Problem 97 - Beginning and Intermediate Algebra

Question

Determine the domain of each function. For labor only, the Arctic Air-Conditioning Company charges \(\$ 40\) to come to the customer's home plus \(\$ 50\) per hour. These labor charges can be described by the function \(L(h)=50 h+40,\) where \(h\) is the time, in hours, and \(L\) is the cost of labor, in dollars. A. Find \(L(1)\) and explain what this means in the context of the problem. B. Find \(L(1.5)\) and explain what this means in the context of the problem. C. Find \(h\) so that \(L(h)=165,\) and explain what this means in the context of the problem.

Step-by-Step Solution

Answer

A. \(L(1) = 90\), meaning the cost of labor for 1 hour is $90. B. \(L(1.5) = 115\), meaning the cost of labor for 1.5 hours is $115. C. \(h=2.5\), meaning the cost of labor is $165 for 2.5 hours of work.

Step 1: A. Finding L(1)

To find L(1), plug in h=1 into the function L(h)=50h+40: L(1) = 50(1) + 40 = 50 + 40 = 90 So, L(1) = \(90. This means that the cost of labor for 1 hour is \)90.

Step 2: B. Finding L(1.5)

To find L(1.5), plug in h=1.5 into the function L(h)=50h+40: L(1.5) = 50(1.5) + 40 = 75 + 40 = 115 So, L(1.5) = \(115. This means that the cost of labor for 1.5 hours is \)115.

Step 3: C. Finding h for L(h) = 165

Now we need to find the duration h such that L(h) = 165. Using the function L(h) = 50h + 40, we have: 165 = 50h + 40 First, subtract 40 from both sides: 125 = 50h Next, divide both sides by 50: h = 2.5 Thus, h = 2.5 hours. This means that for the cost of labor to be $165, the Arctic Air-Conditioning Company has to work for 2.5 hours.

Key Concepts

Linear Functions

A linear function is a type of function where the relationship between the independent variable and the dependent variable is a straight line when graphed. In mathematical terms, a linear function can be written in the form of \( f(x) = mx + b \). Here, \( m \) represents the slope of the line, and \( b \) is the y-intercept, which is where the line crosses the y-axis.

For the Arctic Air-Conditioning Company in our exercise, the labor charges are represented by the function \( L(h) = 50h + 40 \). The variable \( h \) stands for hours worked, while \( L(h) \) is the total labor cost.

• The slope \( m = 50 \) indicates that for every additional hour worked, labor costs increase by \(50.
• The y-intercept \( b = 40 \) shows a fixed charge of \)40 that is applied regardless of the hours worked.

Understanding linear functions helps to easily predict costs based on hours, making budgeting simpler for the client.

Cost Function

A cost function is a fundamental concept in business and economics, representing how total costs change with the level of production or use of resources. It's crucial for businesses to know this to determine pricing and profitability.

In our exercise, the function \( L(h) = 50h + 40 \) acts as the cost function for labor. Here is how it breaks down:

• The term \( 50h \) reflects the variable cost, which directly correlates with the hours worked. More hours lead to higher costs.
• The constant term \( 40 \) is the fixed cost. This cost remains the same no matter how many hours are worked.

By understanding this cost function, the company can estimate labor expenses and clients can anticipate charges, leading to fair pricing strategies.

Solving Equations

Solving equations involves finding the value of variable expressions that make a mathematical statement true. This often requires the application of algebraic techniques to isolate the variable in question.

In Step C of the solution, we aim to find \( h \) when \( L(h) = 165 \). We start with the equation:

\[ 165 = 50h + 40 \]

• First, subtract 40 from both sides to remove the fixed cost:\
  \[ 125 = 50h \]
• Then, divide by 50 to solve for hours \( h \):
  \[ h = \frac{125}{50} = 2.5 \]

The solution shows that it takes 2.5 hours to accumulate a labor cost of $165, providing a clear application of solving equations in real-world scenarios.

Function Application

Function application refers to using mathematical functions to solve real-life problems. This involves plugging in values into a function to get results that have practical significance.

In our example, the function \( L(h) = 50h + 40 \) is applied to calculate labor costs.

• For \( L(1) \), when \( h = 1 \), the labor cost for 1 hour is computed as \(90.
• For \( L(1.5) \), when \( h = 1.5 \), the cost is \)115, showing flexibility in calculations for non-whole numbers.

This process of applying functions helps both service providers and consumers predict expenditures and adjust plans accordingly, showcasing the practical utility of mathematics in everyday finance.