## 8.5 – Applications of Rational Equations

### Learning Objectives

• (8.5.1) – Solve a rational formula for a specified variable
• (8.5.2) – Solving applications of rational equations
• Average cost problems
• Work problems
• Motion problems
• Concentration of a mixture problems

# (8.5.1) – Solve a rational formula for a specified variable

Rational formulas can be useful tools for representing real-life situations and for finding answers to real problems. Equations representing direct, inverse, and joint variation are examples of rational formulas that can model many real-life situations. As you will see, if you can find a formula, you can usually make sense of a situation.

When solving problems using rational formulas, it is often helpful to first solve the formula for the specified variable. For example, work problems ask you to calculate how long it will take different people working at different speeds to finish a task. The algebraic models of such situations often involve rational equations derived from the work formula, $W=rt$. The amount of work done $W$ is the product of the rate of work $r$ and the time spent working $t$. Using algebra, you can write the work formula 3 ways:

$W=rt$

Find the time $t$:

$\displaystyle t=\frac{W}{r}$ (divide both sides by $r$)

Find the rate $r$:

$\displaystyle r=\frac{W}{t}$(divide both sides by $t$)

### Example

The formula for finding the density of an object is $\displaystyle D=\frac{m}{v}$, where $D$ is the density, $m$ is the mass of the object and $v$ is the volume of the object. Rearrange the formula to solve for the mass $m$ and then for the volume $v$.

Now let’s look at an example using the formula for the volume of a cylinder.

### Example

The formula for finding the volume of a cylinder is $V=\pi{r^{2}}h$, where $V$ is the volume, $r$ is the radius and $h$ is the height of the cylinder. Rearrange the formula to solve for the height $h$.

In the following video we give another example of solving for a variable in a formula, or as they are also called, a literal equation.

# (8.5.2) – Solving applications of rational equations

### Average cost problems

Suppose we know that the cost of making a product is dependent on the number of items, $x$, produced. This is given by the equation $C\left(x\right)=15,000+500x$. Note, the cost function consists of a fixed cost ($\15,000$) and a variable cost $(500x)$. The fixed cost doesn’t change when more items are produced, whereas the variable cost increases as more cars are produced.  If we want to know the average cost for producing $x$ items, we would divide the cost function by the number of items, $x$.

The average cost function, which yields the average cost per item for $x$ items produced, is

$\displaystyle \overline{C}\left(x\right)=\frac{15,000 +500x}{x}$

Many other application problems require finding an average value in a similar way, giving us variables in the denominator.

### ExAMPLE

Let $x$ represent the number of cars produced in a factory in a typical week. Suppose that the the cost in dollars to produce $x$ cars is given by $C(x) = 15,000+500x$. How many cars would the factory have to produce in one week, in order for the average production cost per car to be $\1,000$?

### Work problems

Rational equations can be used to solve a variety of problems that involve rates, times and work. Using rational expressions and equations can help you answer questions about how to combine workers or machines to complete a job on schedule.

A Good Day’s Work

A “work problem” is an example of a real life situation that can be modeled and solved using a rational equation. Work problems often ask you to calculate how long it will take different people working at different speeds to finish a task. The algebraic models of such situations often involve rational equations derived from the work formula, $W=rt$. (Notice that the work formula is very similar to the relationship between distance, rate, and time, or $d=rt$.) The amount of work done $W$ is the product of the rate of work $r$ and the time spent working $t$. The work formula has 3 versions.

$\large \begin{array}{l}W=rt\\\\\,\,\,\,\,t=\frac{W}{r}\\\\\,\,\,\,\,r=\frac{W}{t}\end{array}$

Some work problems include multiple machines or people working on a project together for the same amount of time but at different rates. In that case, you can add their individual work rates together to get a total work rate. Let’s look at an example.

### Example

Myra takes 2 hours to plant 50 flower bulbs. Francis takes 3 hours to plant 45 flower bulbs. Working together, how long should it take them to plant 150 bulbs?

Other work problems go the other way. You can calculate how long it will take one person to do a job alone when you know how long it takes people working together to complete the job.

### Example

Joe and John are planning to paint a house together. John thinks that if he worked alone, it would take him 3 times as long as it would take Joe to paint the entire house. Working together, they can complete the job in 24 hours. How long would it take each of them, working alone, to complete the job?

In the video that follows, we show another example of finding one person’s work rate given a combined work rate.

As shown above, many work problems can be represented by the equation $\displaystyle \frac{t}{a}+\frac{t}{b}=1$, where $t$ is the time to do the job together, $a$ is the time it takes person A to do the job, and $b$ is the time it takes person B to do the job. The 1 refers to the total work done—in this case, the work was to paint 1 house.

The key idea here is to figure out each worker’s individual rate of work. Then, once those rates are identified, add them together, multiply by the time $t$, set it equal to the amount of work done, and solve the rational equation. Here is a summary:

### Formula for work problems

$a=$ the time needed for A to complete 1 job alone

$b=$ the time needed for B to complete 1 job alone

$t=$ the time needed for A and B to complete the 1 job together:

$\displaystyle \frac{t}{a}+\frac{t}{b} = 1$

We present another example of two people painting at different rates in the following video.

### Motion problems

We have solved uniform motion problems using the formula $D = rt$ in previous chapters. We used a table like the one below to organize the information and lead us to the equation.

$\begin{array}{|c|c|c|c|} \hline & \,\text{rate}\,&\,\text{time}\,&\,\text{distance}\\ \hline \,\text{First}\,&\,&\,&\\ \hline \,\text{Second}\,&\,&\,&\\ \hline \end{array}$

The formula $D=rt$ assumes we know $r$ and  $t$ and use them to find  $D$. If we know  $D$ and $r$ and need to find  $t$, we would solve the equation for  $t$ and get the formula $\displaystyle t=\frac{D}{r}$.

### ExAMPLE

Greg went to a conference in a city 120 miles away. On the way back, due to road construction he had to drive 10 mph slower which resulted in the return trip taking 2 hours longer. How fast did he drive on the way to the conference?

### Exercises

A man rows downstream for 30 miles then turns around and returns to his original location, the total trip took 8 hours. If the current flows at 2 miles per hour, how fast would the man row in still water?

### Concentration of a mixture problems

Mixtures are made of ratios of different substances that may include chemicals, foods, water, or gases. There are many different situations where mixtures may occur both in nature and as a means to produce a desired product or outcome.  For example, chemical spills, manufacturing and even biochemical reactions involve mixtures.  The thing that can make mixtures interesting mathematically is when components of the mixture are added at different rates and concentrations. In our last example we will define an equation that models the concentration  – or ratio of sugar to water – in a large mixing tank over time. You are asked whether the final concentration of sugar is greater than the concentration at the beginning.

### Example

A large mixing tank currently contains 100 gallons of water into which 5 pounds of sugar have been mixed. A tap will open pouring 10 gallons per minute of water into the tank at the same time sugar is poured into the tank at a rate of 1 pound per minute. Find the concentration (pounds per gallon) of sugar in the tank after 12 minutes. Is that a greater concentration than at the beginning?

In the following video, we show another example of how to use rational functions to model mixing.