Question

The projected fuel cost \(C\) (in millions of dollars per year) for an airline company from 2007 through 2013 is \(C_{1}=568.5+7.15 t\), where \(t=7\) corresponds to \(2007 .\) If the company purchases more efficient airplane engines, fuel cost is expected to decrease and to follow the model \(C_{2}=525.6+6.43 t\). How much can the company save with the more efficient engines? Explain your reasoning.

Short answer

The savings is given by the difference between the total cost from the current model and the total cost from the efficient model for each year from 2007 to 2013.

Step-by-step solution

Calculate cost with current engines

First, calculate the cost for each year from 2007 to 2013 using the current model, \(C_1=568.5+7.15t\). Here t starts from 7 (representing 2007) and goes up to 13 (representing 2013). So, calculate \(C_1\) for each \(t\) from 7 to 13 and sum up the costs.

Calculate cost with efficient engines

Next, calculate the cost for each year from 2007 to 2013 using the efficient model, \(C_2=525.6+6.43t\). As in previous step, \(t\) starts from 7 and ends at 13. Calculate \(C_2\) for each \(t\) within the range and sum up the costs.

Find the difference between the costs

Finally, find the difference between the two sums of costs obtained in steps 1 and 2 to figure out the savings achievable by switching to the efficient engines.

Key concepts

Algebraic Modeling

The concept of algebraic modeling is foundational in understanding varied real-world scenarios, including fuel cost predictions for companies. It involves creating a mathematical representation to emulate a particular situation. In our context, we use algebraic expressions to model the fuel cost for an airline over a series of years.

This kind of modeling translates a verbal description of a problem into an algebraic equation. For the exercise mentioned, the projected fuel cost, represented as function of time, embodies this principle. The equation provided, like the initial model for cost (\(C_1=568.5+7.15t\)) and the subsequent efficient model (\(C_2=525.6+6.43t\)), are examples of algebraic models, where variables and coefficients reflect real-world quantities. Variables in such models provide flexibility and allow for easy adjustment to represent different scenarios – in this case, varying years.

By using algebraic modeling, making predictions and planning for the future becomes more systematic for a company. It incorporates tangible data, such as the annual increase in cost and base costs, to facilitate decision-making, forecasting, and even understanding the impacts of introducing new technologies or practices, such as more efficient airplane engines.

Linear Functions

Linear functions are ubiquitous in everyday calculations, and they form the backbone of simple algebraic modeling. For students tackling problems involving cost over time, mastering the linear function is paramount. A linear function can be intuitively understood as a straight line when plotted on a graph, described by the formula \(y = mx + b\), where \(m\) denotes the slope, and \(b\) is the y-intercept.

In the exercise given, the airline's fuel costs over time are represented through linear functions - \(C_1\) and \(C_2\). The slope in these functions (7.15 and 6.43, respectively) shows the rate at which fuel costs are increasing per year. Meanwhile, the y-intercept (568.5 and 525.6) indicates the basic cost at the beginning of the timeframe when \(t = 0\), which could be seen as an initial fixed cost.

Understanding Linear Functions for Cost Predictions

When using linear functions, the narrow 'straight line' assumption is beneficial. It's important for students to realize this limitation as well. While the model provides simplicity and ease of use, real-world costs might not always increase at a constant rate. Nevertheless, for the purposes of evaluating cost savings as shown in the exercise, linear functions provide excellent approximations and are easy to work with.

Cost-Benefit Analysis

Cost-benefit analysis is a systematic evaluation of the advantages (benefits) and disadvantages (costs) of a project or decision. It’s a critical concept for making business decisions, including those related to investments in technology or infrastructure improvements, like in the exercise with the airline company contemplating the switch to more efficient engines.

By applying this analysis, the company can quantify in monetary terms the expected savings over time. The aim is to determine if the anticipated benefits, such as fuel cost savings, outweigh the costs associated with the new engines. In straightforward exercises like ours, the calculation may only account for direct savings, but in a full-scope cost-benefit analysis, one would include costs like the purchase and installation of the engines, training, downtime, and even potential environmental impact reductions.

Practical Application of Cost-Benefit Analysis

After calculating the costs for each model over the years and finding the difference, the company gains a clear picture of the potential fuel savings. This crucial insight supports a well-informed financial decision. Should the savings be substantial, reasoning would dictate that investing in the new engines is a prudent move. Whereas if savings are minimal or non-existent, the company might look into alternative cost-saving strategies or negotiate better terms for the efficiency upgrade.