91Ó°ÊÓ

This is a continuation of Exercise 8 . The cost of making a can is determined by how much aluminum \(A\), in square inches, is needed to make it. As we saw in Exercise 8, we can express both the height \(h\) and the amount of aluminum \(A\) in terms of the radius \(r\) : $$ \begin{aligned} &h=\frac{15}{\pi r^{2}} \\ &A=2 \pi r^{2}+\frac{30}{r} \end{aligned} $$ a. What is the height, and how much aluminum is needed to make the can, if the radius is 1 inch? (This is a tall, thin can.) b. What is the height, and how much aluminum is needed to make the can, if the radius is 5 inches? (This is a short, fat can.) c. The first two parts of this problem are designed to illustrate that for an aluminum can, different surface areas can enclose the same volume of 15 cubic inches. i. Make a graph of \(A\) versus \(r\) and explain what the graph is showing. ii. What radius should you use to make the can using the least amount of aluminum? iii. What is the height of the can that uses the least amount of aluminum?

Step by step solution

01

Calculate Height and Aluminum for r = 1 inch

Using the formulas from the problem, calculate the height and aluminum area needed when the radius is 1 inch:For height:\[h = \frac{15}{\pi \cdot 1^2} = \frac{15}{\pi} \approx 4.77 \text{ inches}\]For aluminum area:\[A = 2 \pi \cdot 1^2 + \frac{30}{1} = 2\pi + 30 \approx 36.28 \text{ square inches}\]

02

Calculate Height and Aluminum for r = 5 inches

Using the same formulas, calculate the height and aluminum area needed when the radius is 5 inches:For height:\[h = \frac{15}{\pi \cdot 5^2} = \frac{15}{25\pi} = \frac{3}{5\pi} \approx 0.19 \text{ inches}\]For aluminum area:\[A = 2 \pi \cdot 5^2 + \frac{30}{5} = 50\pi + 6 \approx 162.83 \text{ square inches}\]

03

Graph Aluminum Area vs. Radius

Plot the function \( A(r) = 2\pi r^2 + \frac{30}{r} \) against \( r \). This graph will show a trend where, at first, as the radius increases, the surface area decreases, reaches a minimum, and then increases again. This represents the different aluminum areas for the same 15 cubic inch volume.

04

Determine Radius for Minimum Aluminum

Differentiate \( A(r) = 2\pi r^2 + \frac{30}{r} \) with respect to \( r \) and find the critical points by setting the derivative equal to zero:Set \(\frac{dA}{dr} = 0\).\[\frac{dA}{dr} = 4\pi r - \frac{30}{r^2}\]\[\Rightarrow 4\pi r - \frac{30}{r^2} = 0\]\[\Rightarrow 4\pi r^3 = 30\]\[\Rightarrow r^3 = \frac{30}{4\pi}\]\[\Rightarrow r \approx 1.38\]Thus, the radius that minimizes the aluminum is approximately 1.38 inches.

05

Calculate Height for Minimum Aluminum Can

Substitute the radius back into the height formula:\[h = \frac{15}{\pi (1.38)^2} \approx 2.49 \text{ inches}\]Thus, the height of the can using the least amount of aluminum is approximately 2.49 inches.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Surface Area Calculation

Calculating the surface area of three-dimensional objects like cylinders can seem tricky at first, but once you understand the formula, it becomes straightforward. When dealing with a cylinder in optimization problems, you need to calculate the surface area expressed as a function of a variable, such as the radius \( r \). For a cylinder, the total surface area is the sum of the lateral area (the side) and the area of the two circular ends. The lateral area is given by \( 2\pi rh \) and the area of the circle caps is \( 2\pi r^2 \). To find the surface area of a can using a specific formula, substitute values to find exact measurements, as we did for \( r = 1 \) inch and \( r = 5 \) inches in the original equation:

• For height: \( h = \frac{15}{\pi r^2} \).
• For surface area: \( A = 2\pi r^2 + \frac{30}{r} \).

Breaking these tasks into smaller steps helps visualize how each part, such as the height and circumference, contributes to the total surface area. Understanding each segment aids in optimizing the dimensions, finding the smallest possible surface needed for a given volume.

Volume of Cylinders

The volume of a cylinder is easy to calculate if you know the height and radius. The cylinder's volume formula is given by \( V = \pi r^2 h \), where \( r \) is the radius and \( h \) is the height. In our exercise, the volume is constant, precisely 15 cubic inches, demonstrated by substituting the values into the formula. As you can see, even when the volume remains consistent, the configuration of the dimensions can vary:

• When \( r \) is small (e.g.,\( r=1 \) inch), the can becomes tall and thin with a greater height.
• When \( r \) is larger (e.g., \( r=5 \) inches), the can is short and wide with a lesser height.

These variations directly illustrate how changes in the radius affect the overall structure. These understandings guide you toward finding the optimum size for materials used, focusing on how to minimize the surface area while maintaining the same volume.

Graphing Functions

Visualizing mathematical relationships through graphs is an invaluable skill in algebra. It allows you to see how one quantity changes with another visually. In our optimization task, we graph the function \( A(r) = 2\pi r^2 + \frac{30}{r} \), which represents the surface area as a function of the radius. When graphing, observe the characteristic shape:

• The graph starts with a high surface area when \( r \) is small.
• The surface area decreases, reaching its minimum at some optimal \( r \).
• It begins to rise again as \( r \) continues to increase.

This cycle highlights how the smallest radius not only minimizes material usage but also maintains structural integrity for the given volume. To find the optimal radius for the least surface area usage, analyze the graph's critical points, typically where the derivative \( \frac{dA}{dr} \) equals zero. This leads to finding the radius where the slope (rate of change) of this function is zero, indicating a minimum point in cost-effective material application.