## Rational Functions

Solve applied problems involving rational functions.

In Example 2, we shifted a toolkit function in a way that resulted in the function [latex]f\left(x\right)=\frac{3x+7}{x+2}[/latex]. This is an example of a rational function. A rational function is a function that can be written as the quotient of two polynomial functions. Many real-world problems require us to find the ratio of two polynomial functions. Problems involving rates and concentrations often involve rational functions.

## A General Note: Rational Function

A rational function is a function that can be written as the quotient of two polynomial functions [latex]P\left(x\right) \text{and} Q\left(x\right)[/latex].

## Example 3: Solving an Applied Problem Involving a Rational Function

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?

Let t be the number of minutes since the tap opened. Since the water increases at 10 gallons per minute, and the sugar increases at 1 pound per minute, these are constant rates of change. This tells us the amount of water in the tank is changing linearly, as is the amount of sugar in the tank. We can write an equation independently for each:

The concentration, C , will be the ratio of pounds of sugar to gallons of water

The concentration after 12 minutes is given by evaluating [latex]C\left(t\right)[/latex] at [latex]t=\text{ }12[/latex].

This means the concentration is 17 pounds of sugar to 220 gallons of water.

At the beginning, the concentration is

Since [latex]\frac{17}{220}\approx 0.08>\frac{1}{20}=0.05[/latex], the concentration is greater after 12 minutes than at the beginning.

## Analysis of the Solution

To find the horizontal asymptote, divide the leading coefficient in the numerator by the leading coefficient in the denominator:

Notice the horizontal asymptote is [latex]y=\text{ }0.1[/latex]. This means the concentration, C , the ratio of pounds of sugar to gallons of water, will approach 0.1 in the long term.

There are 1,200 freshmen and 1,500 sophomores at a prep rally at noon. After 12 p.m., 20 freshmen arrive at the rally every five minutes while 15 sophomores leave the rally. Find the ratio of freshmen to sophomores at 1 p.m.

- Precalculus. Authored by : Jay Abramson, et al.. Provided by : OpenStax. Located at : http://cnx.org/contents/[email protected] . License : CC BY: Attribution . License Terms : Download For Free at : http://cnx.org/contents/[email protected].

## Rational Functions, Equations, and Inequalities

Rational Functions are just a ratio of two polynomials (expression with constants and/or variables), and are typically thought of as having at least one variable in the denominator (which can never be 0 ).

Note that we talk about how to graph rationals using their asymptotes in the Graphing Rational Functions, including Asymptotes section. Also, since limits exist with Rational Functions and their asymptotes, limits are discussed here in the Limits and Continuity section . Since factoring is so important in algebra, you may want to revisit it first. Remember that we first learned factoring here in the Solving Quadratics by Factoring and Completing the Square section, and more Advanced Factoring can be found here .

## Introducing Rational Expressions

Again, think of a rational expression as a ratio of two polynomials . Here are some examples of expressions that are and are not rational expressions:

## Multiplying, Dividing, and Simplifying Rationals

Frequently, rational expressions can be simplified by factoring the numerator, denominator, or both, and crossing out factors. They can be multiplied and divided like regular fractions.

Here are some examples. Note that these look really difficult, but we’re just using a lot of steps of things we already know. That’s the fun of math! Also, note in the last example, we are dividing rationals , so we flip the second and multiply .

Remember that when you cross out factors, you can cross out from the top and bottom of the same fraction, or top and bottom from different factors that you are multiplying. You can never cross out two things on top, or two things on bottom.

Also remember that at any point in the problem , when variables are in the denominator, we’ll have domain restrictions , since denominators can’t be $ 0$.

