Mathematical Models

A mathematical model is a description of a system using mathematical concepts and language. Mathematical models are used not only in the natural sciences and  engineering disciplines, but also in the social sciences. Linear modeling can include population change, telephone call charges, the cost of renting a bike, weight management, or fundraising. A linear model includes the rate of change $(m)$ and the initial amount, the y-intercept $b$. After the model is written and a graph of the line is made, either one can be used to make predictions about behaviors.

Real Life Linear Model

Many everyday activities require the use of mathematical models, perhaps unconsciously. One difficulty with mathematical models lies in translating the real world application into an accurate mathematical representation. 

Example: Renting a Moving Van  

A rental company charges a flat fee of $$30$ and an additional $$0.25$ per mile to rent a moving van. Write a linear equation to approximate the cost $y$ (in dollars) in terms of $x$, the number of miles driven. How much would a 75 mile trip cost?  

Using the slope-intercept form of a linear equation, with the total cost labeled $y$ (dependent variable) and the miles labeled $x$ (independent variable):

$\displaystyle y=mx+b$

The total cost is equal to the rate per mile times the number of miles driven plus the cost for the flat fee: 

 $\displaystyle y=0.25x+30$

To calculate the cost of a $75 $ mile trip, substitute $75$ for $x$ into the equation:

 $\displaystyle \begin{aligned} y&=0.25x+30\\ &=0.25(75)+30\\ &=18.75+30\\ &=48.75 \end{aligned}$

Real life Model with Multiple Equations

It's also possible to model multiple lines and their equations.

Example  

Initially, trains A and B are $325$ miles away from each other. Train A is traveling towards B at $50$ miles per hour and train B is traveling towards A at $80$ miles per hour.  At what time will the two trains meet? At this time how far did the trains travel?

First, begin with the starting positions of the trains,  ($y$-intercepts, $b$). Train A starts are the origin, $(0,0)$. Since train B is $325$ miles away from train A initially, its position is $(0,325)$. 

Second, in order to write the equations representing each train's total distance in terms of time, calculate the rate of change for each train. Since train A is traveling towards train B, which has a greater $y$ value, train A's rate of change must be positive and equal to its speed of $50$. Train B is traveling towards A, which has a lesser $y$ value, giving B a negative rate of change: $-80$.

The two lines are thus: 

$\displaystyle y_A=50x\\ $

And:

$\displaystyle y_B=−80x+325$

The two trains will meet where the two lines intersect.  To find where the two lines intersect set the equations equal to each other and solve for $x$:

$\displaystyle y_{A}=y_{B}$  

$\displaystyle 50x=-80x+325$

Solving for $x$ gives:

 $\displaystyle x=2.5$

The two trains meet after $2.5$ hours. To find where this is, plug $2.5$ into either equation.

Plugging it into the first equation gives us $50(2.5)=125$, which means it meets after A travels $125$ miles.

Here is the distance versus time graphic model of the two trains:

Trains

Train A (red line) is represented by the equation: $y=50x$, and Train B (blue line) is represented by the equation: $y=-80x+325$.  The two trains meet at the intersections point $(2.5,125)$, which is after $125$ miles in $2.5$ hours.