Problem 75 - College Algebra

Question

For the following exercises, find the dimensions of the box described. The length is 3 inches more than the width. The width is 2 inches more than the height. The volume is 120 cubic inches.

Step-by-Step Solution

Answer

The dimensions are: height = 3 in, width = 5 in, length = 8 in.

Step 1: Define Variables

Begin by defining the variables for the box's dimensions. Let the height of the box be \( h \). Then, the width will be \( w = h + 2 \), and the length will be \( l = w + 3 \), which simplifies to \( l = h + 5 \).

Step 2: Write the Volume Equation

The volume \( V \) of a box is given by the formula \( V = l imes w imes h \). For this box, substituting the expressions for \( l \) and \( w \), we get:\[ V = (h + 5) \times (h + 2) \times h = 120 \].

Step 3: Solve the Equation

Expand and solve the equation from Step 2:\((h + 5) \times (h + 2) \times h = 120 \) becomes:\( h(h^2 + 7h + 10) = 120 \).This simplifies to: \( h^3 + 7h^2 + 10h = 120 \).Rearrange to find roots: \( h^3 + 7h^2 + 10h - 120 = 0 \). Try values of \( h \). If \( h = 3 \),\(3^3 + 7(3)^2 + 10(3) = 27 + 63 + 30 = 120 \).

Step 4: Calculate Width and Length

Once you identify \( h = 3 \), substitute it back to find \( w \) and \( l \):\( w = h + 2 = 3 + 2 = 5 \) inches.\( l = h + 5 = 3 + 5 = 8 \) inches.

Key Concepts

Algebraic Expressions

Algebraic expressions are mathematical phrases that can include numbers, variables, and operations. In our exercise, we used algebraic expressions to represent the dimensions of the box. We defined the height of the box as the variable \( h \). Then, we expressed the width and length of the box in terms of \( h \), using simple arithmetic expressions.

• Width \( w \) was defined as \( h + 2 \), indicating that the width is always 2 inches more than the height.
• Length \( l \) was defined as \( h + 5 \), suggesting that the length is 5 inches more than the height.

These expressions allow us to work with the variable \( h \) while keeping relationships between the dimensions intact. This is crucial when dealing with equations, especially when solving for unknown variables.

Volume of a Box

The volume of a box refers to the amount of space it occupies and is usually measured in cubic units. The formula for the volume of a box is found by multiplying its length, width, and height together. In our example, the volume is given as 120 cubic inches.
To find this, we used the formula:\[V = l \times w \times h\]By substituting our algebraic expressions for length and width, we expanded this to:\[V = (h + 5) \times (h + 2) \times h = 120\]
Calculating the volume requires understanding these relationships and substituting correctly to solve for the unknowns.

Solving Equations

Solving equations involves finding the value of the variables that make the equation true. In this exercise, we needed to find the dimensions of the box that satisfy the given volume of 120 cubic inches. We had the equation:\[(h + 5)(h + 2)h = 120\]We expanded and simplified:\[h^3 + 7h^2 + 10h = 120\]This resulted in a cubic equation \( h^3 + 7h^2 + 10h - 120 = 0 \) that needed solving for \( h \). We tried different values for \( h \) (beginning with simpler integers) to find that \( h = 3 \) satisfied the equation.
Once the height \( h \) was found, determining the remaining dimensions was a straightforward application of previously defined expressions for width and length. Solving equations efficiently requires a systematic approach, often by testing potential solutions or through algebraic manipulation.