Problem 20 - A Graphical Approach to College Algebra

Question

Solve each problem. Geometry \(\quad\) A cylindrical aluminum can is being constructed to have a height \(h\) of 4 inches. If the can is to have a volume of 28 cubic inches, approximate its radius \(r\) (Hint: \(\left.V=\pi r^{2} h .\right)\)

Step-by-Step Solution

Answer

The radius of the can is approximately 1.49 inches.

Step 1: Understand the Problem

We need to find the radius of a cylindrical can with a given height of 4 inches and a volume of 28 cubic inches. We will use the formula for the volume of a cylinder.

Step 2: Recall the Formula

The volume of a cylinder is calculated using the formula \[ V = \pi r^2 h \]where:- \(V\) is the volume,- \(r\) is the radius of the base, and- \(h\) is the height of the cylinder.

Step 3: Substitute Known Values into the Formula

We substitute the known values, volume \(V = 28\) cubic inches and height \(h = 4\) inches, into the formula:\[ 28 = \pi r^2 \times 4 \]

Step 4: Solve for \(r^2\)

First, divide both sides of the equation by 4 to isolate \(\pi r^2\):\[ \frac{28}{4} = \pi r^2 \]which simplifies to:\[ 7 = \pi r^2 \]

Step 5: Solve for \(r\)

Now, divide by \(\pi\) to solve for \(r^2\):\[ r^2 = \frac{7}{\pi} \]Find the square root:\[ r = \sqrt{\frac{7}{\pi}} \]

Step 6: Calculate the Approximate Radius

Using a calculator, approximate the value of \(\sqrt{\frac{7}{\pi}}\). Using \(\pi \approx 3.14159\):\[ r \approx \sqrt{\frac{7}{3.14159}} \approx \sqrt{2.228} \approx 1.49\]

Key Concepts

Cylindrical Volume Calculation

The core concept when calculating the volume of a cylinder involves understanding the geometric properties of this shape. A cylinder's volume is determined by the formula \[ V = \pi r^2 h \]where:

• \( V \) is the volume,
• \( r \) is the radius of the base,
• \( h \) is the height of the cylinder.

This formula originates from the idea that the volume of a solid is the area of its base times its height. For a cylinder, the base is a circle, and the area of a circle is calculated as \( \pi r^2 \). Therefore, multiplying this area by the cylinder's height, \( h \), gives you the total volume.
In practice, if you know two out of the three variables—\( V \), \( r \), or \( h \)—you can rearrange this formula to solve for the missing parameter. For example, if you are given the volume and height, you can isolate \( r \) to find the radius.

Solving Equations

Solving equations is a critical skill in algebra and involves finding the value of an unknown variable. In the context of geometry, like in the cylindrical volume calculation, we often isolate the variable in question through algebraic manipulation.
In our specific problem:

• Start by using the formula for the volume of a cylinder, \( V = \pi r^2 h \), and substitute known values.
• To isolate \( r^2 \), divide both sides by any coefficients associated with it.
• After solving for \( r^2 \), take the square root to find \( r \).
• Finally, compute the numerical approximation using a calculator, which simplifies the mathematics for easier understanding.

By carefully manipulating equations as described, you can find unknown values in various mathematical problems. This skill applies broadly in sciences and everyday situations where determining an unknown is necessary.

Geometry Formulas

Geometry involves various formulas that help to solve real-world problems involving shapes and spaces. In our problem concerning the aluminum can, we use the formula for the volume of a cylinder. It's one among many geometry formulas used to find measurements like volume, area, or perimeter.
Some essential geometry formulas include:

• Circle area: \( A = \pi r^2 \)
• Rectangle area: \( A = l \times w \)
• Triangle area: \( A = \frac{1}{2} b \times h \)
• Volume of a sphere: \( V = \frac{4}{3} \pi r^3 \)

Understanding these fundamental formulas and their derivations is crucial for solving more complex geometry problems. They provide the means to calculate essential characteristics of objects and shapes in a straightforward manner.