Problem 82 - Algebra and Trigonometry

Question

Dimensions of a Box A large plywood box has a volume of \(180 \mathrm{ft}^{3} .\) Its length is 9 \(\mathrm{ft}\) greater than its height, and its width is 4 \(\mathrm{ft}\) less than its height. What are the dimensions of the box?

Step-by-Step Solution

Answer

The dimensions of the box are 15 ft (length), 2 ft (width), and 6 ft (height).

Step 1: Define Variables

Let's define the height of the box as \( h \). Then according to the problem, the length \( l \) can be expressed as \( h + 9 \) and the width \( w \) can be expressed as \( h - 4 \).

Step 2: Set Up the Volume Equation

The volume \( V \) of a box is calculated with the formula \( V = l \times w \times h \). Given the volume is \( 180 \ \mathrm{ft}^3 \), we can substitute the values of length and width to get: \[ (h + 9)(h - 4)h = 180 \].

Step 3: Expand and Simplify Equation

Expand the expression \((h + 9)(h - 4)\) to get \( h^2 + 5h - 36 \). The equation now becomes \( h(h^2 + 5h - 36) = 180 \) or \( h^3 + 5h^2 - 36h = 180 \).

Step 4: Solve the Cubic Equation

Rearrange the equation to \( h^3 + 5h^2 - 36h - 180 = 0 \). Use trial and error or synthetic division to find that \( h = 6 \) is a root of this equation.

Step 5: Calculate Dimensions

Substitute the height \( h = 6 \) back into the expressions for length and width. This gives \( l = h + 9 = 15 \) and \( w = h - 4 = 2 \).

Step 6: Verify Volume

Calculate the volume with the dimensions found: \( 15 \times 2 \times 6 = 180 \ \mathrm{ft}^3 \), which matches the given volume. Thus, the dimensions are correct.

Key Concepts

Cubic Equations

Cubic equations are polynomial equations of degree three. An example of a cubic equation is of the form \(ax^3 + bx^2 + cx + d = 0\), where \(a, b, c,\) and \(d\) are constants, and \(a eq 0\). These equations are used when the relationship between variables involves a cubic term. In our case, the box's volume is expressed as a cubic equation because it involves a product of three expressions: height \((h)\), length \((h + 9)\), and width \((h - 4)\).
To solve a cubic equation, we often look for a potential root through methods like trial and error or through more systematic approaches such as synthetic division.
This leads us to finding the height \(h = 6\), which satisfies the equation \(h^3 + 5h^2 - 36h - 180 = 0\).
Solving cubic equations is crucial in finding unknown dimensions when dealing with volume problems.

Volume Calculation

Volume calculation is an essential part of understanding the size and capacity of three-dimensional objects. Volume, in simple terms, is the amount of space an object occupies. The formula for calculating the volume of a box (rectangular prism) is straightforward: \(V = l \times w \times h\), where \(l\), \(w\), and \(h\) represent the length, width, and height respectively.
For the plywood box problem, the given volume is \(180 \text{ ft}^3\). To determine the correct geometric dimensions, we used the specified relationships between height, length, and width by substituting into the volume formula as \((h + 9)(h - 4)h = 180\).
This type of calculation allows us to discern how different dimensions interact and alter the volume outcome.

• Length is the longest side of a box.
• Width is the shorter side when viewed from the top.
• Height is the vertical dimension when the box is upright.

Understanding how these dimensions multiply together to form volume helps us grasp the significance of each measurement's role in spatial analysis.

Dimension Analysis

Dimension analysis is a powerful technique that helps in solving problems involving physical measurements. It essentially involves ensuring consistency in equations by analyzing the dimensions of various physical quantities.
In the context of our box problem:

• We begin by defining a key variable, which is typically the unknown dimension—in this case, the box's height \((h)\).
• By understanding how each dimension relates to the others (here, length is height plus 9 feet, and width is height minus 4 feet), we establish a clear system of expressions.
• This involves comparing, contrasting, or substituting dimensions based on known relationships and constraints. For instance, if height \(h = 6\), then length can be easily calculated as \(h + 9 = 15\), and width as \(h - 4 = 2\).

Dimension analysis is crucial not only in verifying solutions but also in ensuring that we have a logical approach to setting up equations that reflect real-world scenarios. When utilized correctly, it validates our mathematical modeling, ensuring all expressions align with the problem's physical realities.