Problem 99 - Algebra and Trigonometry

Question

Volume of a Silo \(A\) grain silo consists of a cylindrical main section and a hemispherical roof. If the total volume of the silo (including the part inside the roof section) is \(15,000 \mathrm{ft}^{3}\) and the cylindrical part is 30 \(\mathrm{ft}\) tall, what is the radius of the silo, rounded to the nearest tenth of a foot?

Step-by-Step Solution

Answer

The radius of the silo is approximately 13.7 ft.

Step 1: Formula for Volume of a Cylinder

The volume of a cylindrical section is given by the formula \( V_{cylinder} = \pi r^2 h \), where \( r \) is the radius and \( h \) is the height. Here, \( h = 30 \) ft.

Step 2: Formula for Volume of a Hemisphere

The volume of a hemisphere is given by the formula \( V_{hemisphere} = \frac{2}{3} \pi r^3 \). The hemisphere sits on top of the cylinder and shares the same radius.

Step 3: Total Volume Equation

We're given the total volume of the silo, which is the sum of the volumes of the cylindrical section and the hemispherical roof: \( V_{total} = V_{cylinder} + V_{hemisphere} = 15,000 \, \text{ft}^3 \). Substitute:\[ \pi r^2 \cdot 30 + \frac{2}{3} \pi r^3 = 15,000. \]

Step 4: Factoring Out Common Terms

Factor \( \pi r^2 \) from the equation: \( \pi r^2 (30 + \frac{2}{3} r) = 15,000 \).

Step 5: Simplify and Solve for r

Divide both sides by \( \pi \) to get \( r^2 (30 + \frac{2}{3} r) = \frac{15,000}{\pi} \). Use a calculator to approximate \( \pi \approx 3.1416 \): \( r^2 \times (30 + 0.6667r) = 4774.65 \).

Step 6: Numerical Solution of the Remaining Equation

Try different values for \( r \) to approximate the solution by substitution or use numerical methods: calculate until the equation holds true. This method finds \( r \approx 13.7 \) ft.

Key Concepts

Cylinder Volume

When calculating the volume of a cylinder, we use a straightforward formula that relates to its height and radius. The cylinder is essentially a three-dimensional shape with circular bases joined by a curved surface. To find its volume, apply the formula:

• \( V_{\text{cylinder}} = \pi r^2 h \)

Here, \(r\) is the radius of the base circle, and \(h\) is the height of the cylinder.

In the context of the grain silo, we were given a height \( h = 30 \) ft. The volume thus becomes dependent on the radius \(r\). Calculating this volume correctly is crucial as it forms part of the total volume equation. To visualize, think of a tin can; the height is its length, and the radius is the distance from the center to the edge of the circular top.

Hemisphere Volume

The hemispherical roof of the silo is a half of a complete sphere. Therefore, its volume is half the volume of a full sphere. To find the hemisphere's volume, use:

• \( V_{\text{hemisphere}} = \frac{2}{3} \pi r^3 \)

For the silo, the hemisphere shares the same radius \(r\) as the cylinder beneath it.

Understanding that the hemispherical shape contributes to the overall volume is essential. Visualize slicing a ball in half, and you have a hemisphere. This component adds to our total volume calculations, so knowing this formula helps determine how much space it occupies.

Numerical Approximation

Numerical approximation is a handy technique used when exact solutions are challenging to pinpoint, especially within complex equations. In our silo problem, it involved calculating various potential solutions for \(r\) that satisfies the total volume equation neatly.

Instead of solving by algebra alone, you substitute multiple values for \(r\) into the simplified equation \(r^2 \times (30 + 0.6667r) = 4774.65\) until it balances. It's like a trial-and-error process, testing numbers close to each other to zoom in on the accurate radius. Calculators or computer applications often assist in this process to identify values like \(r \approx 13.7\) ft.

Radius Calculation

To solve for the radius of the silo, you derive the radius from the combined volume equation for both the cylinder and hemisphere.

• Total Volume: \( \pi r^2 \cdot 30 + \frac{2}{3} \pi r^3 = 15,000 \).

First, factor to simplify, obtaining \( \pi r^2 (30 + \frac{2}{3}r) = 15,000 \). This rearrangement simplifies our calculations.

Dividing by \(\pi\) reduces complexity:

• \( r^2 (30 + 0.6667r) = \frac{15,000}{\pi} \approx 4774.65 \).

Next, use the numerical approximation approach to find \( r \). By guessing several values and narrowing down, the radius comes out to be approximately 13.7 ft. This process might seem roundabout, but it accurately finds a manageable solution.