Multiplying, Dividing, and Simplifying Rationals

$ \displaystyle \require{cancel} \frac{{\left( {3{{x}^{2}}-5x-2} \right)}}{{{{x}^{2}}-4}}=\frac{{\left( {3x+1} \right)\cancel{{\left( {x-2} \right)}}}}{{\left( {x+2} \right)\cancel{{\left( {x-2} \right)}}}}=\frac{{3x+1}}{{x+2}};\,\,\,x\ne -2,\,\,2$

$ \displaystyle \require{cancel} \begin{align}\frac{{4{{x}^{2}}+12xy+2x+6y}}{{2{{x}^{2}}+6xy+6x+18y}}\,&=\,\frac{{2\left( {2{{x}^{2}}+6xy+x+3y} \right)}}{{2\left( {{{x}^{2}}+3xy+3x+9y} \right)}}\,=\,\frac{{2\left[ {2x\left( {x+3y} \right)+1\left( {x+3y} \right)} \right]}}{{2\left[ {x\left( {x+3y} \right)+3\left( {x+3y} \right)} \right]}}\,\,\,\\&=\,\frac{{\cancel{2}\left( {2x+1} \right)\cancel{{\left( {x+3y} \right)}}}}{{\cancel{2}\left( {x+3} \right)\cancel{{\left( {x+3y} \right)}}}}\,=\,\frac{{2x+1}}{{x+3}};\,\,\,\,x\ne -3,\,\,-3y\end{align}$

$ \displaystyle \require{cancel} \begin{align}\frac{{\frac{{7{{y}^{2}}}}{{{{y}^{2}}-16}}}}{{\frac{{14{{y}^{2}}+49y}}{{3{{y}^{2}}-10y-8}}}}\,&=\,\frac{{\frac{{7{{y}^{2}}}}{{\left( {y-4} \right)\left( {y+4} \right)}}}}{{\frac{{7y\left( {2y+7} \right)}}{{\left( {y-4} \right)\left( {3y+2} \right)}}}}\,=\,\frac{{\cancel{7}{{{\cancel{{{{y}^{2}}}}}}^{y}}}}{{\cancel{{\left( {y-4} \right)}}\left( {y+4} \right)}}\cdot \frac{{\cancel{{\left( {y-4} \right)}}\left( {3y+2} \right)}}{{\cancel{7}\cancel{y}\left( {2y+7} \right)}}\,\,\,\\&=\,\frac{{y\left( {3y+2} \right)}}{{\left( {y+4} \right)\left( {2y+7} \right)}};\,\,\,\,\,\,y\ne -4,\,-\frac{7}{2},\,\,-\frac{2}{3},\,\,0,\,\,4\end{align}$

## Finding the Common Denominator

When we add or subtract two or more rationals, we need to find the least common denominator (LCD) , just like when we add or subtract regular fractions. If the denominators are the same, we can just add the numerators across, leaving the denominators as they are. We then must be sure we can’t do any further factoring:

$ \require{cancel} \displaystyle \frac{2}{{3x}}+\frac{4}{{3x}}=\frac{{(2+4)}}{{3x}}=\frac{{{{{\cancel{6}}}^{2}}}}{{{{{\cancel{3}}}^{1}}x}}=\frac{2}{x};\,\,\,\,x\ne 0$.

Just like with regular fractions, we want to use the factors in the denominators in every fraction, but not repeat them across denominators. When nothing is common, just multiply the factors.

Let’s find the least common denominators for the following denominators (ignore the numerators for now).

## Adding and Subtracting Rationals

Now let’s add and subtract the following rational expressions. Note that one way to look at finding the LCD is to multiply the top by what’s missing in the bottom . For example, in the first example, the LCD is $ \left( {x+3} \right)\left( {x+4} \right)$, and we need to multiply the first fraction’s numerator by $ \left( {x+4} \right)$, since that’s missing in the denominator.

## Restricted Domains of Rational Functions

As we’ve noticed, since rational functions have variables in denominators, we must make sure that the denominators won’t end up as “ 0 ” at any point of solving the problem .

Thus, the domain of $ \displaystyle \frac{{x+1}}{{2x(x-2)(x+3)}}$ is $ \{x:x\ne -3,0,2\}$. This means if we ever get a solution to an equation that contains rational expressions and has variables in the denominator (which they probably will!), we must make sure that none of our answers would make any denominator in that equation “ 0 ” . These “answers” that we can’t use are called extraneous solutions . We’ll see this in the first example below.

## Solving Rational Equations

When we solve rational equations, we can multiply both sides of the equations by the least common denominator , or LCD (which is $ \displaystyle \frac{{\text{least common denominator}}}{1}$ in fraction form), and not even worry about working with fractions! The denominators will cancel out and we just solve the equation using the numerators. Remember that with quadratics, we need to get everything to one side with 0 on the other side and either factor or use the Quadratic Formula .

