Problem 81 - A Graphical Approach to College Algebra

Question

A box has rectangular sides and a rectangular top and base that are twice as long as they are wide. The volume of the box is 588 cubic inches, and the surface area of the outside of the box is 448 square inches. Find the dimensions of the box.

Step-by-Step Solution

Answer

The dimensions of the box are 14 inches by 7 inches by 6 inches.

Step 1: Understanding the Problem

We have to find the dimensions of a rectangular box given its volume and surface area. The length is twice the width, and we must identify the length (l), width (w), and height (h).

Step 2: Define Variables

Let the width of the box be \( w \), then the length \( l = 2w \). The height is given as \( h \). The two main equations we have are the formula for the volume and the surface area of the box.

Step 3: Volume Equation

The volume \( V \) of a rectangular box is given by \( V = l \times w \times h \). Substituting \( l = 2w \), the volume becomes \( V = 2w \times w \times h = 2w^2h = 588 \).

Step 4: Surface Area Equation

The surface area \( S \) is given by \( S = 2(lw + lh + wh) \). Substituting \( l = 2w \), we get \( S = 2(2w \times w + 2w \times h + w \times h) = 4w^2 + 4wh + 2wh = 6w^2 + 6wh \). So \( 6w^2 + 6wh = 448 \).

Step 5: Simplify and Solve Equations

Use the volume equation to express \( h \) as \( h = \frac{588}{2w^2} = \frac{294}{w^2} \). Substitute this into the surface area equation: \( 6w^2 + 6w\left(\frac{294}{w^2}\right) = 448 \).

Step 6: Solve for Width

This simplifies to \( 6w^2 + \frac{1764}{w} = 448 \). Multiply through by \( w \) to clear the fraction: \( 6w^3 + 1764 = 448w \). Rearrange to find \( 6w^3 - 448w + 1764 = 0 \).

Step 7: Solve the Cubic Equation

Dividing the whole equation by 6 simplifies to \( w^3 - 74.67w + 294 = 0 \). Solving this equation will give the solution for \( w \).

Step 8: Find Width Solution

Trial and error, or a calculator solving cubic equations, finds that \( w \approx 7 \).

Step 9: Calculate Length and Height

Since \( w = 7 \), then \( l = 2w = 14 \). Use \( h = \frac{294}{w^2} = \frac{294}{49} = 6 \).

Key Concepts

Volume of a Box

The volume of a box, particularly one with a rectangular shape, refers to the amount of space inside it. When determining the volume, it's essential to know the dimensions: length, width, and height. The fundamental equation to calculate the volume \[ V = l \times w \times h \]where \( l \) is the length, \( w \) is the width, and \( h \) is the height. This equation allows you to multiply these dimensions to find how much space the box contains.
In our exercise, the box has one side longer than the other based on the condition that the length is twice the width. Thus, this gives the specific form \[ V = 2w^2h \]
knowing the volume is 588 cubic inches helps us find one missing variable when the other two dimensions are known.

Surface Area Calculation

Surface area is the total area covered by the outer surfaces of the box. For a rectangular box, it involves calculating the areas of all six faces. The formula is:\[ S = 2(lw + lh + wh) \]
This formula considers that opposite faces have the same area: two face pairs each having areas of \( lw \), \( lh \), and \( wh \) respectively.
For our specific problem, where the length is twice the width, we substitute to simplify:\[ S = 6w^2 + 6wh \]
Given a total surface area of 448 square inches, this equation connects the dimensions together, facilitating the calculation of missing parameters, when combined with other equations like the volume formula.

Cubic Equation Solving

When calculating dimensions using given conditions, we often encounter cubic equations. These arise in cases where we have multiple terms involving the variable raised to different powers, often leading to complex algebraic challenges. In the exercise, the process simplifies to the equation:\[ 6w^3 - 448w + 1764 = 0 \]
Dividing by 6 to get:\[ w^3 - 74.67w + 294 = 0 \]
This cubic equation represents the real-world geometric constraints of the box and finding the root of it provides the width. Solving cubic polynomials can be intricate, involving techniques such as polynomial division, graphical methods, or applying numerical methods and tools like calculators that handle these types of equations. In this exercise, trial and error or computational aides lead to discovering that \( w \approx 7 \), solving the width of the box.