5.8 Modeling using variation  (Page 17/14)

A used-car company has just offered their best candidate, Nicole, a position in sales. The position offers 16% commission on her sales. Her earnings depend on the amount of her sales. For instance, if she sells a vehicle for $4,600, she will earn $736. She wants to evaluate the offer, but she is not sure how. In this section, we will look at relationships, such as this one, between earnings, sales, and commission rate.

Solving direct variation problems

In the example above, Nicole’s earnings can be found by multiplying her sales by her commission. The formula $e=0.16s$ tells us her earnings, $e,$ come from the product of 0.16, her commission, and the sale price of the vehicle. If we create a table, we observe that as the sales price increases, the earnings increase as well, which should be intuitive. See [link] .

Notice that earnings are a multiple of sales. As sales increase, earnings increase in a predictable way. Double the sales of the vehicle from $4,600 to $9,200, and we double the earnings from $736 to $1,472. As the input increases, the output increases as a multiple of the input. A relationship in which one quantity is a constant multiplied by another quantity is called direct variation . Each variable in this type of relationship varies directly with the other.

[link] represents the data for Nicole’s potential earnings. We say that earnings vary directly with the sales price of the car. The formula $y=k{x}^n$ is used for direct variation. The value $k$ is a nonzero constant greater than zero and is called the constant of variation . In this case, $k=0.16$ and $n=1.$ We saw functions like this one when we discussed power functions.