On many problems, the most challenging aspect is keeping track of the variables and what they represent. When the variables are scaled in hundreds, thousands, or even millions, we need to pay careful attention.
Scaling means that the quantity has been divided by some amount. For instance, we can write the amount $14,400,000 as 14.4 million dollars.
To scale $14,400,000 in millions, divide by 1,000,000 and then add the word “million” after the result. We could also scale in thousands by dividing by 1000. This gives 14,400 thousand dollars.
Let’s look at an application with scaling.
The daily demand for monorail service in a city can be modeled by
D(p) = -5.1p + 45.2 thousand rides
where p is the fare charged in dollars. The city will provide
S(p) = 3p + 16.1 thousand rides
each day at a fare of p dollars.
a. Find the daily demand when the price is $4.
Since the question asks about demand, we need to use the demand function D(p) to find the solution. This demand function takes in the price p and outputs the number of rides. In this case, we need to find
D(4) = -5.1(4) + 45.2 = 24.8 thousand rides
or 24,800 rides. Be careful when answering questions with units that are scaled (like thousand rides). Confusing 24,800 with 24.8 can lead to many mistakes.
b. At what price is the city willing to supply 32,600 rides?
Like the previous problem, we need to be careful of the scaling. The quantity 32,600 rides may be scaled to 32.6 thousand rides by dividing by 1000. The question indicates that we want the price that results in a supply of 32.6 thousand rides. This is found by finding where S(p) = 32.6:
The city is willing to provide 32,600 rides at a price of $5.5.