Problem 71 - College Algebra

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

Width is 4 inches, length is 8 inches, and height is 6 inches.

Step 1: Define the Variables

Let's define the variables for the dimensions of the box. Let \( w \) represent the width of the box. Since the length is twice the width, the length will be \( 2w \). The height is 2 inches greater than the width, so the height will be \( w + 2 \).

Step 2: Write the Volume Formula

The volume of a box is given by the formula: \( V = \, \text{length} \, \times \, \text{width} \, \times \, \text{height} \). Plug in the expressions from Step 1 to get \( V = 2w \times w \times (w + 2) \).

Step 3: Set Up the Equation for Volume

We know the volume of the box is 192 cubic inches. Therefore, we can set up the equation: \[ 2w \times w \times (w + 2) = 192 \] Simplifying, this gives:\[ 2w^2(w + 2) = 192 \]

Step 4: Simplify the Equation

Expand the equation from Step 3:\[ 2w^2(w + 2) = 192 \] \[ 2w^3 + 4w^2 = 192 \] Divide through by 2 to simplify:\[ w^3 + 2w^2 = 96 \]

Step 5: Solve the Equation for Width

Here's where we solve for \( w \):If we trial simple values for a cubic equation such as for \( w \):Trying \( w = 4 \):\[ w^3 + 2w^2 = 4^3 + 2(4^2) = 64 + 32 = 96 \]Thus, \( w = 4 \).

Step 6: Calculate Other Dimensions

Using \( w = 4 \), calculate length and height:- Length = \( 2w = 2(4) = 8 \)- Height = \( w + 2 = 4 + 2 = 6 \).

Key Concepts

Dimensions of a Box

Understanding the dimensions of a box is crucial when trying to find its volume, surface area, or when creating dimensions based on a given volume. For a box, dimensions typically include:

• Length (L)
• Width (W)
• Height (H)

In the exercise, we are given a practical problem where we need to find these dimensions based on some relational information. The width (\( w \)) of the box is our base variable:

• The length is twice the width, so it can be expressed as \( 2w \).
• The height is 2 inches more than the width, represented as \( w + 2 \).

This approach of defining dimensions in terms of one variable simplifies solving the problem, especially when involving calculations of volume based on specified relationship constraints.

Cubic Equations

Cubic equations play a key role when calculating dimensions if we're given volume. A cubic equation is a polynomial of degree three. For any box dimensional problem involving volume, like this one, the resulting equation from the volume formula is a cubic equation.Here, the starting volume equation is:\[V = \text{length} \times \text{width} \times \text{height}\]Substituting the expressions from our relationship gives:\[2w \times w \times (w + 2) = 192\]Simplifying this yields the cubic equation:\[2w^3 + 4w^2 = 192\]To work with it more easily, we divide by 2:\[w^3 + 2w^2 = 96\]Cubic equations like this might seem complex, but they track the relationships of three-dimensional spaces, crucial when seeking unknown lengths, widths, and heights.

Solving Equations

Solving a cubic equation often involves trying different values to find a solution, especially when the equation isn't easily factored. Given the problem's constraint and knowing the equation:\[w^3 + 2w^2 = 96\]You can try potential solutions for \( w \) iteratively. In this exercise, when testing \( w = 4 \), you calculate:\[4^3 + 2(4^2) = 64 + 32 = 96\]Thus solving the equation as true.This method known as 'trial and error' or synthetic division is practical for solving cubic problems in educational settings. Once \( w \) is found, it’s easy to calculate other dimensions using their simple relationships:

• Length = \( 2w \)
• Height = \( w + 2 \)

This efficient approach helps deepen understanding of how dimensions relate and ensures the solutions satisfy the initial conditions of the volume.