Problem 73 - College Algebra

Question

Find the dimensions of the box described. The length is one inch more than the width, which is one inch more than the height. The volume is 86.625 cubic inches.

Step-by-Step Solution

Answer

Height = 4.5 inches, Width = 5.5 inches, Length = 6.5 inches.

Step 1: Define Variables

Let's define the variables for the dimensions. Let the height of the box be \( h \) inches. Then, the width will be \( h + 1 \) inches, and the length will be \( h + 2 \) inches because it is stated that the length is one inch more than the width, which is one inch more than the height.

Step 2: Write the Volume Equation

The volume \( V \) of a box is given by the formula \( V = \, \text{length} \times \text{width} \times \text{height} \). Substituting the expressions for length, width, and height, we get: \[ V = (h + 2)(h + 1)(h) \]We know the volume is 86.625 cubic inches, so we set the equation equal to the given volume: \[ (h + 2)(h + 1)(h) = 86.625 \]

Step 3: Simplify the Equation

Expand the left-hand side of the equation:\[ (h + 2)(h + 1)(h) = (h^2 + 3h + 2)(h) = h^3 + 3h^2 + 2h \]

Step 4: Solve the Equation

Now, set up the equation:\[ h^3 + 3h^2 + 2h = 86.625 \]Use algebraic or numerical methods to find the value of \( h \). You can try values for \( h \) or use a calculator that can handle solving cubic equations. You may estimate \( h \) by trying different values. After testing, \( h \approx 4.5 \) inches.

Step 5: Find Width and Length

With \( h = 4.5 \), calculate the width and length:- Width = \( h + 1 = 4.5 + 1 = 5.5 \) inches.- Length = \( h + 2 = 4.5 + 2 = 6.5 \) inches.

Key Concepts

Algebraic Expressions

Algebraic expressions are mathematical phrases that can contain numbers, variables, and operation signs. In this exercise, we use algebraic expressions to represent the dimensions of a box. By defining a variable for the height as \( h \), we can write expressions for the width and length as \( h + 1 \) and \( h + 2 \), respectively. This shows how algebraic expressions help simplify and generalize math problems. By establishing these expressions, we can easily manipulate the variables to establish a connection to the problem's requirements.
They allow us to express unknowns conveniently and make solving the problem systematic. With expressions for each box dimension, we can create an equation that relates these expressions to the box's volume, which is a key step in solving the problem.

Volume Calculation

Volume calculation is an essential concept that helps us determine the capacity of three-dimensional objects. Here, we're targeting the volume of a box, which can be found using the formula:

• Volume (\( V \)) = length × width × height

In our specific example, we calculated the box's volume by substituting the algebraic expressions for the dimensions into the formula. This gives us the equation:

• \( V = (h + 2)(h + 1)h \)

We're given that the volume of the box is 86.625 cubic inches, so we set our expression equal to this value to find the unknown dimension. Understanding how to calculate volume is crucial in solving geometry problems, especially when dealing with cubic equations in algebra.

Solving Equations

Solving equations is the process of finding the values of variables that make the equation true. In this exercise, we approach a cubic equation, which is an equation where the highest exponent of the variable is a cube. This means the equation takes the form \( ax^3 + bx^2 + cx + d = 0 \).
Here, our equation is:

• \( h^3 + 3h^2 + 2h = 86.625 \)

To solve for \( h \), we might estimate potential values using a method of trials or utilize a calculator capable of handling cubic equations. Solving this equation correctly determines the value of \( h \) and by extension, the dimensions of our box. Once we find \( h \approx 4.5 \), we can easily find the other dimensions using previously defined algebraic expressions. Solving such equations improves understanding of algebra and the practical application of mathematical principles in real-world scenarios.