Langara College - Department of Mathematics and Statistics

## Applications of Linear Functions

A car travelling at a constant speed of 80km/h covers a distance of 80km in each hour. So after 2hours it has gone 160km, and so on. On the other hand, after just 30 minutes it will have gone just half as far as in a full hour, ie 40km. In general, for any time period of length t hours, the number, d, of kilometers covered will be t times the distance covered in one hour. So in this case, d and t are related by the formula d=80t. And of course if we had used x in place of t for the number of hours and y in place of d for the number of km, then the equation would be y=80x.
This is of the form y=mx+b with m=80 and b=0.

Now consider a car travelling say from Vancouver to Abbotsford along Hwy1. If we measure time from when it crosses the Port Mann bridge, then at time zero (when it's on the bridge) the distance that it still has to travel is about   80km, so at a speed of 80km/h, it should take just about one hour to complete its trip. If we now let y be the number of km left to travel (as opposed to the distance travelled so far), and x be the number of minutes since the car crossed the bridge, then when x=0 we have y=80, and when x=60 we have y=0. More generally, after x minutes, the distance travelled will be x/60 times the 80km distance travelled in an hour, and the distance remaining will be 80km minus that, so now (with our new definitions of x and y) y and x will be related by  y = 80 - (x/60)80 = -(3/4)x+80, (which is mx+b with m=-3/4 and b=80).

When people are selling things, they often find that the number of customers goes down if they raise the price (and vice versa). Imagine that a seller of gizmos makes on average 50 sales a week at a price of  \$40 per gizmo, but that for each \$4 price increase she loses 3 sales per week (and for each \$4 price reduction she increases her sales by the same amount). So if she charges \$44 per gizmo she might expect to sell only 47 gizmos per week, but if she charges just \$36 each she might sell 53 a week. What would you expect to happen if she raised the price by just \$2? Since this is just half of the \$4 we would expect the number of sales to drop just half as much, ie from 50 down to 48.5. (Although it may not make sense to talk about a fractional number of sales in any particular week, it is perfectly possible for the average weekly sales volume to be a fractional number). Similarly, for a price increase of just \$1, we'd expect a sales drop of 3/4 or 0.75 sales per week.

Now consider how many sales to expect if the price charged is \$x.  This represents an increase of \$(x-40) from the \$40 price that us gave 50 sales, and each \$1 of increase should lose us 0.75 sales, so altogether we'd expect the sales to be reduced from the original 50 by x-40 times 0.75. In other words, if we let y be the expected average number of sales per week, then when the price is \$x we should get y=50-(x-40)(0.75)=80-0.75x= (-0.75)x+80

(Of course this all depends on the number of sales lost per \$ of price increase being independent of the actual price, which might not always be the case, but is a reasonable assumption if the price change is not too great.)

In each of the above examples one variable is given in terms of another by multiplying by a constant and adding a constant (though in the first case the added term is just zero). The same kind of formula applies (at least approximately) in many other applications.