Problem 74 - College Algebra

Question

For the following exercises, find the dimensions of the box described. The length is three times the height and the height is one inch less than the width. The volume is 108 cubic inches.

Step-by-Step Solution

Answer

The box dimensions are: length = 9 inches, width = 4 inches, height = 3 inches.

Step 1: Define variables

Let's define the variables for the dimensions of the box:- Let \( h \) be the height of the box.- The length \( l \) of the box is given as three times the height, so \( l = 3h \).- The width \( w \) is one inch more than the height, so \( w = h + 1 \).

Step 2: Set up the volume equation

The volume \( V \) of a box is calculated as the product of length, width, and height: \[ V = l \times w \times h \]We know the volume is 108 cubic inches, so substituting the expressions for \( l \) and \( w \) gives:\[ 108 = (3h) \times (h + 1) \times h \].

Step 3: Simplify the equation

First, simplify the volume equation by multiplying:\[ 108 = 3h^2(h + 1) \]This simplifies to:\[ 108 = 3h^3 + 3h^2 \].

Step 4: Solve for height

Rearrange the equation to solve for \( h \):\[ 3h^3 + 3h^2 - 108 = 0 \]Divide through by 3:\[ h^3 + h^2 - 36 = 0 \]We solve this equation for \( h \) using trial-and-error or factoring methods. Let's try \( h = 3 \):\[ 3^3 + 3^2 - 36 = 27 + 9 - 36 = 0 \]So, \( h = 3 \).

Step 5: Find the other dimensions

Now that we know \( h = 3 \):- The length \( l = 3h = 3 \times 3 = 9 \).- The width \( w = h + 1 = 3 + 1 = 4 \).

Step 6: Verify the solution

Check that these dimensions give the correct volume:\[ V = l \times w \times h = 9 \times 4 \times 3 = 108 \]The calculated volume is indeed 108 cubic inches, which matches the problem description. The solution satisfies all conditions.

Key Concepts

Volume of a Box

To grasp the concept of the volume of a box, let’s begin with understanding what volume itself means in geometry. Volume is a measure of the space that an object occupies. For a box, which is also known as a rectangular prism, the volume is found by multiplying its length, width, and height together. This is often expressed with the formula:

• Volume (V) = Length (l) \( \times \) Width (w) \( \times \) Height (h)

For instance, if we have a box with a length of 3 units, a width of 4 units, and a height of 5 units, its volume would be calculated as:\[V = 3 \times 4 \times 5 = 60 \text{ cubic units}\]This tells us how many cubic units can fit inside the box. Relating this to our problem, we know the volume is 108 cubic inches. The goal is to find dimensions that fulfill this condition.

Solving Equations

Solving equations is a vital skill in mathematics that helps us find unknown values. Equations are essentially mathematical statements indicating that two expressions are equal, and it is our task to determine the value of the variable that makes this statement true.
In the box problem, we start by expressing the dimensions in terms of a single variable. Knowing the volume formula and substituting our expressions for length and width introduces an equation. Thus, we have:

• Volume equation: \( 108 = (3h) \times (h + 1) \times h \)

To solve this, we simplify it step by step. Initially, we get:\[108 = 3h^3 + 3h^2\]Simplification usually involves factoring or solving by substitution or elimination until you find the value for the unknown variable. In this case, trial and error was used to find that \( h = 3 \) satisfies the equation. Understanding different methods of solving equations can help tackle problems in multiple ways.

Algebraic Expressions

Algebraic expressions are combinations of numbers, variables, and operators (like addition or multiplication) that represent mathematical ideas. They are essential in forming equations and solving problems like the one involving the box.In our example, we used algebraic expressions to define the dimensions of the box:

• Length: \( l = 3h \)
• Width: \( w = h + 1 \)

These expressions allowed us to replace \( l \) and \( w \) in the volume formula. Manipulating expressions through operations like distributing, combining like terms, and factoring is key to solving the volume equation. By substituting clear expressions for \( l \) and \( w \), we connect the relationships between dimensions and the given volume, creating a pathway to find the values that solve the problem.