Problem 71 - College Algebra with Corequisite Support

Question

For the following exercises, find the dimensions of the box described. The length is twice as long as the width. The height is 2 inches greater than the width. The volume is 192 cubic inches.

Step-by-Step Solution

Answer

The dimensions are: width = 4 inches, length = 8 inches, height = 6 inches.

Step 1: Define Variables

Let's define variables to represent the dimensions of the box. Let the width of the box be \( x \) inches, the length be \( 2x \) inches (since the length is twice the width), and the height be \( x + 2 \) inches (as the height is 2 inches more than the width).

Step 2: Write the Volume Equation

The volume of a box is given by the product of its length, width, and height. Therefore, the volume \( V \) of the box can be expressed as:\[ V = \text{length} \times \text{width} \times \text{height} = 2x \times x \times (x + 2) \]Set this equation equal to the given volume of 192 cubic inches:\[ 2x^2 (x + 2) = 192 \]

Step 3: Expand the Equation

Distribute and simplify the equation:\[ 2x^2 (x + 2) = 192 \]\[ 2x^3 + 4x^2 = 192 \]

Step 4: Rearrange and Formulate a Polynomial

Bring all terms to one side to form a polynomial:\[ 2x^3 + 4x^2 - 192 = 0 \]

Step 5: Factor the Polynomial

Factor out the greatest common factor, which is 2:\[ 2(x^3 + 2x^2 - 96) = 0 \]Divide both sides by 2:\[ x^3 + 2x^2 - 96 = 0 \]To find the roots, we test potential rational root candidates using synthetic division or plug suitable values.Trying \( x = 4 \) results in:\[ 4^3 + 2\times4^2 - 96 = 64 + 32 - 96 = 0 \]Hence, \( x = 4 \) is a solution.

Step 6: Solve for Dimensions

Use \( x = 4 \) to solve for other dimensions:Width = 4 inches,Length = \( 2x = 2(4) = 8 \) inches,Height = \( x + 2 = 4 + 2 = 6 \) inches.

Key Concepts

Volume Calculation

Volume calculation is an important concept in understanding the relationship between size and space within a three-dimensional object. In this context, volume refers to the amount of space that the box occupies, measured in cubic inches. To calculate this, we multiply the three dimensions: length, width, and height.

• The formula for the volume of a box is given by: \[ \text{Volume} = \text{Length} \times \text{Width} \times \text{Height} \]
• In the exercise, the volume was set to 192 cubic inches, guiding us to solve for any unknown dimensions.

Knowing how to handle these calculations helps not only in academics but also in real-life scenarios, where determining the volume of storage or products is required. Always ensure units are in sync for accuracy.

Polynomial Equations

Polynomial equations are mathematical expressions that involve variables raised to positive integer powers. These equations play a crucial role in various fields like physics, economics, and engineering.
In our exercise, a polynomial equation is formed when performing volume calculations by multiplying the dimensions:

• The equation to determine the volume of the box is: \( 2x^2(x + 2) = 192 \).
• Upon expansion and simplification, it turns into a cubic polynomial: \( 2x^3 + 4x^2 - 192 = 0 \).

Working with polynomial equations requires understanding their degree and finding values that satisfy these equations, known as roots. Solving them often requires factoring or other algebraic methods. This knowledge is foundational for advanced mathematical concepts.

Factoring Polynomials

Factoring polynomials is a technique used to simplify polynomials and find their roots. It involves expressing a polynomial as a product of its factors and is an invaluable skill in algebra.To factor the polynomial from the exercise, the strategy involved:

• Recognizing and factoring out the greatest common factor (GCF), here it was 2, which helped simplify the cubic equation: \( 2(x^3 + 2x^2 - 96) = 0 \).
• After dividing by 2, we sought potential values for \( x \), identifying \( x = 4 \) as a valid root when plugged into the simplified polynomial \( x^3 + 2x^2 - 96 = 0 \).

By focusing on potential rational roots or using synthetic division when appropriate, one can efficiently pinpoint the solutions needed. Learning to factor polynomials aids not only in solving equations but also in understanding the properties of algebraic expressions.