Again, think of multiplying the top by what’s missing in the bottom from the LCD .

Notice that sometimes you’ll have to solve literal equations , which just means that you have to solve an equation for a variable, but you’ll have other variables in the answer. The last example shows this.

## Rational Inequalities, including Absolute Values

Solving rational inequalities are a little more complicated since we are typically multiplying or dividing by variables, and we don’t know whether these are positive or negative. Remember that we have to change the direction of the inequality when we multiply or divide by negative numbers . When we solve these rational inequalities, our answers will typically be a range of numbers .

## Rational Inequalities from a Graph

It’s not too bad to see inequalities of rational functions from a graph . Look at this graph to see where $ y<0$ and $ y\ge 0$. Notice that we have ranges of $ x$ values in the two cases:

## Solving Rational Inequalities Algebraically Using a Sign Chart

The easiest way to solve rational inequalities algebraically is using the sign chart method , which we saw here in the Quadratic Inequalities section . A sign chart or sign pattern is simply a number line that is separated into partitions (intervals or regions), with boundary points (called “ critical values “) that you get by setting the factors of the rational function, both in the numerator and denominator, to 0 and solving for $ x$.

With sign charts, you can pick any point in between the critical values , and see if the whole function is positive or negative . Then you just pick that interval (or intervals) by looking at the inequality. Generally, if the inequality includes the $ =$ sign, you have a closed bracket, and if it doesn’t, you have an open bracket. But any factor that’s in the denominator must have an open bracket for the values that make it 0 , since you can’t have 0 in the denominator.

The first thing you have to do is get everything on the left side (if it isn’t already there) and 0 on the right side , since we can see what intervals make the inequality true. We can only have one term on the left side , so sometimes we have to find a common denominator and combine terms.

Also, it’s a good idea to put open or closed circles on the critical values to remind ourselves if we have inclusive points (inequalities with equal signs, such as $ \le $ and $ \ge $) or exclusive points (inequalities without equal signs, or factors in the denominators ).

Let’s do some examples; y ou can always use your graphing calculator to check your answers, too. Put in both sides of the inequalities and check the zeros, and make sure your ranges are correct!

And here’s another one where we have to do a lot of “organizing” first, including moving terms to one side, and then factoring:

Sometimes we get a funny interval with a single $ x$ value as part of the interval:

And here’s one where we have a removable discontinuity in the rational inequality, so we have to make sure we skip over that point:

Here are a couple that involves solving radical inequalities with absolute values . (You might want to review Solving Absolute Value Equations and Inequalities before continuing on to this topic.)

Here are more complicated ones, where the absolute value may need to be multiplied by other variables (think of if you had to cross multiply). Notice how it’s best to separate the inequality into two separate inequalities: one case when $ x$ is positive, and the other when $ x$ is negative. Notice also how we had to use the Quadratic Formula to get the critical points when $ x$ is negative:

The problem calls for $ >0$, so we look for the plus sign intervals. But we have to throw away any intervals where the sign doesn’t agree with our conditions of $ x$ (positive or negative), such as the interval $ \left( {.434,1} \right)$ ($ x$ is supposed to be negative). Tricky!

Look at the negative part of the sign chart when $ x$ is negative, and the positive part of the sign chart when $ x$ is positive.

When $ x$ is positive, the numerator yields no “real” critical points. When $ x$ is negative, use the Quadratic Formula to get the roots (critical points) of the numerator:

$ \displaystyle \frac{{-b\pm \sqrt{{{{b}^{2}}-4ac}}}}{{2a}}=\frac{{1\pm \sqrt{{{{{\left( {-1} \right)}}^{2}}-4\left( {-3} \right)\left( 1 \right)}}}}{{-6}}$

$ \displaystyle =\frac{{1\pm \sqrt{{13}}}}{{-6}}\approx -.768,\,\,\,.434$

The answer is $ \displaystyle \left( {.768,-\frac{1}{2}} \right)\cup \left( {1,\infty } \right)$.

Put the inequality in your graphing calculator to check your answer!

Here’s one more that’s a bit tricky, since we have two expressions with absolute value in it. In this case, we have to separate in four cases , just to be sure we cover all the possibilities.

