Problem 73 - College Algebra with Corequisite Support

Question

For the following exercises, 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

The dimensions are height: 3.5 inches, width: 4.5 inches, length: 5.5 inches.

Step 1: Define the variables

Let's define the dimensions of the box. Let \( h \) represent the height of the box in inches. Then, the width \( w \) will be \( h + 1 \) inches, and the length \( l \) will be \( w + 1 = h + 2 \) inches.

Step 2: Set up the volume equation

The volume of a box is given by the formula: \( V = l \times w \times h \). We know that the volume is 86.625 cubic inches. Substitute the expressions for length and width in terms of height: \[(h + 2)(h + 1)h = 86.625\]

Step 3: Expand the equation

Expand the equation from Step 2. First, multiply \( (h + 2)(h + 1) \) to get: \[(h^2 + 3h + 2)\]Then, multiply this with \( h \):\[h^3 + 3h^2 + 2h = 86.625\]

Step 4: Solve the cubic equation

We need to find the value of \( h \) that satisfies the equation \( h^3 + 3h^2 + 2h = 86.625 \). Testing values, we find that \( h = 3.5 \) satisfies the equation.

Step 5: Find width and length

Once we have \( h = 3.5 \), calculate the width: \[w = h + 1 = 3.5 + 1 = 4.5\] Then, find the length: \[l = w + 1 = 4.5 + 1 = 5.5\]

Key Concepts

Cubic Equations

Cubic equations are polynomial equations of degree three, meaning they involve an unknown raised to the third power. They take the general form \( ax^3 + bx^2 + cx + d = 0 \), where \( a \), \( b \), \( c \), and \( d \) are constants, and \( a eq 0 \). In our box problem, the volume equation \( h^3 + 3h^2 + 2h = 86.625 \) is a cubic equation with terms set up in descending order of \( h \). Understanding cubic equations is essential because they often appear when dealing with volumes or other three-dimensional problems.
To solve a cubic equation like this one, you often need to use trial and error or graphing to find reasonable values that satisfy the equation. Sometimes, they might have more than one real solution, so comprehensive testing is essential. Briefly, cubic equations are more complex than quadratic or linear equations but provide deeper insights into geometric problems.

Solving Equations

Solving equations involves finding the solution that makes an equation true. In the context of our box problem, we tackle the cubic equation formed by substituting the expressions for length and width into the volume formula. Finding the correct value of \( h \) that satisfies \( h^3 + 3h^2 + 2h = 86.625 \) is key to solving the problem.
When dealing with cubic equations like this, you can:

• Use substitution to simplify variables around other given conditions.
• Test potential solutions through evaluation or graphing to find the most suitable value.
• Utilize mathematical software or a calculator for more precise results.

By testing several values of \( h \), we discovered that \( h = 3.5 \) satisfies the cubic equation. This process involves substituting \( h \) values into the equation until the left-hand side equals 86.625.

Geometry

Geometry in this context is all about understanding the three-dimensional shape of a box, described using terms like length, width, and height. Geometric concepts play an essential role in calculating the volume of the box by multiplying these three dimensions. The formula for volume, \( V = l \times w \times h \), gives us a straightforward way to grasp how these dimensions interact and contribute to the overall size of the box.
In geometry problems involving boxed calculations:

• Recognize the relationships between dimensions, such as how length relates to width and height.
• Use geometric formulas to calculate properties like volume or surface area.
• See the practical applications in solving real-world problems involving packaging or storage.

This understanding of geometric principles ensures that you can accurately find values like volume, even with intricate equations in play.

Box Dimensions

Box dimensions refer to the length, width, and height of a box, key factors that determine the box's shape and size. In this problem, understanding the relationships between these dimensions is crucial.
We defined these dimensions as follows:

• Height \( h \)
• Width \( w = h + 1 \)
• Length \( l = h + 2 \)

When given the volume, which is 86.625 cubic inches, we used these relationships to establish a cubic equation corresponding to the volume formula. Understanding these dimensions allows you to effectively rearrange them in various algebraic forms using predefined relationships, ultimately helping identify each specific measurement through calculation.
Thus, the problem exemplifies how box dimensions are foundational in problem solving and essential in shaping the method used to reach the solution.