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.

Solve 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.

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.

The graph below represents the data for Nicole’s potential earnings. We say that earnings vary directly with the sales price of the car. The formula [latex]y=k{x}^{n}[/latex] 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.

Figure 1

Example 1: Solving a Direct Variation Problem

The quantity y varies directly with the cube of x. If y = 25 when x = 2, find y when x is 6.

Show Solution

The general formula for direct variation with a cube is [latex]y=k{x}^{3}[/latex]. The constant can be found by dividing y by the cube of x.

[latex]\begin{align} k&=\frac{y}{{x}^{3}} \\[1mm] &=\frac{25}{{2}^{3}}\\[1mm] &=\frac{25}{8}\end{align}[/latex]

Now use the constant to write an equation that represents this relationship.

[latex]y=\frac{25}{8}{x}^{3}[/latex]

Substitute x = 6 and solve for y.

[latex]\begin{align}y&=\frac{25}{8}{\left(6\right)}^{3} \\[1mm] &=675 \end{align}[/latex]

Analysis of the Solution

The graph of this equation is a simple cubic, as shown below.

Figure 2

Solve inverse variation problems

Water temperature in an ocean varies inversely to the water’s depth. Between the depths of 250 feet and 500 feet, the formula [latex]T=\frac{14,000}{d}[/latex] gives us the temperature in degrees Fahrenheit at a depth in feet below Earth’s surface. Consider the Atlantic Ocean, which covers 22% of Earth’s surface. At a certain location, at the depth of 500 feet, the temperature may be 28°F.

If we create a table we observe that, as the depth increases, the water temperature decreases.

We notice in the relationship between these variables that, as one quantity increases, the other decreases. The two quantities are said to be inversely proportional and each term varies inversely with the other. Inversely proportional relationships are also called inverse variations.

For our example, the graph depicts the inverse variation. We say the water temperature varies inversely with the depth of the water because, as the depth increases, the temperature decreases. The formula [latex]y=\frac{k}{x}[/latex] for inverse variation in this case uses k = 14,000.

Figure 3

Example 3: Solving an Inverse Variation Problem

A quantity y varies inversely with the cube of x. If y = 25 when x = 2, find y when x is 6.

Show Solution

The general formula for inverse variation with a cube is [latex]y=\frac{k}{{x}^{3}}[/latex]. The constant can be found by multiplying y by the cube of x.

[latex]\begin{align}k&={x}^{3}y \\ &={2}^{3}\cdot 25 \\ &=200 \end{align}[/latex]

Now we use the constant to write an equation that represents this relationship.

[latex]\begin{align}&y=\frac{k}{{x}^{3}},k=200 \\ &y=\frac{200}{{x}^{3}} \end{align}[/latex]

Substitute x = 6 and solve for y.

[latex]\begin{align}y&=\frac{200}{{6}^{3}} \\ &=\frac{25}{27} \end{align}[/latex]

Analysis of the Solution

The graph of this equation is a rational function.

Figure 4

Solve problems involving joint variation

Many situations are more complicated than a basic direct variation or inverse variation model. One variable often depends on multiple other variables. When a variable is dependent on the product or quotient of two or more variables, this is called joint variation. For example, the cost of busing students for each school trip varies with the number of students attending and the distance from the school. The variable c, cost, varies jointly with the number of students, n, and the distance, d.

Key Equations