Problem 79 - College Algebra with Corequisite Support

Question

For the following exercises, write the polynomial function that models the given situation. A right circular cone has a radius of \(3 x+6\) and a height 3 units less. Express the volume of the cone as a polynomial function. The volume of a cone is \(V=\frac{1}{3} \pi r^{2} h\) for radius \(r\) and height \(h\).

Step-by-Step Solution

Answer

The volume is expressed as \( V(x) = \pi (9x^3 + 45x^2 + 72x + 36) \).

Step 1: Identify the expressions for radius and height

The problem states that the radius of the cone is \( r = 3x + 6 \). The height is 3 units less than the radius, so we express the height as \( h = (3x + 6) - 3 = 3x + 3 \).

Step 2: Plug expressions into the volume formula

The volume of the cone is given by the formula \( V = \frac{1}{3} \pi r^{2} h \). Substitute \( r = 3x + 6 \) and \( h = 3x + 3 \) into this formula.

Step 3: Simplify the expression \( r^2 \)

Calculate \( r^2 \) by squaring the expression for the radius: \((3x + 6)^2 = (3x + 6)(3x + 6) = 9x^2 + 36x + 36 \).

Step 4: Multiply \( r^2 \) by \( h \)

Multiply the expression for \( r^2 = 9x^2 + 36x + 36 \) by \( h = 3x + 3 \):\((9x^2 + 36x + 36)(3x + 3) = 9x^2(3x + 3) + 36x(3x + 3) + 36(3x + 3)\).Expanding this gives:\(27x^3 + 27x^2 + 108x^2 + 108x + 108x + 108\).Combining like terms, we obtain:\(27x^3 + 135x^2 + 216x + 108\).

Step 5: Complete the volume expression

Substitute \( r^2 h = 27x^3 + 135x^2 + 216x + 108 \) back into the volume formula: \(V = \frac{1}{3} \pi (27x^3 + 135x^2 + 216x + 108)\).Distribute the \( \frac{1}{3} \pi \) over each term:\(V = \pi (9x^3 + 45x^2 + 72x + 36)\). So the polynomial function for the volume is \(V(x) = \pi (9x^3 + 45x^2 + 72x + 36)\).

Key Concepts

Volume of a Cone

Understanding the volume of a cone is crucial in geometry and everyday situations involving cones. A cone is a three-dimensional shape with a circular base and a single vertex point, often described as having a pointed end.
To find the volume, we use the formula:

• \( V = \frac{1}{3} \pi r^2 h \)

Here, \( V \) represents the volume, \( r \) is the radius of the base, while \( h \) stands for the height of the cone.
The formula shows that the volume is proportional to both the square of the radius and the height of the cone. Doubling the radius greatly increases the volume, as it affects the squared term \( r^2 \).
Understanding this formula helps in solving practical problems, such as filling up a conical tank with liquid, where knowing the volume is essential.

Algebraic Expressions

Algebraic expressions are mathematical phrases that can include numbers, variables, and operations. They are fundamental for modeling real-life situations mathematically.
When we refer to an expression like \( 3x + 6 \), we are considering a linear equation where \( x \) is a variable. This formula represents a set of operations to be performed.

• Variables like \( x \) represent unknown or changeable numbers.
• Constants like 6 are fixed values within the expression.

Algebraic equations allow us to articulate complex relationships or conditions. In our cone example, expressions define the cone’s dimensions.
By understanding and manipulating these expressions, we can translate real-world problems into mathematical language, making them easier to solve and interpret.

Polynomial Simplification

Simplifying polynomials is a process of making algebraic expressions more concise. This involves combining like terms and arranging them in standard form.
In our example, we had to simplify a polynomial generated by multiplying algebraic expressions for a cone's dimensions. Here are the steps followed:

• Calculate \( r^2 \): Square the expression \((3x + 6)\) to form \(9x^2 + 36x + 36\).
• Multiply \( r^2 \) by \( h \): Expand \((9x^2 + 36x + 36)(3x + 3)\) into a new polynomial, combine like terms, and simplify.

After simplification, we obtain the polynomial \(27x^3 + 135x^2 + 216x + 108\).
Ultimately, simplifying polynomials reduces complexity, paving the way for easier handling and integration into larger formulas, such as the volume of a cone in this scenario.