Problem 90 - College Algebra with Modeling and Visualization

Question

A cylindrical aluminum can is being manufactured so that its height \(h\) is 8 centimeters more than its radius \(r\). Estimate values for the radius (to the nearest hundredth) that result in the can having a volume between 1000 and 1500 cubic centimeters inclusive.

Step-by-Step Solution

Answer

The radius is between approximately 5.06 cm and 5.87 cm.

Step 1: Understanding the Problem

We have a cylindrical can with a height, which is 8 centimeters more than its radius. We need to find the radius values that make the volume of the cylinder between 1000 and 1500 cubic centimeters.

Step 2: Volume Formula for Cylinder

The volume of a cylinder is given by the formula \( V = \pi r^2 h \), where \( r \) is the radius and \( h \) is the height. Here, \( h = r + 8 \), so the volume becomes \( V = \pi r^2 (r + 8) \).

Step 3: Set up Inequality

Since the volume must be between 1000 and 1500 cubic centimeters, set up the inequality: \[1000 \leq \pi r^2 (r + 8) \leq 1500\]

Step 4: Solve the Inequality

We'll solve the inequality using the endpoints:1. Find the roots of \(\pi r^2 (r + 8) = 1000\)2. Find the roots of \(\pi r^2 (r + 8) = 1500\)For both equations:\[ r^3 + 8r^2 = \frac{1000}{\pi} \text{ and } r^3 + 8r^2 = \frac{1500}{\pi} \] We'll solve these cubic equations to find the possible values for \( r \).

Step 5: Approximate Solutions Using Technology

Using numerical methods or graphing calculators, approximately solve the equations:- For \( \pi r^2 (r + 8) = 1000 \), \( r \approx 5.06 \).- For \( \pi r^2 (r + 8) = 1500 \), \( r \approx 5.87 \).

Step 6: Determine Radius Range

The possible radius values that give a volume between 1000 and 1500 cubic centimeters are approximately from \( r = 5.06 \) to \( r = 5.87 \). Round these values to the nearest hundredth.

Key Concepts

Cylindrical Geometry

Understanding cylindrical geometry is essential for solving problems related to volumes of cylinders, like in this exercise. A cylinder is a 3D shape with two parallel bases connected by a curved surface. These bases are congruent circles, giving the cylinder its unique look. The key measurements of a cylinder include:

• Radius (\( r \)), which is the distance from the center to the edge of the base circle.
• Height (\( h \)), which is the perpendicular distance between the two bases.

For this problem, the height of the cylinder is 8 centimeters more than the radius. Therefore, it is represented as \( h = r + 8 \).
Once the height and radius are known, the volume (\( V \)) of the cylinder is calculated using the formula \( V = \pi r^2 h \). Understanding this geometry helps in setting up and solving the problem effectively.

Inequality Solving

Inequalities are mathematical expressions that determine the range within which a certain condition holds true. For this problem, inequalities help to find the range of radius values that will result in a volume between 1000 and 1500 cubic centimeters. The inequality is given by:

• \( 1000 \leq \pi r^2 (r + 8) \leq 1500 \).

Solving this involves breaking it down into separate equations \( \pi r^2 (r + 8) = 1000 \) and \( \pi r^2 (r + 8) = 1500 \), then finding the roots.
These roots represent potential values for the radius that satisfy the volume condition set by the inequality. Working through inequalities requires understanding how each component of the inequality interacts to form a valid solution range for the radius.

Numerical Methods

Numerical methods are techniques used for finding approximate solutions to mathematical problems that cannot be solved precisely by algebraic means. In this problem, when dealing with cubic equations derived from the given inequality, numerical methods like graphing calculators or software tools are employed.
This approach helps in quickly approximating roots of the equations:

• For equation \( \pi r^2 (r + 8) = 1000 \), solutions give \( r \approx 5.06 \).
• For equation \( \pi r^2 (r + 8) = 1500 \), solutions yield \( r \approx 5.87 \).

Using numerical methods facilitates accurate approximations, especially when precise calculations are complex or impossible. These tools help visualize and verify potential solutions, providing a practical approach to solving such real-world exercise problems.