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The cost of making a can is determined by how much aluminum \(A\), in square inches, is needed to make it. This in turn depends on the radius \(r\) and the height \(h\) of the can, both measured in inches. You will need some basic facts about cans. See Figure \(2.110\). The surface of a can may be modeled as consisting of three parts: two circles of radius \(r\) and the surface of a cylinder of radius \(r\) and height \(h\). The area of these circles is \(\pi r^{2}\) each, and the area of the surface of the cylinder is \(2 \pi r h\). The volume of the can is the volume of a cylinder of radius \(r\) and height \(h\), which is \(\pi r^{2} h\). In what follows, we assume that the can must hold 15 cubic inches, and we will look at various cans holding the same volume. a. Explain why the height of any can that holds a volume of 15 cubic inches is given by $$ h=\frac{15}{\pi r^{2}} \text {. } $$ b. Make a graph of the height \(h\) as a function of \(r\), and explain what the graph is showing. c. Is there a value of \(r\) that gives the least height \(h\) ? Explain. d. If \(A\) is the amount of aluminum needed to make the can, explain why $$ A=2 \pi r^{2}+2 \pi r h $$ e. Using the formula for \(h\) from part a, explain why we may also write \(A\) as $$ A=2 \pi r^{2}+\frac{30}{r} $$

Step by step solution

01

Deriving the formula for height

Given the can's volume is a constant 15 cubic inches, use the formula for the volume of a cylinder, which is \( V = \pi r^{2} h = 15 \). Rearranging gives: \( h = \frac{15}{\pi r^{2}} \). This formula shows the relationship between the radius and height needed to maintain a volume of 15 cubic inches.

02

Graphing height as a function of radius

Plot the equation from Step 1, \( h = \frac{15}{\pi r^{2}} \), with \( r \) on the x-axis and \( h \) on the y-axis. The graph is a decreasing curve, showing that as the radius \( r \) increases, the height \( h \) decreases. This illustrates the inverse relationship between radius and height for a fixed volume.

03

Analyzing graph for minimum height

Since \( h \) decreases continuously as \( r \) increases, there is no minimum value for \( h \) that is greater than zero. However, physically, both \( r \) and \( h \) must be positive, so the least practical height occurs as \( r \to \infty \), which is not feasible, implying no specific minimum height for finite values of \( r \).

04

Deriving the formula for aluminum area

The surface area of the can includes the top and bottom circle areas, \( 2\pi r^{2} \), and the side area, \( 2\pi rh \). Combining these, the total surface area \( A \) is \( A = 2 \pi r^{2} + 2 \pi r h \). This sum accounts for all the aluminum needed for the given radius and height.

05

Expressing area using height formula

Substitute the height formula from Step 1 into the aluminum area formula: \( h = \frac{15}{\pi r^{2}} \) into \( A = 2 \pi r^{2} + 2 \pi r h \). This gives \( A = 2 \pi r^{2} + 2 \pi r \left(\frac{15}{\pi r^{2}}\right) = 2 \pi r^{2} + \frac{30}{r} \), simplifying the expression.

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Key Concepts

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

Surface Area

Understanding the surface area of a cylinder is key to solving problems related to can manufacturing. The surface area encompasses the entire outer region of the can, including the top and bottom covers plus the lateral surface area (the part that wraps around the sides). The formula for the surface area of a cylinder involves a combination of areas:

• Two circles at the top and bottom: Each of these has an area of \( \pi r^2 \), where \( r \) is the radius.
• The side or lateral surface: This part can be unwound into a rectangle with height \( h \) and width equivalent to the circumference of the circles, yielding an area of \( 2 \pi r h \).

So, the total surface area \( A \) is expressed as \( A = 2 \pi r^2 + 2 \pi r h \). Understanding how these parts come together helps us comprehend the material calculations for making a can.

Radius-Height Relationship

The radius and height of a cylinder have a particular relationship when the volume is kept constant. For a can with a fixed volume of 15 cubic inches, the height \( h \) changes in relation to the radius \( r \). This relationship is derived from the volume formula for a cylinder, which is \( V = \pi r^2 h \). By substituting 15 for \( V \), we can rearrange to solve for \( h \):

• \( h = \frac{15}{\pi r^2} \)

This means that as the radius increases, the height must decrease to maintain the same volume. Understanding this inverse relationship helps in determining the can's proportions for different situations.

Aluminum Material Usage

Calculating how much material is needed is crucial in production. The amount of aluminum required for a can involves calculating its surface area. As explained, a can's total surface area \( A \) is given by:

• \( A = 2 \pi r^2 + 2 \pi r h \)

When we replace the height \( h \) with the expression \( h = \frac{15}{\pi r^2} \), we derive a new formula for \( A \):

• \( A = 2 \pi r^2 + \frac{30}{r} \)

This formula helps simplify the calculation of aluminum needed by expressing \( A \) solely in terms of \( r \), eliminating direct dependence on \( h \). This breakdown ensures efficient use of materials by understanding how changes in \( r \) affect material use.

Mathematical Modeling

Mathematical modeling plays a vital role in visually illustrating relationships and patterns between diverse variables. By deriving formulas and expressions, such as:

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

We create representations of practical situations like constructing a can. Graphing these expressions shows how variables like radius and height interact. When the height \( h \) is plotted against the radius \( r \), a curve appears, demonstrating how one decreases as the other increases while maintaining a constant volume. This visual form helps both in understanding theoretical aspects and applying them to real-world problem-solving. Understanding and implementing mathematical modeling are fundamental skills in analyzing physical and production-related scenarios efficiently.