Use sign charts to determine what intervals the rational will be positive or negative. The problem calls for $ >0$, so look for the plus sign intervals, and make sure they work in the original inequality .

The answer is $ \displaystyle \left( {-1,-\frac{1}{3}} \right)\cup \left( {-\frac{1}{3},0} \right)$. We have to “skip over” (asymptote) $ \displaystyle -\frac{1}{3}$ (so we don’t divide by 0 ), and use soft brackets, since the inequality is $ >$ and not $ \ge $.

There’s one more that we did here in the Compositions of Functions, Even and Odd, and Increasing and Decreasing section, when we worked on domains of composites.

## Applications of Rationals

There are certain types of word problems that typically use rational expressions. These tend to deal with rates , since rates are typically fractions (such as distance over time). We also see problems dealing with plain fractions or percentages in fraction form.

Here is a rational “fraction” problem:

Here is a rational “percentage” problem:

## Distance/Rate/Time Problems

With rational rate problems, we must always remember: $ \text{Distance}=\text{Rate }\times \,\,\text{Time}$ . It seems most of the problems deal with comparing times or adding times .

Shalini can run 3 miles per hour faster than her sister Meena can walk. If Shalini ran 12 miles in the same time it took Meena to walk 8 miles, what is the speed of each sister in this case ?

This is a “$ \text{Distance}=\text{Rate }\times \,\,\text{Time}$” problem, and let’s use a table to organize this information like we did in the Algebra Word Problems and Systems of Linear Equations and Word Problems sections. Let $ x=$ Meena’s speed, since Shalini runs faster and it’s easier to add than subtract:

Put it all together:

## Work Problems:

Work problems typically have to do with different people or things working together and alone, at different rates. Instead of distance, we work with jobs (typically, 1 complete job).

I find that usually the easiest way to work these problems is to remember:

$ \displaystyle \frac{{\operatorname{time} \,\text{together}}}{{\text{time alone}}}\,\,+\,…\,\,+\,\,\frac{{\text{time together}}}{{\text{time alone}}}\,\,=\,\,1$ (or whatever part of the job or jobs is done; if they do half the job, this equals $ \displaystyle \frac{1}{2}$; if they do a job twice, this equals 2 ).

(Use a “$ +$” between the terms if working towards the same goal, such as painting a room, and “$ -$” if working towards opposite goals, such as filling and emptying a pool.)

“Proof” : For work problems, $ \text{Rate }\times \,\text{Time = }1\text{ Job}$. Add up individuals’ portions of a job with this formula, using the time working with others (time together):

$ \displaystyle \begin{align}\left( {\text{individual rate }\times \text{ time }} \right)\text{ + }…\text{ +}\left( {\text{individual rate }\times \text{ time }} \right)&=1\\\left( {\frac{1}{{\text{individual time to do 1 job}}}\text{ }\times \text{ time}} \right)\text{ + }…\text{ +}\left( {\frac{1}{{\text{individual time to do 1 job}}}\text{ }\times \text{ time}} \right)&=1\\\left( {\frac{{\text{time working with others}}}{{\text{individual time to do 1 job}}}} \right)\text{ + }…\text{ +}\left( {\frac{{\text{time working with others}}}{{\text{individual time to do 1 job}}}} \right)&=1\end{align}$

Also, as explained after the first example below, often you see this formula as $ \displaystyle \frac{\text{1}}{{\text{time alone}}}\,+…+\,\frac{\text{1}}{{\text{time alone}}}\,=\,\frac{\text{1}}{{\text{time together}}}$.

Let’s do some problems:

Note : The formula above can also be derived by using the concept that you can figure out how much of the job the girls do per hour (or whatever the time unit is), both together and alone . Then you can add the individual “rates” to get the “rate” of their painting together.

We are actually adding the Work they complete (together and alone) using formula $ \text{Rate }\times \text{ }\text{Time }=\text{ Work}$, where the Time is 1 hour (or whatever the unit is).

