A cylindrical water heater in your garage is 4 feet tall and 30 inches in diameter. To save money on the electric bill, you have decided to cover the lateral surface and the top with an insulation material (bottom cannot be covered because tank too heavy to lift).
A cylindrical water heater in your garage is 4 feet tall and 30 inches in diameter. To save money on the electric bill, you have decided to cover the lateral surface and the top with an insulation material (bottom cannot be covered because tank too heavy to lift).
Elementary Geometry For College Students, 7e
7th Edition
ISBN:9781337614085
Author:Alexander, Daniel C.; Koeberlein, Geralyn M.
Publisher:Alexander, Daniel C.; Koeberlein, Geralyn M.
ChapterP: Preliminary Concepts
SectionP.CT: Test
Problem 1CT
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![### Problem Statement
A cylindrical water heater in your garage is \(4\) feet \(6\) inches (or \(4 \frac{1}{2}\) feet) tall and has a diameter of \(30\) inches.
To save money on the electric bill, you have decided to cover the lateral surface and the top with an insulation material (the bottom cannot be covered because the tank is too heavy to lift).
**Given**:
- Insulation is sold by the square foot and costs \($0.50\) per square foot.
**Question**:
What is the cost of insulating the water heater?
### Solution Step by Step
**1. Conversion and Formulas Needed**:
First, convert all measurements to feet for consistency:
- Diameter \(= 30\) inches \(\div 12 = 2.5\) feet.
- Radius \(= \frac{Diameter}{2} = 1.25\) feet.
- Height \(= 4 \frac{1}{2}\) feet \(= 4.5\) feet.
**Formulas**:
- Lateral Surface Area of Cylinder \( = 2\pi rh \)
- Area of the Top Circle \(= \pi r^2\)
**2. Calculate Lateral Surface Area**:
\[
Lateral\ Surface\ Area = 2 \pi rh = 2 \pi (1.25\ \text{feet})(4.5\ \text{feet})
\]
\[
= 2 \pi (5.625)\ \text{square feet} \approx 35.343\ \text{square feet} \ (using\ \pi \approx 3.14).
\]
**3. Calculate Top Surface Area**:
\[
Top\ Surface\ Area = \pi r^2 = \pi (1.25\ \text{feet})^2 \approx 4.91\ \text{square feet}
\]
**4. Total Surface Area to Cover**:
\[
Total\ Surface\ Area = Lateral\ Surface\ Area + Top\ Surface\ Area
\]
\[
\approx 35.343\ \text{square feet} + 4.91\ \text{square feet} \approx 40.253\ \text{square feet}
\]
**5. Calculate the Cost**:
Since insulation is sold by the square foot, round the](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F604cf164-1f6b-4733-bc3a-b747e5cf0678%2F4af49c19-ffaf-4197-ac42-58bead2d53b9%2F9z3lqfm_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Problem Statement
A cylindrical water heater in your garage is \(4\) feet \(6\) inches (or \(4 \frac{1}{2}\) feet) tall and has a diameter of \(30\) inches.
To save money on the electric bill, you have decided to cover the lateral surface and the top with an insulation material (the bottom cannot be covered because the tank is too heavy to lift).
**Given**:
- Insulation is sold by the square foot and costs \($0.50\) per square foot.
**Question**:
What is the cost of insulating the water heater?
### Solution Step by Step
**1. Conversion and Formulas Needed**:
First, convert all measurements to feet for consistency:
- Diameter \(= 30\) inches \(\div 12 = 2.5\) feet.
- Radius \(= \frac{Diameter}{2} = 1.25\) feet.
- Height \(= 4 \frac{1}{2}\) feet \(= 4.5\) feet.
**Formulas**:
- Lateral Surface Area of Cylinder \( = 2\pi rh \)
- Area of the Top Circle \(= \pi r^2\)
**2. Calculate Lateral Surface Area**:
\[
Lateral\ Surface\ Area = 2 \pi rh = 2 \pi (1.25\ \text{feet})(4.5\ \text{feet})
\]
\[
= 2 \pi (5.625)\ \text{square feet} \approx 35.343\ \text{square feet} \ (using\ \pi \approx 3.14).
\]
**3. Calculate Top Surface Area**:
\[
Top\ Surface\ Area = \pi r^2 = \pi (1.25\ \text{feet})^2 \approx 4.91\ \text{square feet}
\]
**4. Total Surface Area to Cover**:
\[
Total\ Surface\ Area = Lateral\ Surface\ Area + Top\ Surface\ Area
\]
\[
\approx 35.343\ \text{square feet} + 4.91\ \text{square feet} \approx 40.253\ \text{square feet}
\]
**5. Calculate the Cost**:
Since insulation is sold by the square foot, round the
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