Problem 79 - College Algebra with Modeling and Visualization

Question

Find possible dimensions for a box with a volume of 196 cubic inches, a surface area of 280 square inches, and a length that is twice the width.

Step-by-Step Solution

Answer

The width is approximately 3.5 inches, length is 7 inches, and height is 8 inches.

Step 1: Define Variables

Let's define the variables for the box's dimensions. Let the width be \( w \), the height be \( h \), and the length be \( l \). Given that the length is twice the width, we have \( l = 2w \).

Step 2: Set Up Volume Equation

The volume of the box is given by the equation \( V = lwh = 196 \). Substituting \( l = 2w \) in the volume equation, we get \( (2w)wh = 196 \). This simplifies to \( 2w^2h = 196 \). Then \( w^2h = 98 \).

Step 3: Set Up Surface Area Equation

The surface area of the box is given by the equation \( S = 2(lw + lh + wh) = 280 \). Substituting \( l = 2w \), we have \( S = 2(2w \cdot w + 2w \cdot h + wh) = 280 \). This simplifies to \( 4w^2 + 4wh + 2wh = 140 \), or \( 4w^2 + 6wh = 140 \).

Step 4: Solve System of Equations

We have two equations, \( w^2h = 98 \) and \( 4w^2 + 6wh = 140 \). First, solve for \( h \) from \( w^2h = 98 \) to get \( h = \frac{98}{w^2} \). Substitute \( h = \frac{98}{w^2} \) into \( 4w^2 + 6wh = 140 \) to get \( 4w^2 + 6w\left(\frac{98}{w^2}\right) = 140 \). Simplify this equation to find \( w \).

Step 5: Simplify and Solve for Width

The substitution gives \( 4w^2 + \frac{588}{w} = 140 \). Multiply through by \( w \) to clear the fraction: \( 4w^3 + 588 = 140w \). Rearrange to \( 4w^3 - 140w + 588 = 0 \). Solve this cubic equation to find possible values for \( w \).

Step 6: Calculate Dimensions

Once you find a valid value for \( w \) from solving the cubic equation, use \( w^2h = 98 \) to find \( h \) and \( l = 2w \) to find \( l \). Check that these dimensions satisfy both the volume and surface area constraints.

Key Concepts

Volume of a Box

To understand how to find the volume of a box, imagine filling the box with tiny cubes. The space inside the box is what we call the volume. It measures how much stuff we can fit inside. For a box, the volume is calculated using the formula:
\[ V = lwh \]
Where:

• \( l \): the length of the box
• \( w \): the width of the box
• \( h \): the height of the box

In our problem, the volume is given as 196 cubic inches. To solve for the box's dimensions, we set up the equation using the given length relation \( l = 2w \). Substituting, we form the equation \( 2w^2h = 196 \), which simplifies our task to solving for the unknowns. Understanding this formula is essential in algebraic problem solving, especially in spatial calculations like volume.

Surface Area of a Box

The surface area of a box is the total area of all the surfaces (or faces) of the box. Imagine painting or wrapping the box; the surface area is the amount of paint or wrapping needed. It can be calculated using the formula:
\[ S = 2(lw + lh + wh) \]
This formula adds up the areas of all the six faces of a box:

• Two faces of size \( lw \)
• Two faces of size \( lh \)
• Two faces of size \( wh \)

In this exercise, the surface area is given as 280 square inches. Using the length relation \( l = 2w \), we substitute into the surface area equation, simplifying it to \( 4w^2 + 6wh = 140 \). This equation, when used along with the volume equation, helps us deduce the box dimensions. Knowing how to manipulate such equations is critical in algebraic problem solving.

System of Equations

A system of equations involves solving two or more equations at the same time, finding values which satisfy all the equations in the system. In this exercise, we have two such equations relating to the box:

• The volume equation: \( w^2h = 98 \)
• The surface area equation: \( 4w^2 + 6wh = 140 \)

By solving these equations as a system, we can find the values of \( w \), \( h \), and subsequently \( l \) (since we know \( l = 2w \)).
A key step is using substitution. For instance, solve \( w^2h = 98 \) for \( h \):
\[ h = \frac{98}{w^2} \]
Substitute \( h \) into the second equation to get:
\[ 4w^2 + \frac{588}{w} = 140 \]
Then, by clearing fractions and rearranging, you form a cubic equation to solve for \( w \). Understanding systems of equations allows us to handle multiple constraints and find practical solutions in geometric problems.