# 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 $\text{\hspace{0.17em}}y.$ $\begin{array}{ccc}\hfill y& =& \frac{25}{8}{\left(6\right)}^{3}\hfill \\ & =& 675\hfill \end{array}$ #### Questions & Answers what is the coefficient of -4× Mehri Reply -1 Shedrak the operation * is x * y =x + y/ 1+(x × y) show if the operation is commutative if x × y is not equal to -1 Alfred Reply An investment account was opened with an initial deposit of$9,600 and earns 7.4% interest, compounded continuously. How much will the account be worth after 15 years?
lim x to infinity e^1-e^-1/log(1+x)
given eccentricity and a point find the equiation
12, 17, 22.... 25th term
12, 17, 22.... 25th term
Akash
College algebra is really hard?
Absolutely, for me. My problems with math started in First grade...involving a nun Sister Anastasia, bad vision, talking & getting expelled from Catholic school. When it comes to math I just can't focus and all I can hear is our family silverware banging and clanging on the pink Formica table.
Carole
I'm 13 and I understand it great
AJ
I am 1 year old but I can do it! 1+1=2 proof very hard for me though.
Atone
hi
Not really they are just easy concepts which can be understood if you have great basics. I am 14 I understood them easily.
Vedant
find the 15th term of the geometric sequince whose first is 18 and last term of 387
I know this work
salma
The given of f(x=x-2. then what is the value of this f(3) 5f(x+1)
hmm well what is the answer
Abhi
If f(x) = x-2 then, f(3) when 5f(x+1) 5((3-2)+1) 5(1+1) 5(2) 10
Augustine
how do they get the third part x = (32)5/4
make 5/4 into a mixed number, make that a decimal, and then multiply 32 by the decimal 5/4 turns out to be
AJ
how
Sheref
can someone help me with some logarithmic and exponential equations.
20/(×-6^2)
Salomon
okay, so you have 6 raised to the power of 2. what is that part of your answer
I don't understand what the A with approx sign and the boxed x mean
it think it's written 20/(X-6)^2 so it's 20 divided by X-6 squared
Salomon
I'm not sure why it wrote it the other way
Salomon
I got X =-6
Salomon
ok. so take the square root of both sides, now you have plus or minus the square root of 20= x-6
oops. ignore that.
so you not have an equal sign anywhere in the original equation?
hmm
Abhi
is it a question of log
Abhi
🤔.
Abhi
I rally confuse this number And equations too I need exactly help
salma
But this is not salma it's Faiza live in lousvile Ky I garbage this so I am going collage with JCTC that the of the collage thank you my friends
salma
Commplementary angles
hello
Sherica
im all ears I need to learn
Sherica
right! what he said ⤴⤴⤴
Tamia
hii
Uday
hi
salma
hi
Ayuba
Hello
opoku
hi
Ali
greetings from Iran
Ali
salut. from Algeria
Bach
hi
Nharnhar
what is a good calculator for all algebra; would a Casio fx 260 work with all algebra equations? please name the cheapest, thanks.
a perfect square v²+2v+_
kkk nice