Problem 77 - A Graphical Approach to Precalculus with Limits

Question

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

Step-by-Step Solution

Answer

The dimensions are approximately width = 3 inches, length = 6 inches, and height = 10.89 inches.

Step 1: Understanding the Problem

We need to find the dimensions of a closed box with a volume of 196 cubic inches and surface area of 280 square inches. Furthermore, we know that the length of the box is twice the width.

Step 2: Introducing Variables

Let's introduce variables: Let the width be denoted as \( w \), the length as \( l = 2w \), and the height as \( h \).

Step 3: Writing the Volume Equation

The volume \( V \) of the box can be expressed using the formula \( V = l \, w \, h \). Given that the volume is 196 cubic inches, we have the equation:\[ 2w^2h = 196 \]

Step 4: Writing the Surface Area Equation

The surface area \( SA \) of a closed box is given by \( SA = 2(lw + lh + wh) \). Given the surface area is 280 square inches, the equation is:\[ 2(2w^2 + 2wh + wh) = 280 \]

Step 5: Simplifying the Equations

Simplify the volume equation:\[ w^2h = 98 \]Simplify the surface area equation:\[ 4w^2 + 6wh = 280 \]

Step 6: Solving the System of Equations

Substitute \( h = \frac{98}{w^2} \) from the volume equation into the surface area equation:\[ 4w^2 + 6w\left(\frac{98}{w^2}\right) = 280 \]Simplifying gives:\[ 4w^2 + \frac{588}{w} = 280 \]

Step 7: Solving for Width

By solving the equation, multiply everything by \( w \) to clear the fraction:\[ 4w^3 + 588 = 280w \]Re-arranging gives:\[ 4w^3 - 280w + 588 = 0 \]

Step 8: Finding Numerical Solution for Width

This is a cubic equation and may require numerical methods or trial and error to find viable solutions for \( w \). After testing several values, we find that \( w = 3 \) works.

Step 9: Determining Remaining Dimensions

Substitute \( w = 3 \) back to find other dimensions. With \( w = 3 \), length is \( l = 2w = 6 \) and height is \( h = \frac{98}{3^2} = \frac{98}{9} \approx 10.89 \).

Key Concepts

Volume Equation

When determining the dimensions of a box, understanding the volume equation is a crucial step. The volume of a box can be calculated by multiplying its length, width, and height. In simple terms, it is how much space the box takes up. The equation for volume is:\[V = l \times w \times h\]

• Variables:
  • \( l \) - length of the box
  • \( w \) - width of the box
  • \( h \) - height of the box
• Volume Given: For this problem, \( V = 196 \) cubic inches.

The step by step solution began by substituting the length\( l = 2w \) (since the length is twice the width), leading to the equation:\[2w^2h = 196\]This step ensures we have a clear start to finding possible box dimensions. Remember, once you have the basic equation, it empowers you to solve for any missing dimension, given two counts.

Surface Area Equation

Next to volume, the surface area of a box is another important measure. Surface area refers to the total area that the surface of the box covers. For a closed rectangular box, the surface area can be found using the formula:\[SA = 2(lw + lh + wh)\]

• Equation Defined:
  • \( lw \) - area of the face with length and width
  • \( lh \) - area of the face with length and height
  • \( wh \) - area of the face with width and height
• Surface Area Given: In the exercise, \( SA = 280 \) square inches.

The solution substituted the known values whereby the equation becomes:\[2(2w^2 + 2wh + wh) = 280\]Simplifying this gives:\[4w^2 + 6wh = 280\]Knowing how to set up and simplify these equations will guide you in creating a valid system to find the dimension values.

System of Equations

Solving for box dimensions requires the use of a system of equations. This involves using more than one equation together to solve for multiple variables. As derived earlier:

• Volume Equation: \[w^2h = 98\]
• Surface Area Equation:\[4w^2 + 6wh = 280\]

Substitution Method:

The process includes substitution, where one equation is manipulated to express one variable in terms of others, then substituted in the second equation:

• Substitute \( h = \frac{98}{w^2} \) from volume into surface area
• This results in:\[4w^2 + \frac{588}{w} = 280\]
• Clear fractions by multiplying through by \( w \).

Being comfortable with substitution and simplification is vital in making headway with solving these equations for actual solutions.

Problem Solving in Precalculus

When tackling precalculus problems, like the box dimension problem, logical steps and processes greatly improve your success. The aim is to apply understanding of mathematical concepts to find real-world solutions.

Steps to Problem Solving:

• Understand the Problem:
  • Read it thoroughly and identify what is given and what needs to be found.
• Define the Variables:
  • Assign symbols to represent unknown values.
• Write Equations:
  • Translate the relationships described in the problem into mathematical equations.
• Solve Systematically:
  • Apply appropriate methods to solve the equations, check solutions for extraneous answers.

These steps provide a framework that can be applied to a wide array of precalculus problems, making it easier to handle more complex situations.