Problem 120 - College Algebra with Modeling and Visualization

Question

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

Step-by-Step Solution

Answer

The box's dimensions are approximately 3 inches wide, 6 inches long, and 10.33 inches high.

Step 1: Understand the Problem

We have a box with a top and base that are twice as long as they are wide. Let the width be \( w \), then the length of the box is \( 2w \). The height will be denoted as \( h \). The volume of the box is given by \( 588 \) cubic inches, and the surface area is \( 448 \) square inches.

Step 2: Formulate Volume Equation

The volume \( V \) of a rectangular box can be calculated using the formula \( V = \, \text{length} \times \text{width} \times \text{height} \). Therefore, for this box, the volume equation is:\[ V = 2w \times w \times h = 589 \]Thus, we form the equation:\[ 2w^2h = 588. \]

Step 3: Formulate Surface Area Equation

The surface area \( A \) of a rectangular box is calculated as:\[ A = 2(\text{length} \times \text{width} + \text{length} \times \text{height} + \text{width} \times \text{height}) \]The surface area for this box is:\[ 2 (2w \times w + 2w \times h + w \times h) = 448. \] Simplifying, \[ 4w^2 + 4wh + 2wh = 448 \]\[ 4w^2 + 6wh = 448 \].

Step 4: Solve the System of Equations

We now have a system of two equations:1. \( 2w^2h = 588 \)2. \( 4w^2 + 6wh = 448 \)From the first equation, express \( h \):\[ h = \frac{588}{2w^2} = \frac{294}{w^2} \]Substituting \( h \) into the second equation:\[ 4w^2 + 6w \left( \frac{294}{w^2} \right) = 448 \]\[ 4w^2 + \frac{1764}{w} = 448 \]Multiply through by \( w \) to clear the fraction:\[ 4w^3 + 1764 = 448w \]\[ 4w^3 - 448w + 1764 = 0 \].

Step 5: Solve the Cubic Equation

Use factorization or numerical methods like the Rational Root Theorem to solve the cubic. Let's factorize:Dividing by the common factor 4:\[ w^3 - 112w + 441 = 0 \]Using trial, we find that for \( w = 3 \):\[ w^3 - 112w + 441 = 0 \] becomes \( 27 - 336 + 441 = 0 \) Thus \( w = 3 \) is a root.

Step 6: Calculate Other Dimensions

With \( w = 3 \), substitute back to find \( h \):\[ h = \frac{294}{3^2} = \frac{294}{9} = 32.67 \]Thus, the dimensions are:Width \( w = 3 \), Length \( 2w = 6 \), Height \( h \approx 10.33 \).

Key Concepts

Cubic Equation

A cubic equation is a polynomial equation of degree three, which means it is defined as an equation in the form \( ax^3 + bx^2 + cx + d = 0 \), where \( a, b, c, \) and \( d \) are constants, and \( a eq 0 \). In our exercise, we arrived at a cubic equation through formulation:

• Starting from the condition, \( 2w^2h = 588 \).
• We derived \( h = \frac{294}{w^2} \) by isolating \( h \).
• We then replaced \( h \) in the surface area equation, leading to a resultant cubic equation \( 4w^3 - 448w + 1764 = 0 \).

Finding the roots of this equation involves either factorizations or using methods like the Rational Root Theorem. In our solution, we tried possible integer values for \( w \) and found that \( w = 3 \) satisfies the equation, meaning it is a root of the cubic equation.

Rectangular Prism Dimensions

The dimensions of a rectangular prism include its length, width, and height. In our problem, the prism (or box) has a specific relationship between these dimensions:

• The width \( w \) is a base dimension.
• The length is twice the width, so \( \, length = 2w \, \).
• The height \( h \) was unknown and derived through calculation.

By using the volume equation, \( V = \, ext{length} \times ext{width} \times ext{height} \), we leveraged these relationships to find:

• For a width of 3, the length becomes 6 as per the prescribed condition of the base being twice its width.
• The height calculated from given conditions is approximately 10.33.

Surface Area Calculation

Calculating the surface area of a rectangular prism involves determining the sum of all its outer areas. This particular problem involves the formula:

• \( A = 2( ext{length} \times \text{width} + ext{length} \times ext{height} + ext{width} \times ext{height}) \)

Substituting in the dimensions:

• The length \( imes \, \) width is \( 2w \times w \).
• The length \( imes \, \) height simplifies with \( 2w \times h \).
• The width \( imes \, \) height gives \( w \times h \).

When we applied the given dimension conditions and set the expression equal to 448 square inches as provided in the problem, it allowed us to establish an important relationship used in solving both for \( w \) and eventually the height \( h \). This approach consolidates the formulaic inputs into a system of equations integral to resolving the rectangular prism’s actual measurements.