Question

A cylindrical oil tank holds 10,000 barrels of oil. If the diameter of the tank is 50 feet, what is its height? Round to the nearest tenth of a foot. Hint: 1 oil barrel \(\approx 5.61458 \mathrm{ft}^{3}\)

Short answer

The height of the tank is approximately 28.6 feet.

Step-by-step solution

- Convert barrels to cubic feet

First, convert the volume of the oil from barrels to cubic feet. Given that 1 barrel is approximately 5.61458 cubic feet, multiply the number of barrels by this conversion factor: \[ 10,000 \text{ barrels} \times 5.61458 \frac{\text{ft}^3}{\text{barrel}} = 56,145.8 \text{ ft}^3 \]

- Understand the cylinder volume formula

The volume (V) of a cylinder is given by the formula: \[ V = \pi r^2 h \] where \( r \) is the radius and \( h \) is the height.

- Calculate the radius of the tank

Given the diameter of the tank is 50 feet, the radius (r) is half of that: \[ r = \frac{50 \text{ feet}}{2} = 25 \text{ feet} \]

- Plug the values into the formula

Now, use the volume formula for the cylinder. We know the volume (V) is 56,145.8 cubic feet and the radius (r) is 25 feet. Substitute these values into the formula: \[ 56,145.8 = \pi (25)^2 h \]

- Isolate the height (h)

Solve for the height (h) by isolating it on one side. First, calculate \( \pi (25)^2 \): \[ \pi (25)^2 = \pi \times 625 = 1,963.5 \text { ft}^2 \]Now, divide the volume by this value: \[ h = \frac{56,145.8}{1,963.5} \approx 28.6 \text{ feet} \]

- Round the height to the nearest tenth

Finally, round the height to the nearest tenth of a foot: \[ h = 28.6 \text{ feet} \]

Key concepts

Cylinder Geometry

A cylinder is a three-dimensional shape with two parallel circular bases connected by a curved surface. Understanding the geometry of a cylinder is crucial for volume calculation. The key components are:

• Height (h): The perpendicular distance between the bases.
• Radius (r): The distance from the center to the edge of the base.
• Diameter (d): Twice the radius, or the distance across the circular base through the center.

The formula for the volume of a cylinder incorporates both the radius and the height, expressed as: \( V = \pi r^2 h \), where \( V \) is the volume. This formula indicates that the volume depends on the square of the radius, emphasizing the significant impact of the radius on the overall volume. Understanding these basic geometrical properties allows for better comprehension of how to work with cylinders, particularly in applications involving volume conversion, radius and diameter relationships, and solving for height.

Volume Conversion

Volume conversion is often essential when dealing with units that are more convenient for specific applications. In the given exercise, the need to convert from barrels to cubic feet arises because the formula for the volume of a cylinder uses cubic feet.
Here’s how you can do it:

• Identify the conversion factor (1 barrel \( \approx 5.61458 \text{ ft}^3 \)).
• Multiply the number of barrels by this factor to convert to cubic feet.

As shown in the calculation:
\( 10,000 \text{ barrels} \times 5.61458 \frac{\text{ft}^{3}}{\text{barrel}} = 56,145.8 \text{ ft}^3 \). This step is crucial for accurate volume calculations in different unit systems, ensuring consistency and correctness when applying the cylinder volume formula. Understanding volume conversion helps bridge different measurement systems, facilitating easier comprehension and application in real-world scenarios.

Radius and Diameter Relationship

Knowing the relationship between the radius and diameter is essential for solving cylinder-related problems. The diameter (d) is twice the length of the radius (r), and this relationship is expressed as:

• \( r = \frac{d}{2} \)
• or conversely \( d = 2r \)

In this exercise, given the diameter is 50 feet, the radius is:
\( r = \frac{50 \text{feet}}{2} = 25 \text{feet} \).
This simplification is vital because many formulas, including the volume of a cylinder, require the radius, not the diameter. Effortlessly switching between these two measurements ensures smooth progress in solving geometric problems. Recognizing how the radius influences the calculation of volume (since volume depends on \( r^2 \)) allows for better overall understanding and accuracy in computations.

Solving for Height in Geometric Shapes

One common task in geometry is solving for unknown dimensions, such as height, given other known values. For a cylinder, this involves manipulating the volume formula: \( V = \pi r^2 h \). To isolate height (h), follow these steps:

• First, plug in the known values for volume (V) and radius (r).
• Next, compute the area of the base (\( \pi r^2 \)).
• Finally, divide the volume by this base area to find the height.

Using the provided values in our problem:
1. Plug in: \( 56,145.8 = \pi (25)^2 h \)
2. Compute the base area: \( \pi (25)^2 = 1,963.5 \text{ ft}^2 \)
3. Solve for height: \( h = \frac{56,145.8}{1,963.5} \approx 28.6 \text{ feet} \)
This method systematically breaks down the problem, allowing for clear and precise solution-finding. Understanding how to isolate variables within equations is a fundamental skill that aids in solving various geometric problems more efficiently.