# 5.8 Modeling using variation

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In this section, you will:
• Solve direct variation problems.
• Solve inverse variation problems.
• Solve problems involving joint variation.

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 $\text{\hspace{0.17em}}e=0.16s\text{\hspace{0.17em}}$ tells us her earnings, $\text{\hspace{0.17em}}e,\text{\hspace{0.17em}}$ 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] .

$\text{\hspace{0.17em}}s\text{\hspace{0.17em}}$ , sales price $e=0.16s$ Interpretation
$4,600 $e=0.16\left(4,600\right)=736$ A sale of a$4,600 vehicle results in $736 earnings.$9,200 $e=0.16\left(9,200\right)=1,472$ A sale of a $9,200 vehicle results in$1472 earnings.
$18,400 $e=0.16\left(18,400\right)=2,944$ A sale of a$18,400 vehicle results in $2944 earnings. 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 $\text{\hspace{0.17em}}y=k{x}^{n}\text{\hspace{0.17em}}$ is used for direct variation. The value $\text{\hspace{0.17em}}k\text{\hspace{0.17em}}$ is a nonzero constant greater than zero and is called the constant of variation . In this case, $\text{\hspace{0.17em}}k=0.16\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}n=1.\text{\hspace{0.17em}}$ We saw functions like this one when we discussed power functions.

## Direct variation

If $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}\text{and}\text{\hspace{0.17em}}y\text{\hspace{0.17em}}$ are related by an equation of the form

$\text{\hspace{0.17em}}y=k{x}^{n}\text{\hspace{0.17em}}$

then we say that the relationship is direct variation    and $\text{\hspace{0.17em}}y\text{\hspace{0.17em}}$ varies directly    with, or is proportional to, the $\text{\hspace{0.17em}}n\text{th}\text{\hspace{0.17em}}$ power of $\text{\hspace{0.17em}}x.\text{\hspace{0.17em}}$ In direct variation relationships, there is a nonzero constant ratio $\text{\hspace{0.17em}}k=\frac{y}{{x}^{n}},\text{\hspace{0.17em}}$ where $\text{\hspace{0.17em}}k\text{\hspace{0.17em}}$ is called the constant of variation    , which help defines the relationship between the variables.

Given a description of a direct variation problem, solve for an unknown.

1. Identify the input, $\text{\hspace{0.17em}}x,$ and the output, $\text{\hspace{0.17em}}y.\text{\hspace{0.17em}}$
2. Determine the constant of variation. You may need to divide $\text{\hspace{0.17em}}y\text{\hspace{0.17em}}$ by the specified power of $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ to determine the constant of variation.
3. Use the constant of variation to write an equation for the relationship.
4. Substitute known values into the equation to find the unknown.

## Solving a direct variation problem

The quantity $\text{\hspace{0.17em}}y\text{\hspace{0.17em}}$ varies directly with the cube of $\text{\hspace{0.17em}}x.\text{\hspace{0.17em}}$ If $\text{\hspace{0.17em}}y=25\text{\hspace{0.17em}}$ when $\text{\hspace{0.17em}}x=2,\text{\hspace{0.17em}}$ find $\text{\hspace{0.17em}}y\text{\hspace{0.17em}}$ when $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ is 6.

The general formula for direct variation with a cube is $\text{\hspace{0.17em}}y=k{x}^{3}.\text{\hspace{0.17em}}$ The constant can be found by dividing $\text{\hspace{0.17em}}y\text{\hspace{0.17em}}$ by the cube of $\text{\hspace{0.17em}}x.\text{\hspace{0.17em}}$

$\begin{array}{ccc}\hfill k& =& \frac{y}{{x}^{3}}\hfill \\ & =& \frac{25}{{2}^{3}}\hfill \\ & =& \frac{25}{8}\hfill \end{array}$

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

$y=\frac{25}{8}{x}^{3}$

Substitute $\text{\hspace{0.17em}}x=6\text{\hspace{0.17em}}$ and solve for $\text{\hspace{0.17em}}y.$

$\begin{array}{ccc}\hfill y& =& \frac{25}{8}{\left(6\right)}^{3}\hfill \\ & =& 675\hfill \end{array}$

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