Problem 34 - Algebra 2

Question

A restaurant orders spaghetti sauce in cylindrical metal cans. The volume of each can is about 160\(\pi\) cubic inches, and the height of the can is 6 inches more than the radius. Find the dimensions of the can.

Step-by-Step Solution

Answer

Radius is 4 inches, height is 10 inches.

Step 1: Set up the formula for the volume of a cylinder

The volume of a cylinder is calculated using the formula \( V = \pi r^2 h \), where \( r \) is the radius and \( h \) is the height of the cylinder. We are given that the volume is \( 160\pi \) cubic inches.

Step 2: Relate height to radius

We are told that the height \( h \) is 6 inches more than the radius \( r \), so we can express height as \( h = r + 6 \).

Step 3: Substitute the height expression into the volume formula

Substitute \( h = r + 6 \) into the volume formula: \[ \pi r^2 (r + 6) = 160\pi \] We can simplify by cancelling \( \pi \) from both sides, which gives: \[ r^2 (r + 6) = 160 \]

Step 4: Simplify and solve the equation

Expand the equation: \[ r^3 + 6r^2 = 160 \] Re-arrange to form a polynomial equation: \[ r^3 + 6r^2 - 160 = 0 \] We solve this using trial and error or factorization techniques.

Step 5: Solve the polynomial equation

Assume a value of \( r \) and test: Trying \( r = 4 \):\[ 4^3 + 6(4)^2 - 160 = 64 + 96 - 160 = 0 \] Therefore, \( r = 4 \) is a solution.

Step 6: Calculate the height using the found radius

Since \( h = r + 6 \) and we found \( r = 4 \), the height \( h = 4 + 6 = 10 \) inches.

Key Concepts

Radius and Height Relationship

In problems involving cylinders, one common scenario is the relationship between the radius and the height. Often, these two dimensions are connected, making it essential to express one variable in terms of the other. In our exercise, the height of the can is described as 6 inches more than the radius. This forms the equation for height as:

• Height, \( h = r + 6 \)

Here, \( r \) represents the radius of the can, and knowing how height relates to the radius is crucial for solving the volume equation accurately. This relationship allows us to simplify problems to a single variable, making them easier to manage and solve. Remember that every physical characteristic mentioned, such as this relationship, directly affects how we structure our equations.

Polynomial Equation

A polynomial equation often emerges when calculating dimensions in geometrical shapes. In our specific exercise, after substituting the given height relationship into the cylinder volume formula, we derived:

• \( r^3 + 6r^2 - 160 = 0 \)

This equation is cubic because the highest power of the variable \( r \) is 3. Solving polynomial equations can sometimes be challenging, as they involve more than one term and can contain various powers of the variable.
However, understanding the degree of the polynomial guides us on which solving techniques, such as factorization, to employ. The aim is to isolate the variable by reducing the equation to its simplest form and finding values of \( r \) that satisfy the equation fully.

Factorization Techniques

Factorization turns complex algebraic expressions into simpler products or expressions. Here, we are solving the polynomial \( r^3 + 6r^2 - 160 = 0 \). One straightforward method for factorization is trying potential roots through trial and error.

• If we consider \( r = 4 \), substituting it back gives \( 4^3 + 6(4)^2 - 160 = 0 \).

This satisfies the equation, confirming \( r = 4 \) as a solution. Factorization can also involve using the Rational Root Theorem or synthetic division, especially if the polynomial is more complex.

• Once a root is identified, the polynomial can be decomposed further into factors if needed for more roots.

This technique simplifies finding the solution significantly by systematically reducing the degrees until a solution is visible.

Volume Formula for Cylinder

The volume of a cylinder is a pivotal concept in geometry and is computed using the formula:

• \( V = \pi r^2 h \)

This formula relies on finding the cross-sectional area (a circle, which is \( \pi r^2 \)) and then multiplying it by the height \( h \). For our exercise, the volume is specified as \( 160\pi \) cubic inches. Knowing this simplifies calculations because

• it leads to easier manipulation of terms, such as removing \( \pi \) from both sides when equating the formula.

Substituting the height expression \( h = r + 6 \), we effectively engage the volume equation, helping us progress to find the cylinder's dimensions.

• This formula remains a fundamental aspect of understanding various calculations involving cylinders and is widely utilized in both academic and practical fields.

It's essential to manipulate and make calculations accurately to deduce the correct physical properties of cylindrical objects.