In this example, Erica’s rate per hour is $ \displaystyle \frac{1}{5}$ (she can do $ \displaystyle \frac{1}{5}$ of the job in 1 hour); Rachel’s rate per hour is $ \displaystyle \frac{1}{R}$; we can add their rates to get the rate of their painting together: $ \displaystyle \frac{1}{5}+\frac{1}{R}=\frac{1}{3}$. If we multiply all the terms by 3 , we come up with the equation above! (Sometimes you see this as $ \displaystyle \frac{\text{1}}{{\text{time alone}}}\,\,+…+\,\frac{\text{1}}{{\text{time alone}}}\,=\,\frac{\text{1}}{{\text{time together}}}$.)

Here is a systems problem and also a work problem at the same time. We did this problem without using rationals here in the Systems of Linear Equations and Word Problems section (and be careful, since the variables we assigned were different).

## Cost Problems

A third type of problem you might get while studying rationals has to do with average cost , or possibly costs per person (or unit cost) problems. Both of these types of costs can still be thought as rates , as they are an amount over time. Here are some examples:

## Rational Inequality Word Problem

Here is a rational inequality word problem that we saw in the Algebra Word Problems section .

Note that there is a Rational Asymptote Application Problem here in the Graphing Rational Functions, including Asymptotes section .

Understand these problems, and practice, practice, practice!

For Practice : Use the Mathway widget below to try a Rational Function problem. Click on Submit (the blue arrow to the right of the problem) and click on Solve for x to see the answer.

You can also type in your own problem, or click on the three dots in the upper right hand corner and click on “Examples” to drill down by topic.

If you click on Tap to view steps , or Click Here , you can register at Mathway for a free trial , and then upgrade to a paid subscription at any time (to get any type of math problem solved!).

On to Graphing Rational Functions, including Asymptotes – you’re ready!

## Rational Function Problems

These lessons are compiled to help PreCalculus students learn about rational function problems and applications.

Related Pages Simplifying Rational Expressions Graphing Rational Functions PreCalculus Lessons

The following figure shows how to solve rational equations . Scroll down the page for examples and solutions on how to solve rational function problems and applications.

## Rational Function Problems - Work And Tank

The video explains application problems that use rational equations. Part 1 of 2.

- Martin can pour a concrete walkway in 6 hours working alone. Victor has more experience and can pour the same walkway in 4 hours working alone. How long will it take both people to pour the concrete walkway working together?
- An inlet pipe can fill a water tank in 12 hours. An outlet pipe can drain the tank in 20 hours. If both pipes are mistakenly left open, how long will it take to fill the tank?

## Rational Function Applications - Work And Rate

The video explains application problems that use rational equations. Part 2 of 2.

- One person can complete a task 8 hours sooner than another person. Working together, both people can perform the task in 3 hours. How many hours does it take each person to complete the task working alone?
- The speed of a passenger train is 12 mph faster than the speed of the freight train. The passenger train travels 330 miles in the same time it takes the freight train to travel 270 miles. Find the speed of each train.

## Rational Functions Word Problems - Work, Tank And Pipe

Here are a few examples of work problems that are solved with rational equations.

- Sam can paint a house in 5 hours. Gary can do it in 4 hours. How long will it take the two working together?
- Joy can file 100 claims in 5 hours. Stephen can file 100 claims in 8 hours. If they work together, how long will it take to file 100 claims?
- A water tank is emptied through two drains in 50 minutes. If only the larger drain is used, the tank will empty in 85 minutes. How long would it take to empty if only the smaller drain is used?
- One computer can run a sorting algorithm in 24 minutes. If a second computer is used together with the first, it takes 13 minutes. How long would it take the second computer alone?
- Two pipes are filling a tank. One pipe fills three times as fast as the other. With both pipes working, the tank fills in 84 minutes. How long would each pipe take working alone?

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## General Mathematics Quarter 1 – Module 22: Solving Real-life Problems Involving Exponential Functions, Equations, and Inequalities

This module was designed and written with you in mind. It is here to help you solve real-life problems involving exponential functions, equations, and inequalities. Most of the time, students like you ask why you need to study Mathematics. Even though you know the answer, still you keep on asking this question because perhaps you did not realize how important it is to real-life situations.

This module hopes to help you make a wise decision in the future because it involves money matter problems.

After going through this module, you are expected to:

1. recall how to solve exponential functions, equations, and inequalities; and

2. solve real-life problems involving exponential functions, equations, and inequalities.

