Problem 451 - Precalculus

Question

Find the dimensions of the right circular cylinder described. The height is one less than one half the radius. The volume is 72\(\pi\) cubic meters

Step-by-Step Solution

Answer

The dimensions are: radius = 6 meters, height = 2 meters.

Step 1: Understand the Problem

We are given a cylinder where the height is one less than one half of the radius. Mathematically, this can be written as: \( h = \frac{1}{2}r - 1 \). Also, the volume of the cylinder is provided as \( 72\pi \) cubic meters.

Step 2: Write the Volume Formula for Cylinder

The volume \( V \) of a cylinder is given by the formula: \( V = \pi r^2 h \). In our case, it is known that \( V = 72\pi \). Substituting the value, we have: \( \pi r^2 h = 72\pi \). This simplifies to \( r^2 h = 72 \) after canceling \( \pi \) from both sides.

Step 3: Substitute for Height

From the problem, we know that \( h = \frac{1}{2}r - 1 \). Substitute this expression into the volume equation to replace \( h \): \[ r^2 \left( \frac{1}{2}r - 1 \right) = 72 \].

Step 4: Simplify the Equation

Distribute \( r^2 \) into the terms inside the parenthesis: \[ r^2 \cdot \frac{1}{2}r - r^2 = 72 \], which simplifies to \[ \frac{1}{2}r^3 - r^2 = 72 \].

Step 5: Clear Fractions

Multiply the entire equation by 2 to clear the fraction: \[ r^3 - 2r^2 = 144 \].

Step 6: Solve the Cubic Equation

Rearrange the equation to the standard form: \[ r^3 - 2r^2 - 144 = 0 \]. Using trial and error or a calculator, find the real root of this cubic equation. After testing possible rational roots, we find \( r = 6 \).

Step 7: Find the Corresponding Height

Plug \( r = 6 \) back into the original height equation: \[ h = \frac{1}{2}(6) - 1 = 3 - 1 = 2 \].

Step 8: Solution Validation

Check the solution by substituting \( r = 6 \) and \( h = 2 \) back into the volume formula: \[ V = \pi (6)^2 (2) = \pi \times 36 \times 2 = 72\pi \], which is correct by the problem's statement.

Key Concepts

Volume of a Cylinder

To understand the volume of a cylinder, think of the cylinder as a three-dimensional shape with circular bases.
The volume measures how much space the cylinder occupies and is expressed in cubic units.
The formula for the volume of a right circular cylinder is:

• \( V = \pi r^2 h \)

In this formula, \( r \) is the radius of the base, \( h \) is the height of the cylinder, and \( \pi \) is a constant (approximately 3.14159).
This formula derives from multiplying the area of the base circle (\( \pi r^2 \)) by the height of the cylinder when you "stack" the base area up through the height.
This is why understanding how to calculate the area of a circle is key to solving volume problems for cylinders.
In application, if you know any two of these three values (volume, radius, height), you can find the missing one using algebraic manipulation of the formula.

Cubic Equation

A cubic equation represents a polynomial equation of degree three.
In this context, it's an equation that you obtain when solving for one of the cylinder dimensions, often the radius, using the volume formula.
Given our exercise, the cubic equation solved was \( r^3 - 2r^2 - 144 = 0 \).
Solving a cubic equation can be more challenging compared to linear (first-degree) or quadratic (second-degree) equations.
There are various methods to solve cubic equations:

• Factoring: Attempts to express the cubic equation as a product of lower degree polynomials.
• Rational Root Theorem: Helps in guessing rational solutions through possible roots calculations by testing factors of the constant term.
• Graphing or using a calculator: Technology can help visualize the roots as points where the graph meets the x-axis.

In our problem, finding a whole number solution (like \( r = 6 \)) greatly simplified solving and verifying the equation.

Cylinder Dimensions

Determining the dimensions of a cylinder involves knowing or identifying two measurements: the radius and the height.
The original problem provides a scenario to explore how these dimensions are connected and influence each other.
In our example, the relationship was given that the height is "one less than one half the radius," or mathematically, \( h = \frac{1}{2}r - 1 \).
This kind of expression that relates height to radius is critical in solving many geometry problems because it limits the number of unknowns.

You can visualize a cylinder as a soup can where you know its inside height by the depth of the can, and the radius is half of the diameter (diameter being the full width across the base circle).

Knowing these dimensions is vital when planning real-world applications like construction design, packaging, or container manufacturing.

Geometry Problem Solving

Geometry problem solving encompasses using mathematical reasoning to find solutions in geometric contexts.
Problems often involve determining lengths, areas, volumes, or other properties of shapes.

• Understanding Relationships: Grasp the given relationships between different parts of the shape, like the cylinder's height expressed as a function of its radius.
• Logical Steps: Take logical and sequential steps, like using the volume formula. Substitute known values, simplify the equations when necessary.
• Testing & Verifying Solutions: After finding possible solutions like dimensions, substitute back to verify, ensuring calculations match the problem statement.
• Utilizing Technology: Tools like calculators or algebra software help in visualizing and solving complex equations faster.

This process ensures that you're not just trying numbers randomly but making informed decisions based on mathematical principles.
Geometry problem solving in particular strengthens your ability to visualize shapes and understand their properties, which is crucial in various scientific and engineering disciplines.