Problem 452 - Precalculus

Question

Find the dimensions of the right circular cylinder described. The radius and height differ by one meter. The radius is larger and the volume is 48\(\pi\) cubic meters.

Step-by-Step Solution

Answer

The cylinder has a radius of 4 meters and a height of 3 meters.

Step 1: Understand the Problem

We need to find both the radius and height of the cylinder with given conditions. The radius is one meter longer than the height, and the volume is given as 48\(\pi\) cubic meters.

Step 2: Use Formula for Cylinder Volume

The formula for the volume of a cylinder is \(V = \pi r^2 h\), where \(r\) is the radius and \(h\) is the height. Set this equal to the given volume: \(\pi r^2 h = 48\pi\).

Step 3: Substitute for Height

Since the radius is one meter longer than the height, we can express the height as \(h = r - 1\). Substitute this into the volume equation: \(\pi r^2 (r - 1) = 48\pi\).

Step 4: Solve for Radius

Cancel \(\pi\) from both sides and expand: \(r^3 - r^2 = 48\). Rearrange to find \(r\): \(r^3 - r^2 - 48 = 0\).

Step 5: Factor the Equation

Factor the cubic equation \(r^3 - r^2 - 48 = 0\). Check possible rational roots or use synthetic division. After factoring, you find \((r - 4)(r^2 + r + 12) = 0\).

Step 6: Solve Factored Equation

The quadratic factor \(r^2 + r + 12\) has no real solutions (discriminant \(b^2 - 4ac = 1 - 48 = -47\)), so \(r = 4\) is the only real solution.

Step 7: Find Height

With \(r = 4\), use the relationship \(h = r - 1\) to find the height: \(h = 4 - 1 = 3\).

Key Concepts

Right Circular Cylinder

A right circular cylinder is a 3-dimensional shape that has two parallel and congruent circular bases and a curved surface that connects the bases. This shape is quite common in real-world applications and can be visualized by thinking of objects like a can or a battery. The key features of a cylinder include its height, which is the perpendicular distance between the bases, and its radius, which is the distance from the center of the base to its edge.
To calculate the volume of a right circular cylinder, you use 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. This equation helps determine how much space is inside the cylinder. It's important to note that both the radius and height must be in the same unit for this equation to be correct.
Understanding the basic geometry and formulas related to cylinders is essential for solving problems involving these shapes.

Radius and Height Relationship

In the context of the given problem, the radius and height of the cylinder have a specific relationship: the radius is one meter longer than the height. This information can be quite helpful as it allows us to express one variable in terms of the other, simplifying the solving process. By stating that the radius, \(r\), is one meter longer than the height, \(h\), we can write this relationship as:

• \[ h = r - 1 \]

This equation allows us to substitute for \(h\) in the volume formula, reducing the number of variables and making it possible to solve for \(r\) directly. Once we know \(r\), finding \(h\) becomes straightforward using the relationship \(h = r - 1\).
Understanding and utilizing such relationships between variables is a crucial skill in algebra and geometric problem solving.

Factoring Cubic Equations

Factoring cubic equations can be a bit tricky but is crucial for finding solutions to specific problems, like the one in this exercise. The original equation after substituting for height was:

• \[ r^3 - r^2 - 48 = 0 \]

Solving for \(r\) involves factoring this cubic equation. A practical approach for factoring is to find rational roots using the Rational Root Theorem or performing synthetic division. Once a root is found, it can be used to simplify the equation further.
The equation factors into:

• \[ (r - 4)(r^2 + r + 12) = 0 \]

The quadratic \(r^2 + r + 12\) has no real roots because its discriminant (\(b^2 - 4ac\)) is negative. Thus, the only real solution comes from \(r - 4 = 0\), giving \(r = 4\).
Factoring is a powerful algebraic tool that helps in finding solutions to polynomial equations by breaking them down into simpler parts.