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## 4.3: Rational Inequalities and Applications

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- Carl Stitz & Jeff Zeager
- Lakeland Community College & Lorain County Community College

In this section, we solve equations and inequalities involving rational functions and explore associated application problems. Our first example showcases the critical difference in procedure between solving a rational equation and a rational inequality.

## Example 4.3.1

- Solve \(\dfrac{x^3-2x+1}{x-1} = \dfrac{1}{2}x-1\).
- Solve \(\dfrac{x^3-2x+1}{x-1} \geq \dfrac{1}{2}x-1\).
- Use your calculator to graphically check your answers to 1 and 2.

\[\begin{array}{rclr} \dfrac{x^3-2x+1}{x-1} & = & \dfrac{1}{2}x-1 & \\[4pt] \left(\dfrac{x^3-2x+1}{x-1}\right) \cdot 2(x-1) & = & \left( \dfrac{1}{2}x-1 \right) \cdot 2(x-1) & \\[4pt] 2x^3 - 4x + 2 & = & x^2-3x+2 & \mbox{expand} \\ 2x^3 -x^2 - x & = & 0 & \\ x(2x+1)(x-1) & = & 0 & \mbox{factor}\\ x & = & -\frac{1}{2}, \, 0, \, 1 & \\ \end{array}\nonumber\]

Since we cleared denominators, we need to check for extraneous solutions. Sure enough, we see that \(x=1\) does not satisfy the original equation and must be discarded. Our solutions are \(x=-\frac{1}{2}\) and \(x=0\).

\[\begin{array}{rclr} \dfrac{x^3-2x+1}{x-1} & \geq & \dfrac{1}{2}x-1 & \\[4pt] \dfrac{x^3-2x+1}{x-1} - \dfrac{1}{2} x + 1& \geq & 0& \\[4pt] \dfrac{2\left(x^3-2x+1\right)-x(x-1)+1(2(x-1))}{2(x-1)} & \geq & 0 & \mbox{get a common denominator} \\[4pt] \dfrac{2x^3-x^2-x}{2x-2} & \geq & 0 & \mbox{expand} \\ \end{array}\nonumber\]

Viewing the left hand side as a rational function \(r(x)\) we make a sign diagram. The only value excluded from the domain of \(r\) is \(x=1\) which is the solution to \(2x-2=0\). The zeros of \(r\) are the solutions to \(2x^3-x^2-x=0\), which we have already found to be \(x=0\), \(x=-\frac{1}{2}\) and \(x=1\), the latter was discounted as a zero because it is not in the domain. Choosing test values in each test interval, we construct the sign diagram below.

We are interested in where \(r(x) \geq 0\). We find \(r(x) > 0\), or \((+)\), on the intervals \(\left(-\infty, -\frac{1}{2}\right)\), \((0,1)\) and \((1, \infty)\). We add to these intervals the zeros of \(r\), \(-\frac{1}{2}\) and \(0\), to get our final solution: \(\left( - \infty, -\frac{1}{2} \right] \cup [0,1) \cup (1, \infty)\).

The ‘Intersect’ command confirms that the graphs cross when \(x=-\frac{1}{2}\) and \(x=0\). It is clear from the calculator that the graph of \(y=f(x)\) is above the graph of \(y=g(x)\) on \(\left(-\infty, -\frac{1}{2}\right)\) as well as on \((0,\infty)\). According to the calculator, our solution is then \(\left(-\infty, -\frac{1}{2}\right] \cup [0, \infty)\) which almost matches the answer we found analytically. We have to remember that \(f\) is not defined at \(x=1\), and, even though it isn’t shown on the calculator, there is a hole 1 in the graph of \(y=f(x)\) when \(x=1\) which is why \(x=1\) is not part of our final answer.

Next, we explore how rational equations can be used to solve some classic problems involving rates.

## Example 4.3.2

Carl decides to explore the Meander River, the location of several recent Sasquatch sightings. From camp, he canoes downstream five miles to check out a purported Sasquatch nest. Finding nothing, he immediately turns around, retraces his route (this time traveling upstream), and returns to camp 3 hours after he left. If Carl canoes at a rate of 6 miles per hour in still water, how fast was the Meander River flowing on that day?

