Problem 29 - Algebra and Trigonometry with Analytic Geometry

Question

A large grain silo is to be constructed in the shape of a circular cylinder with a hemisphere attached to the top (see the figure). The diameter of the silo is to be 30 feet, but the height is yet to be determined. Find the height \(h\) of the silo that will result in a capacity of \(11,250 \pi \mathrm{ft}^{3}\).

Step-by-Step Solution

Answer

The height of the silo is 40 feet.

Step 1: Understand the Silo Structure

The silo consists of a circular cylinder with a hemisphere on top. The diameter of both shapes is 30 feet: hence, the radius is \(r = 15\) feet.

Step 2: Express the Volume of the Cylinder

The volume \(V_{c}\) of a cylinder is given by the formula \(V_{c} = \pi r^{2} h\). As the diameter is 30 feet, the radius is 15 feet, so the formula becomes \(V_{c} = \pi \times 15^{2} \times h = 225\pi h\).

Step 3: Express the Volume of the Hemisphere

The volume \(V_{h}\) of a hemisphere is half of the volume of a sphere: \(V_{h} = \frac{2}{3} \pi r^{3}\). Using the radius 15 feet, \(V_{h} = \frac{2}{3} \pi \times 15^{3} = 2250 \pi\).

Step 4: Set Up the Total Volume Equation

The total volume must equal \(11,250\pi\). Sum the volumes of the cylinder and hemisphere to form the equation: \(225\pi h + 2250\pi = 11,250\pi\).

Step 5: Solve for the Height h

Subtract the volume of the hemisphere from the total volume equation: \(225\pi h = (11,250\pi - 2250\pi)\). Simplifying, \(225\pi h = 9,000\pi\). Divide both sides by \(225\pi\) to find \(h\):\[h = \frac{9,000\pi}{225\pi} = 40.\]

Key Concepts

Cylinder Volume

The volume of a cylinder is crucial for many geometry problems, especially those dealing with composite shapes like a silo. A cylinder is a three-dimensional solid with two parallel circular bases and a curved surface connecting them. To calculate its volume, you need the formula:

• \( V_c = \pi r^2 h \)

This formula tells us that the volume depends on the circle's radius, \( r \), squared, and the height, \( h \), of the cylinder.
For our silo, the diameter is given as 30 feet, so the radius is half of that, which is 15 feet. Our volume formula becomes:

• \( V_c = \pi \times 15^2 \times h = 225\pi h \)

The cylinder volume is essential because it makes up a significant portion of the silo's capacity. Understanding this concept helps you set up the right equations when dealing with composite shapes.

Hemisphere Volume

A hemisphere is half of a sphere. Knowing how to calculate its volume is fundamental when it's part of a composite solid, such as a silo with a hemispherical top. The volume of a full sphere is given by:

• \( V = \frac{4}{3} \pi r^3 \)

Since a hemisphere is half of a sphere, its volume is:

• \( V_h = \frac{1}{2} \times \frac{4}{3} \pi r^3 = \frac{2}{3} \pi r^3 \)

Using the radius of 15 feet in our problem, the volume of the hemisphere is calculated as:

• \( V_h = \frac{2}{3} \pi \times 15^3 = 2250 \pi \)

This hemispherical volume is a fixed component in the total volume of the silo, making it necessary to determine the remaining cylindrical volume that achieves the desired total capacity.

Silo Geometry

Understanding silo geometry involves comprehending how different shapes work together in composite solids. A silo in this context is a combination of a cylinder and a hemisphere. Together, they form a unique structure with distinct geometric properties. Working with silo geometry requires:

• A thorough grasp of the formulas applicable to each shape, i.e., cylinders and hemispheres.
• The ability to combine their volumes accurately.

By adding the volume of the cylinder to that of the hemisphere, we derive the total volume of the silo. Setting up the equation using the desired capacity:

• \( 225\pi h + 2250\pi = 11,250\pi \)

is crucial in solving for certain parameters such as the height of the cylindrical section. Understanding this composite form leads to correct calculations and practical applications in real-world scenarios.

Calculus Applications

Some geometry problems, especially those involving volumes, intersect with calculus concepts. Although this problem is handled algebraically, it introduces a calculus mindset: decomposing complex forms into manageable parts. In volumes of composite solids:

• Volume integrals can be used for calculating volumes of irregular shapes.
• Understanding solid revolutions might be applied for more complex forms.

While basic geometry suffices for simple silo questions, calculus offers more tools for complicated designs and optimizations. This mindset aids in moving beyond simple calculations to comprehend areas, rates of change, and spatial transformations. For students advancing in mathematics, being aware of such intersections enriches their analytical skills and problem-solving toolkit.