Problem 117 - College Algebra with Modeling and Visualization

Question

Find the radius and height of a cylindrical container with a volume of 50 cubic inches and a lateral surface area of 65 square inches.

Step-by-Step Solution

Answer

Radius is approximately 1.54 inches and height is approximately 6.71 inches.

Step 1: Understand the formulas

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. The lateral surface area of a cylinder is given by \( A = 2 \pi r h \). We need to solve these equations for \( r \) and \( h \) given \( V = 50 \) and \( A = 65 \).

Step 2: Express height in terms of radius using volume formula

From the volume formula, solve for \( h \): \( 50 = \pi r^2 h \). Rearrange to get \( h = \frac{50}{\pi r^2} \).

Step 3: Substitute height in surface area formula

Substitute \( h = \frac{50}{\pi r^2} \) into the surface area formula: \( 65 = 2 \pi r \left(\frac{50}{\pi r^2}\right) \). Simplify to \( 65 = \frac{100}{r} \).

Step 4: Solve for radius

From \( 65 = \frac{100}{r} \), solve for \( r \) by multiplying both sides by \( r \) and then dividing by 65 to obtain \( r = \frac{100}{65} \approx 1.54 \text{ inches} \).

Step 5: Solve for height using radius

Substitute \( r = 1.54 \) back into the equation for \( h \): \( h = \frac{50}{\pi (1.54)^2} \). Calculate \( h \approx 6.71 \text{ inches} \).

Step 6: Verify the solution

Using \( r = 1.54 \) and \( h = 6.71 \), check if both the volume \( V = \pi r^2 h \approx 50 \) and surface area \( A = 2 \pi r h \approx 65 \) match the original conditions. They do, confirming the solutions are correct.

Key Concepts

Cylinder Volume Formula

The volume of a cylinder expresses the space it occupies within its circular base and perpendicular height. To calculate this, we use the formula: \[V = \pi r^2 h\]where:

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

It is important to understand that the term \(\pi r^2\) calculates the area of the base, while multiplying by the height \(h\) stretches this area vertically, filling it up with volume. This formula is fundamental in many real-world applications allowing us to determine how much space a cylindrical container can hold.

Lateral Surface Area of Cylinder

The lateral surface area of a cylinder refers to the area of the curved surface wrapping around the cylinder. It excludes the top and bottom circles. To find this area, use the formula:\[A = 2 \pi r h\]where:

• \(A\) denotes the lateral surface area,
• \(r\) is the radius of the base,
• \(h\) is the height of the cylinder.

This formula works by essentially "unrolling" the cylinder's side into a rectangular shape. The height of the rectangle is \(h\), while the length is the circumference of the base, expressed as \(2\pi r\). Therefore, this calculation gives the complete boundary area minus the circular caps. Understanding this helps provide insights into questions involving surface coating, labeling, or covering cylindrical materials.

Algebraic Solving Techniques

In problems involving cylinders with given volume and surface area, we often use algebraic solving techniques to find unknown dimensions like radius and height. In such scenarios, using formulas as equations is crucial.First, you isolate one variable. For instance, from the volume formula \(V = \pi r^2 h\), solve for \(h\) to get:\[h = \frac{V}{\pi r^2}\]This allows substitution into another equation, like the lateral surface area \(A = 2\pi r h\). Replacing \(h\) here simplifies the problem to solving for \(r\) alone.Mastering these rearrangements and substitutions is essential:

• It allows us to express one dimension in terms of another.
• It converts complex problems into simpler calculations.
• It ensures that all conditions given in a problem are logically and mathematically satisfied.

By using these algebraic techniques, you clarify and streamline problem-solving, ensuring reliable solutions, which is particularly useful in real-world engineering and design scenarios.