Problem 74 - College Algebra

Question

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

Width: 4 inches, Height: 3 inches, Length: 9 inches.

Step 1: Define Variables

Let's define the variables first. Let the width be denoted as \( w \) (in inches). Then the height is given as \( h = w - 1 \). The length of the box is three times the height, which gives \( l = 3h = 3(w - 1) \).

Step 2: Set Up the Volume Equation

The volume of a box is given by the product of its dimensions: \( V = l \times w \times h \). Here, \( V = 108 \) cubic inches. Substitute the expressions for \( l \) and \( h \) to get:\[ V = (3(w - 1)) \times w \times (w - 1) = 108 \]

Step 3: Simplify the Volume Equation

Expand the equation: \[ 3(w - 1) \times w \times (w - 1) = 108 \]This becomes:\[ 3w(w - 1)^2 = 108 \]

Step 4: Solve for Width \( w \)

First, distribute \( (w - 1) \):\[ 3w(w^2 - 2w + 1) = 108 \]Expanding gives:\[ 3w^3 - 6w^2 + 3w = 108 \]Now, solve this equation. Divide through by 3 to simplify:\[ w^3 - 2w^2 + w = 36 \]

Step 5: Factor the Polynomial

We need to solve \( w^3 - 2w^2 + w = 36 \). This equation can be trial-factored or solved using potential rational roots. Testing \( w = 3 \):\[ 3^3 - 2(3)^2 + 3 = 27 - 18 + 3 = 12 eq 36 \]Actually, after checking, a small calculation mistake was earlier. Correcting gives us:Finding roots correctly (such as a numerical/graphical method) yields \( w = 4 \).

Step 6: Calculate Remaining Dimensions

With the width \( w = 4 \), substitute back:Height \( h = w - 1 = 3 \).Length \( l = 3h = 9 \).

Step 7: Verify Solution with Volume

Check the solution by calculating the volume:\[ V = l \times w \times h = 9 \times 4 \times 3 = 108 \]The calculation confirms the given volume of 108 cubic inches.

Key Concepts

Volume Equation

In this problem, we're tasked to find the dimensions of a box, and one crucial aspect is the volume equation. The volume of a rectangular box is computed by multiplying its dimensions: length, width, and height. In formula form, it is written as:

• \( V = l \times w \times h \)

Given that the volume of the box is 108 cubic inches, we set that equal to the product of the dimensions. Remember this rule anytime you're dealing with volume calculations involving a box or cuboid. It helps us find missing dimensions when given the volume and some other relationships between dimensions.

Factoring Polynomials

Solving for an unknown often requires manipulating algebraic expressions, like factoring polynomials. In this context, we first set up a polynomial expression from our volume equation in terms of the width, \( w \). We derived:

• \( w^3 - 2w^2 + w = 36 \)

To solve this, the equation needs to be equalized to zero, leading to:

• \( w^3 - 2w^2 + w - 36 = 0 \)

Factoring helps simplify expressions and solve equations. In this, checking for rational roots or using synthetic division can sometimes assist in finding the value of \( w \). Choose the strategy that fits your current level and ensure you factor properly for accurate results.

Variable Definition

Defining variables is about assigning letters or symbols to represent quantities we are dealing with. At the start of this problem:

• Width is chosen as \( w \)
• Height is defined as \( h = w - 1 \)
• Length becomes \( l = 3h = 3(w - 1) \)

By expressing these dimensions in terms of \( w \), we simplify the problem into a single variable equation, making it easier to solve. Always begin with establishing clear relationships among variables to streamline and solve your problem efficiently.

Dimensional Analysis

Dimensional analysis is a fancy term for checking that your calculations make sense regarding the units you're working with. In this problem, we're dealing with inches and cubic inches. When we calculate our dimensions:

• Check the units of each dimension (e.g., width, height, and length are in inches)
• Ensure their product equals the volume in cubic inches

Dimensional analysis helps catch errors in equations or calculations by ensuring the units align correctly. This practice is vital in any mathematics or physics problem dealing with dimensions.