Problem 76 - College Algebra and Calculus: An Applied Approach

Question

A company weighs each 16-ounce bag of flour it produces. After production, any bag that does not weigh within \(0.4\) ounce of 16 ounces cannot be sold. Solve the equation \(|x-16|=0.4\) to find the least and greatest acceptable weights of a 16 -ounce bag of flour.

Step-by-Step Solution

Answer

The least and greatest acceptable weights of a 16-ounce bag of flour are 15.6 ounces and 16.4 ounces, respectively.

Step 1: Understanding the absolute value equation

An absolute value equation is of the form \(|f(x)| = a\), where \(f(x)\) is a function of \(x\), and \(a\) is a real number. This equation means the distance of \(f(x)\) from zero is \(a\). So in this problem, \(|x-16| = 0.4\) means the weight of the bag of flour deviates at most 0.4 ounces from 16 ounces.

Step 2: Solving the absolute value equation

Since the distance of \(x\) from 16 is 0.4, this means \(x\) can be 0.4 more or less than 16. Thus the solution to \(|x-16| = 0.4\) can be found by setting \(x-16 = 0.4\) and \(x-16 = -0.4\) and solving for \(x\). Here, \(x = 16 + 0.4 = 16.4\) (the greatest acceptable weight) and \(x = 16 - 0.4 = 15.6\) (the least acceptable weight).

Step 3: Check the results

To check these solutions, substitute \(x = 16.4\) and \(x = 15.6\) into the original equation \(|x-16| = 0.4\). Both of them make the equation true, so they are both solutions to the equation.

Key Concepts

Problem Solving

When tackling absolute value equations, problem solving skills are essential to ensure a clear understanding of the problem. In this exercise, the task is to determine the acceptable weight range for a 16-ounce bag of flour that deviates by no more than 0.4 ounces.
Absolute value represents a distance from zero on a number line, so the equation \(|x-16| = 0.4\) tells us the weight can either be 0.4 ounces more or less than 16 ounces.
The key steps in problem-solving involve:

• Understanding the concepts involved in the problem, like absolute value in this case.
• Setting up the equation based on the described scenario.
• Solving the equation logically, ensuring results match the given conditions.

These steps guide us through breaking down the absolute value statement into understandable components, helping us to solve for the maximum and minimum acceptable weights.

Algebraic Solutions

Solving absolute value equations using algebra involves understanding that such an equation, \(|x-a|=b\), indicates two possible scenarios: \(x-a = b\) or \(x-a = -b\).
Let's dive into how this applies to our exercise: \(|x-16| = 0.4\).
This situation can be broken down as:

• \(x - 16 = 0.4\): Solving gives the equation \(x = 16.4\), indicating the greatest acceptable weight.
• \(x - 16 = -0.4\): Solving here results in \(x = 15.6\), which is the least acceptable weight.

By addressing each scenario separately, we find the complete solution. Using algebra simplifies the process of translating absolute values into solvable equations, making it easier to find the numerical answers.

Inequalities

The concept of inequalities is crucial in understanding absolute value equations because they set boundaries. The equation \(|x-16| = 0.4\) inherently defines an inequality for acceptable weights.
When broken into two parts, \(x-16 = 0.4\) and \(x-16 = -0.4\), this leads to the inequality \(15.6 \leq x \leq 16.4\).
Here's why inequalities matter:

• They help describe a range of possible solutions rather than a single solution.
• We can visualize the allowed values, helping us to easily identify and set quality control measures.
• Inequalities are utilized in determining what deviations are permissible within a specific context, such as manufacturing standards.

Converting absolute value statements into inequalities helps us precisely determine acceptable ranges, thereby ensuring compliance with set standards or conditions.