Problem 109 - College Algebra with Modeling and Visualization

Question

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: \(V=\pi r^{2} h\).)

Step-by-Step Solution

Answer

The radius is approximately 1.49 inches.

Step 1: Understanding the Formula

The volume formula for a cylinder is given by the equation \(V = \pi r^2 h\). This represents how to calculate the volume based on the can's radius \(r\) and height \(h\). In this problem, \(V=28\) cubic inches and \(h=4\) inches. We need to find \(r\).

Step 2: Rearranging the Formula

We need to solve for \(r\) in the equation \(28 = \pi r^2 \times 4\). First, isolate \(r^2\) by dividing both sides of the equation by \(4\pi\).

Step 3: Solving for r^2

After dividing both sides by \(4\pi\), we have the equation \(r^2 = \frac{28}{4\pi}\). Simplify this to \(r^2 = \frac{7}{\pi}\).

Step 4: Finding the Radius

Take the square root of both sides of the equation to solve for \(r\). This gives \(r = \sqrt{\frac{7}{\pi}}\).

Step 5: Calculating Numerical Approximation

Calculate \(r = \sqrt{\frac{7}{\pi}}\) using a calculator for a more precise value. After computing, we find that \(r \approx 1.49\) inches.

Key Concepts

Radius Calculation

In order to find the radius of a cylinder, we need to have a grasp of its geometric properties. A cylinder can be visualized as a solid shape with two identical circular bases and a certain height. The radius is the distance from the center of any circle in the cylinder to its edge.
For this problem, we start with the formula for the volume of a cylinder, which is: \[ V = \pi r^2 h \] where

• \( V \) is the volume,
• \( r \) is the radius of the base,
• \( h \) is the height of the cylinder,
• \( \pi \) is a mathematical constant approximately equal to 3.14159.

From the given problem, we know the volume \( V = 28 \) cubic inches and the height \( h = 4 \) inches. We must calculate the radius \( r \). To find the radius, we manipulate the formula to solve for \( r \) by rearranging it.

Algebraic Manipulation

Algebraic manipulation is an important skill in solving equations involving geometric shapes like cylinders. When rearranging equations, we aim to isolate the variable of interest. In this case, we want to isolate \( r^2 \) from the volume formula.
Starting with the equation: \[ 28 = \pi r^2 \times 4 \] We perform the following steps:

• Divide both sides of the equation by \( 4 \pi \) to isolate \( r^2 \).
• This results in: \[ r^2 = \frac{28}{4 \pi} \]

From here, we simplify the fraction on the right:

• The fraction \( \frac{28}{4} \) simplifies to \( 7 \),
• leading to: \[ r^2 = \frac{7}{\pi} \].

Step-by-Step Solution

Solving for the radius involves taking the square root of both sides of the equation. Once we have isolated \( r^2 \) as \( r^2 = \frac{7}{\pi} \), the final step is easy.
To find \( r \), calculate the square root:

• \( r = \sqrt{\frac{7}{\pi}} \)

Now, this gives us a mathematical expression. To find a numerical approximation:

• Use a calculator to perform \( \sqrt{\frac{7}{\pi}} \),
• which results in approximately \( r \approx 1.49 \) inches.

This value provides a practical measure of the radius and completes the solution to the problem. By following a step-by-step approach, we've made sure each step logically follows from the previous one, ensuring clarity in the solution process.