Problem 77 - Algebra for College Students

Question

Set up an equation and solve each of the following problems. The total surface area of a right circular cylinder is \(54 \pi\) square inches. If the altitude of the cylinder is twice the length of a radius, find the altitude of the cylinder.

Step-by-Step Solution

Answer

The altitude of the cylinder is 6 inches.

Step 1: Understand the Problem

The total surface area (SA) of a cylinder is given as \(54\pi\) square inches, and the altitude (height) of the cylinder is twice the radius. We need to find the height.

Step 2: Recall the Surface Area Formula

The formula for the surface area of a right circular cylinder is: \[ SA = 2\pi r^2 + 2\pi r h \] where \( r \) is the radius and \( h \) is the height of the cylinder.

Step 3: Express the Height in Terms of the Radius

Since the height \( h \) is twice the radius \( r \), we substitute \( h = 2r \) into the surface area formula.

Step 4: Substitute and Set Up Equation

Substituting \( h = 2r \) into the surface area formula yields:\[ 2\pi r^2 + 2\pi r (2r) = 54\pi \]Simplify this to:\[ 2\pi r^2 + 4\pi r^2 = 54\pi \]

Step 5: Simplify the Equation

Combine like terms:\[ 6\pi r^2 = 54\pi \]

Step 6: Solve for the Radius \( r \)

Cancel out \( \pi \) from both sides:\[ 6r^2 = 54 \]Divide both sides by 6:\[ r^2 = 9 \]Take the square root of both sides:\[ r = 3 \]

Step 7: Find the Altitude \( h \)

Since \( h = 2r \), substitute \( r = 3 \) to find:\[ h = 2(3) = 6 \]

Key Concepts

Understanding Cylinder Surface Area

The surface area of a cylinder is a key concept in geometry that helps us measure how much space the surface of the cylinder occupies. For a right circular cylinder, like the one in the exercise, the surface area consists of two parts: the area of the two circular bases, and the area of the curved surface that connects these bases.

The formula for calculating the surface area of a cylinder is:

• The area of the two circular bases: \(2\pi r^2\) (where \(r\) is the radius)
• The area of the curved surface: \(2\pi rh\) (where \(h\) is the height/altitude)

By combining these two components, the total surface area \(SA\) is given by the equation: \[ SA = 2\pi r^2 + 2\pi rh \]
In the problem, we know that this total surface area equals \(54\pi\) square inches. Understanding how to apply this formula allows us to find unknown dimensions, like the radius or the height, if some other relation between the cylinder's parameters is given.

Applying Geometric Problem Solving

Geometric problem solving often requires us to visualize the problem and translate conditions into equations we can solve. In the given exercise, we started with the surface area equation, already set in terms of radius \(r\) and height \(h\). The problem becomes more manageable by scrutinizing the relationships between these measurements.

Given that the height is twice the radius (\(h = 2r\)), substituting this into the formula helps break down the problem step by step. Importantly, identifying these relationships is key to transform complex-looking problems into simple algebraic expressions.

• Recognize the relationships (\(h = 2r\))
• Substitute these relationships into known formulas
• Simplify the resulting expressions

This systematic approach ensures we solve for a single variable at a time, making even challenging problems easier to handle.

Working with Algebraic Expressions

Algebraic expressions form the backbone of solving mathematical problems involving unknown quantities. In this exercise, the expression derived from the cylinder's surface area equation helps us solve for the radius.

Once the substitution \(h = 2r\) is made into the equation, we're left with \[ 2\pi r^2 + 4\pi r^2 = 54\pi \],which simplifies further as \[ 6\pi r^2 = 54\pi \].
The next step is to solve for \(r\) by simplifying algebraic expressions.

• First, cancel by dividing both sides by \(\pi\).
• Then, divide by 6 to isolate \(r^2\).
• Finally, solve for \(r\) by taking the square root on both sides.

This process illustrates how algebraic techniques break down complex expressions to derive values, leading to the discovery of the cylinder's dimensions effectively.