91Ó°ÊÓ

The weekly cost of running a small firm is a function of the number of employees. Every week there is a fixed cost of \(\$ 2500\), and each employee costs the firm \(\$ 350\). For example, if there are 10 employees, then the weekly cost is \(2500+350 \times 10=6000\) dollars. a. What is the weekly cost if there are 3 employees? b. Find a formula for the weekly cost as a function of the number of employees. (You need to choose variable and function names. Be sure to state the units.) c. Make a graph of the weekly cost as a function of the number of employees. Include values of the variable up to 10 employees. d. For what number of employees will the weekly cost be \(\$ 4250\) ?

Step by step solution

01

- Determine the weekly cost with 3 employees

To find the weekly cost with 3 employees, we'll use the given cost structure: a fixed cost of \(2500 plus \)350 per employee. For 3 employees, the cost is calculated as follows:\[\text{Total Cost} = 2500 + 350 \times 3 = 2500 + 1050 = 3550\] The weekly cost with 3 employees is $3550.

02

- Create the function for weekly cost

Let \( C(n) \) represent the weekly cost based on \( n \) employees. The function is expressed as:\[C(n) = 2500 + 350n\] where \( n \) is the number of employees, and \( C(n) \) returns the weekly cost in dollars.

03

- Graph the weekly cost function

To graph \( C(n) = 2500 + 350n \), plot the value of \( n \) on the x-axis (number of employees) and \( C(n) \) on the y-axis (cost in dollars). Calculate \( C(n) \) for \( n = 0 \) to \( n = 10 \):- \( n = 0 \), \( C(0) = 2500 \)- \( n = 1 \), \( C(1) = 2850 \)- \( n = 2 \), \( C(2) = 3200 \)- \( n = 3 \), \( C(3) = 3550 \)- ...- \( n = 10 \), \( C(10) = 6000 \)Plot these points to form a linear graph.

04

- Solve for number of employees with given weekly cost

We need to find \( n \) such that \( C(n) = 4250 \). Set up the equation:\[2500 + 350n = 4250\]Subtract 2500 from both sides:\[350n = 1750\]Divide by 350:\[n = 5\]There are 5 employees when the weekly cost is $4250.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Linear Functions

Linear functions are a significant mathematical concept that represents a straight line relationship between two variables. In the context of our exercise, the weekly cost in a firm is linearly related to the number of employees.

This means that as the number of employees increases, the total cost increases at a constant rate. This kind of function is expressed in the standard form of a linear equation:

• Slope-intercept form: \( y = mx + b \)
• Here, \( m \) is the slope and \( b \) is the y-intercept.

In the exercise, the function can be rewritten as \( C(n) = 350n + 2500 \), where \( C(n) \) is the weekly cost, \( n \) stands for the number of employees, and:

• The slope \(m = 350\) represents the additional cost per employee.
• The y-intercept \(b = 2500\) is the fixed cost per week, regardless of the number of employees.

This setup creates a steadily increasing straight line when graphed on a coordinate plane.

Cost Functions

Cost functions are mathematical models that describe how costs change with different variables, like number of employees in a business. They play a crucial role in business planning and management, allowing companies to predict and control expenses.

The cost function in the exercise is \( C(n) = 2500 + 350n \). It consists of:

• A fixed cost: \(2500\) dollars, representing the basic expenses the firm incurs regardless of employee count.
• A variable cost: \(350n\), where each employee adds \(350\) dollars to the weekly total.

This means that for any given number of employees, you can calculate the expected weekly expenses. Understanding cost functions can help businesses make strategic decisions about hiring or optimizing their operations.

Graphing

Graphing linear equations allows us to visualize relationships between variables. It's a powerful tool to understand how two things, like costs and employees, correlate in real life. The graphical representation of our cost function will look like a straight line.

To graph the function \( C(n) = 2500 + 350n \):

• Start by plotting points; for example, when \( n = 0, C(0) = 2500 \), and when \( n = 10, C(10) = 6000 \).
• These points show the cost for having zero and ten employees, respectively.
• Connect these points with a straight line.

The line demonstrates how the cost rises as more employees are hired. Each point on this line represents the cost for a different number of employees, making it easy to predict costs for any given number of employees up to the maximum plotted.

Algebraic Equations

Algebraic equations are the foundation of solving problems involving numbers and unknowns. In this exercise, they allow us to create and understand the function for costs, and solve for unknowns such as determining the number of employees required for a specific cost.

To find out how many employees lead to a weekly cost of \(4250\) dollars, set the equation:

• \(2500 + 350n = 4250\)
• First, subtract \(2500\) from \(4250\) to isolate the term \(350n\): \(350n = 1750\).
• Then, divide each side by \(350\) to solve for \(n\): \(n = 5\).

This means at \(5\) employees, the weekly cost will be \(4250\) dollars. Algebraic methods are key in translating real-world scenarios into solvable problems.