Question

Monthly Cost A company determines that the average monthly cost \(C\) (in dollars) for staffing temporary positions can be modeled by \(C=135.47 t^{2}+13,702, \quad t \geq 0\) where \(t\) represents the year, with \(t=0\) corresponding to 2000 . Use the model to predict the year in which the average monthly cost is about \(\$ 25,000\).

Short answer

Solving for \(t\) gives \(t ≈ \sqrt{11298 / 135.47}\). Thus, the year when the average monthly cost is about $25,000 is approximately 2000 + \(t ≈ \sqrt{11298 / 135.47}\)

Step-by-step solution

Expressing the Problem Statement in Equation Form

The equation from the problem is \(C=135.47 t^{2}+13,702\). We need to solve for \(t\) when \(C = 25,000\). Let's set up the equation as follows: \(25000 = 135.47 t^{2}+13,702\).

Rearrange the Equation and Simplify

We start by moving \(13,702\) from the right-hand side (RHS) to the left-hand side (LHS) by subtracting \(13,702\) from both sides. The equation becomes \(25000 - 13702 = 135.47 t^{2}\), which simplifies to \(11298 = 135.47 t^{2}\).

Solving for \(t\)

To find the value of \(t\), divide both sides of the equation by \(135.47\). This gives \(t^{2} = 11298 / 135.47\). Taking the square root of both sides to solve for \(t\) gives \(t = \sqrt{11298 / 135.47}\).

Calculating the Year

This gives the approximate value for \(t\). Summing this value with 2000 gives the year in which the average cost will be $25,000.

Key concepts

Algebraic Equations

To explore the world of algebraic equations, let's relate them to real-life applications, such as predicting costs in a business. Equations serve as tools to describe mathematical relationships succinctly. In our textbook problem, the average monthly cost for staffing temporary positions is given by an equation that directly relates the cost to the square of the time since the year 2000.

An algebraic equation can look daunting at first, but by understanding that each component has a specific role, the equation becomes much clearer. For instance, our equation, \( C = 135.47 t^2 + 13,702 \), consists of variables and constants. Here, \( C \) represents the monthly cost, and \( t \) stands for time in years since the year 2000. The numbers 135.47 and 13,702 are constants that have been determined by the company through some statistical analysis or historical data to best fit their cost model.

When using these equations to solve real problems, we often rearrange them to isolate the variable of interest—like solving for \( t \) when \( C \) is known. This is done through inverse operations: subtracting or adding to both sides, multiplying or dividing, and even taking roots when necessary. The power of algebraic equations lies in their flexibility and the in-depth understanding they provide into the relationships between different quantities.

Quadratic Models

Understanding the Shape and Properties

Quadratic models are a specific type of algebraic equation characterized by the presence of a squared variable—hence the term 'quadratic', which originates from 'quad' meaning square. The general form of a quadratic equation is \( ax^2 + bx + c = 0 \), where \( a \), \( b \), and \( c \) are constants and \( x \) is the variable.

In the context of our average monthly cost problem, the equation \( C = 135.47 t^2 + 13,702 \) is a quadratic model where the variable \( t \) is squared. One of the most crucial properties of quadratic models is their graph; it's a parabola. Parabolas have a distinct 'U' shape, which can open up or down depending on the sign of the \( a \) constant, which in our case is positive, indicating that the costs will rise over time.

Real-World Application of Quadratics

Quadratic models are incredibly useful in various fields such as physics, economics, engineering, and biology. They can model trajectories, economic profit curves, and growth rates, among others. The ability to predict future values by fitting a quadratic model to known data points allows businesses to plan and make decisions with a better understanding of potential outcomes.

Mathematical Prediction

Mathematical predictions enable us to project future outcomes based on current or past information. This practice is integral across disciplines, from meteorology to finance. In our exercise, we use mathematical prediction to determine a future year when the average monthly cost reaches a particular value.

By manipulating the given quadratic equation, we make a prediction about the year. It's important to recognize that predictions based on mathematical models, like the one used for the staffing costs, rely on the assumption that the factors influencing the model remain constant over time. This means that changes in the market, the economy, or internal company policies could affect the accuracy of our prediction. Nevertheless, mathematical predictions are powerful because they can offer quantifiable insights and help guide strategic planning and decision-making processes.

In any prediction task, the careful interpretation of results is crucial. While the model gives us an estimated time, it's essential to consider the margin of error and the model's limitations. This encourages a sensible approach to decision-making, informed by data but cognizant of its potential variability.