\ {\text { (b) What is the domain of } V ? \text { (Use the fact that length and }} \\ {\text { volume must be positive.) }} \\\ {\text { (c) Draw a graph of the function } V \text { and use it to estimate }} \\ {\text { the maximum volume for such a box. }}\end{array} $$

Step-by-Step Solution

Answer

The maximum volume occurs around \( x = 3.33 \) cm with a peak volume near 592.59 cm³.

Step 1: Understand the Box Construction

The box is made by cutting squares of side length \( x \) from each corner of a 20 cm by 40 cm piece of cardboard and folding up the sides. The new dimensions of the box will be \((40-2x)\) cm in length, \((20-2x)\) cm in width, and \(x\) cm in height.

Step 2: Express the Volume Function

The volume \( V \) of a box is calculated by multiplying its length, width, and height. Here, the volume is given by: \[ V(x) = x (40-2x)(20-2x) \]. This equation represents the volume of the box as a function of \( x \).

Step 3: Determine the Domain of the Function

Since \( x \) represents a length, it must be positive. Moreover, \( x \) cannot be greater than half of the smallest original dimension (otherwise negative dimensions result post-cuts), which is 20 cm. Thus, \( 0 < x < 10 \).

Step 4: Graph the Volume Function

Plot the function \( V(x) = x(40-2x)(20-2x) \) within the domain \( 0 < x < 10 \). The graph is a cubic parabola opening downwards, and it can help visualize the maximum volume.

Step 5: Estimate the Maximum Volume from the Graph

By observing the graph plotted, estimate the value of \( x \) that gives the maximum volume. This is the value at the peak of the curve. Alternatively, calculate the derivative of \( V(x) \) and solve \( V'(x) = 0 \) to find the critical points for exact calculations.

Key Concepts

Function of x

The volume of the open box is described as a mathematical function of the variable \( x \). This function represents how the volume changes as \( x \) changes. When we cut squares of side \( x \) from the cardboard, the resulting box has new dimensions:

• Length: \( (40-2x) \)
• Width: \( (20-2x) \)
• Height: \( x \)

Therefore, the volume \( V \) as a function of \( x \) is defined as: \[ V(x) = x (40-2x)(20-2x) \]This function relates the side length of the squares we cut (\( x \)) to the total volume \( V \) of the box created.

Domain of the Function

The domain of a function is the set of all possible values of \( x \) that can be plugged into the function to produce a valid output. For our volume function, we have specific conditions to consider:

• The value of \( x \) must be positive, as it represents a physical dimension.
• \( x \) must also be less than half of the smallest dimension of the cardboard, which is 20 cm, because a larger \( x \) leads to impossible negative dimensions. Thus, \( x < 10 \).

Putting these conditions together, the domain of our volume function \( V(x) \) is:0 < x < 10This means \( x \) can take any value between 0 and 10, not including the endpoints.

Maximum Volume Estimation

To find the maximum volume of the box, we need to analyze how the volume changes with different values of \( x \). The key tool here is the cubic function \( V(x) = x(40-2x)(20-2x) \).A practical approach is to graph the function and look for the highest point, known as the peak. This highest point will visually show the maximum volume. Another precise method is to find the derivative of \( V(x) \), set it to zero, and solve for \( x \). This critical point calculation mathematically pinpoints the maximum volume along the curve in the graph.

Graph of Function V

Visualizing mathematical functions is greatly aided by graphing. For the volume function \( V(x) = x(40-2x)(20-2x) \), graphing within the domain 0 < x < 10 provides a clear picture of volume behavior.The graph of this cubic function is a downward-opening parabola. This shape is typical for a volume function of this type, where volume starts small, increases to a maximum, and then decreases back down as \( x \) continues to increase. By plotting this graph, we can easily identify the maximum volume. The peak of the graph indicates this highest volume and guides us in determining the best \( x \) to use for achieving it.