Problem 2
Question
An open box of maximum volume is to be made from a square piece of material, 24 inches on a side, by cutting equal squares from the corners and turning up the sides (see figure). (a) Analytically complete six rows of a table such as the one below. (The first two rows are shown.) Use the table to guess the maximum volume. $$ \begin{array}{|c|c|c|} \hline \text { Height } & \begin{array}{c} \text { Length and } \\ \text { Width } \end{array} & \text { Volume } \\ \hline 1 & 24-2(1) & 1[24-2(1)]^{2}=484 \\ \hline 2 & 24-2(2) & 2[24-2(2)]^{2}=800 \\ \hline \end{array} $$ (b) Write the volume \(V\) as a function of \(x\). (c) Use calculus to find the critical number of the function in part (b) and find the maximum value. (d) Use a graphing utility to graph the function in part (b) and verify the maximum volume from the graph.
Step-by-Step Solution
Verified Answer
The maximum volume of the box is obtained at the critical number found by setting the derivative of the volume function \(f(x)=x(24-2x)^2\) equal to zero.
1Step 1: Complete the Table
Proceed to complete the table for the next four heights, that is, x=3, x=4, x=5, and x=6. For every row, calculate the Length and Width by using \(24-2x\) and the volume by using \(x[24-2x]^2\).
2Step 2: Formulate the Volume function
Formulate the volume function as a function of \(x\). The volume \(V\) is given by length * width * height. Since the length and width are the same, the volume function \(V\) is given by \(V=x(24-2x)^2\).
3Step 3: Find the Critical Number
Utilize calculus to find the critical number for the function defined in the prior step. The critical number of a function is where the derivative equals zero or is undefined. Derive \(V=x(24-2x)^2\), set it equal to 0, and solve for \(x\). That gives the critical number.
4Step 4: Use Calculus to Find the Maximum Value
To find the maximum value, use the second derivative test. Calculate the second derivative of the volume function and substitute the critical number. If the result is less than zero, the volume function has a maximum at the critical number. Calculate the maximum volume by substituting the critical number in the volume function.
5Step 5: Graph the Volume Function
Utilize a graphing tool to represent the function \(f(x)=x(24-2x)^2\) and confirm the maximum volume visually. This graph will peak at the critical number and the height of the graph at this point is the maximum volume.
Key Concepts
Critical PointsVolume FunctionSecond Derivative Test
Critical Points
When solving optimization problems in calculus, finding the **critical points** of a function is a crucial step. Critical points help identify where potential maxima and minima occur. These are the values where the derivative of the function is either zero or undefined. In simpler terms, if you think of the function as a landscape, the critical points would be the flat areas and peaks.
To find them, you take the derivative of the function. For example, for the volume function \[V = x(24 - 2x)^2,\] first, differentiate it with respect to \(x\). This differentiating process gives you an expression that indicates how volume changes as \(x\) changes.
Setting this derivative equal to zero helps locate critical points. By solving \[dV/dx = 0,\] you identify the \(x\) values where these points occur. These are the possible candidates for maximum or minimum volumes of the box, highlighting areas to focus on further analysis.
To find them, you take the derivative of the function. For example, for the volume function \[V = x(24 - 2x)^2,\] first, differentiate it with respect to \(x\). This differentiating process gives you an expression that indicates how volume changes as \(x\) changes.
Setting this derivative equal to zero helps locate critical points. By solving \[dV/dx = 0,\] you identify the \(x\) values where these points occur. These are the possible candidates for maximum or minimum volumes of the box, highlighting areas to focus on further analysis.
Volume Function
Understanding the **volume function** is essential in solving optimization problems where you aim to maximize or minimize space or capacity. In our specific problem, we are interested in maximizing the volume of an open box.
The volume of our box is calculated using the formula:\[V = x(24 - 2x)^2,\]where \(x\) is the height of the box after cutting out squares from the corners of the sheet.
Here's how it works:
The volume of our box is calculated using the formula:\[V = x(24 - 2x)^2,\]where \(x\) is the height of the box after cutting out squares from the corners of the sheet.
Here's how it works:
- The initial side length of the square material is 24 inches.
- For each corner, you cut out a square with side length \(x\).
- After cutting, the new size of the base becomes \(24 - 2x\) for both length and width.
Second Derivative Test
The **second derivative test** is a powerful calculus tool for identifying whether a critical point is a maximum or a minimum. After locating the critical points using the first derivative, as explained earlier, the next step is applying this secondary test to determine if these points will yield the desired outcome.
The second derivative of the volume function **V** is found by differentiating the first derivative.
If the second derivative, denoted as \(d^2V/dx^2\), is less than zero at a critical point, that point is a local maximum. In simple terms, if you plot the volume changes, this would look like a peak in the graph, indicating a maximum volume.
For our box volume problem, use the test on the critical values found. Substitute the \(x\) values back into the second derivative:
The second derivative of the volume function **V** is found by differentiating the first derivative.
If the second derivative, denoted as \(d^2V/dx^2\), is less than zero at a critical point, that point is a local maximum. In simple terms, if you plot the volume changes, this would look like a peak in the graph, indicating a maximum volume.
For our box volume problem, use the test on the critical values found. Substitute the \(x\) values back into the second derivative:
- If \(d^2V/dx^2 < 0\), it confirms that the box’s volume is maximized at those specific \(x\) values.
- This lets you confidently calculate the maximum achievable volume using the volume function formula \(V = x(24-2x)^2\) for any confirmed critical point.
Other exercises in this chapter
Problem 1
In Exercises \(1-4,\) explain why Rolle's Theorem does not apply to the function even though there exist \(a\) and \(b\) such that \(f(a)=f(b)\). $$ f(x)=1-|x-1
View solution Problem 2
Determine the open intervals on which the graph is concave upward or concave downward. \(y=-x^{3}+3 x^{2}-2\)
View solution Problem 2
In Exercises 1 and \(2,\) describe in your own words what the statement means. $$ \lim _{x \rightarrow-\infty} f(x)=2 $$
View solution Problem 2
Explain why Rolle's Theorem does not apply to the function even though there exist \(a\) and \(b\) such that \(f(a)=f(b)\). $$ f(x)=\cot \frac{x}{2}, \quad[\pi,
View solution