We are given information about distances, rates (speeds) and times. The basic principle relating these quantities is: \[\text{distance} = \text{rate} \cdot \text{time}\nonumber\] The first observation to make, however, is that the distance, rate and time given to us aren’t ‘compatible’: the distance given is the distance for only part of the trip, the rate given is the speed Carl can canoe in still water, not in a flowing river, and the time given is the duration of the entire trip. Ultimately, we are after the speed of the river, so let’s call that \(R\) measured in miles per hour to be consistent with the other rate given to us. To get started, let’s divide the trip into its two parts: the initial trip downstream and the return trip upstream. For the downstream trip, all we know is that the distance traveled is \(5\) miles.

\[\begin{array}{rcl} \text{distance downstream} & = & \text{rate traveling downstream} \cdot \text{time traveling downstream} \\ 5 \, \text{miles} & = & \text{rate traveling downstream} \cdot \text{time traveling downstream} \\ \end{array}\nonumber\]

Since the return trip upstream followed the same route as the trip downstream, we know that the distance traveled upstream is also 5 miles.

\[\begin{array}{rcl} \text{distance upstream} & = & \text{rate traveling upstream} \cdot \text{time traveling upstream} \\ 5 \, \text{miles} & = & \text{rate traveling upstream} \cdot \text{time traveling upstream} \\ \end{array}\nonumber\]

We are told Carl can canoe at a rate of \(6\) miles per hour in still water. How does this figure into the rates traveling upstream and downstream? The speed the canoe travels in the river is a combination of the speed at which Carl can propel the canoe in still water, 6 miles per hour, and the speed of the river, which we’re calling \(R\). When traveling downstream, the river is helping Carl along, so we add these two speeds:

\[\begin{array}{rcl} \text{rate traveling downstream} & = & \text{rate Carl propels the canoe} + \text{speed of the river} \\ & = & 6 \frac{\text{miles}}{\text{hour}} + R \frac{\text{miles}}{\text{hour}} \\ \end{array}\nonumber\]

So our downstream speed is \((6+R) \frac{\text{miles}}{\text{hour}}\). Substituting this into our ‘distance-rate-time’ equation for the downstream part of the trip, we get:

\[\begin{array}{rcl} 5 \, \text{miles} & = & \text{rate traveling downstream} \cdot \text{time traveling downstream} \\ 5 \, \text{miles} & = & (6+R) \frac{\text{miles}}{\text{hour}} \cdot \text{time traveling downstream} \\ \end{array}\nonumber\]

When traveling upstream, Carl works against the current. Since the canoe manages to travel upstream, the speed Carl can canoe in still water is greater than the river’s speed, so we subtract the river’s speed from Carl’s canoing speed to get:

\[\begin{array}{rcl} \text{rate traveling upstream} & = & \text{rate Carl propels the canoe} - \text{river speed} \\ & = & 6 \frac{\text{miles}}{\text{hour}} - R \frac{\text{miles}}{\text{hour}} \\ \end{array}\nonumber\]

Proceeding as before, we get

\[\begin{array}{rcl} 5 \, \text{miles} & = & \text{rate traveling upstream} \cdot \text{time traveling upstream} \\ 5 \, \text{miles} & = & (6 - R) \frac{\text{miles}}{\text{hour}} \cdot \text{time traveling upstream} \\ \end{array}\nonumber\]

The last piece of information given to us is that the total trip lasted \(3\) hours. If we let \(t_{\text{down}}\) denote the time of the downstream trip and \(t_{\text{up}}\) the time of the upstream trip, we have: \(t_{\text{down}} + t_{\text{up}} = 3 \, \text{hours}\). Substituting \(t_{\text{down}}\) and \(t_{\text{up}}\) into the ‘distance-rate-time’ equations, we get (suppressing the units) three equations in three unknowns: 2 \[\left\{\begin{array}{lrcl} E1 & (6+R) \, t_{\text{down}} & = & 5 \\ E2 & (6-R) \, t_{\text{up}} & = & 5 \\ E3 & t_{\text{down}} + t_{\text{up}} & = & 3 \end{array} \right.\nonumber\]

Since we are ultimately after \(R\), we need to use these three equations to get at least one equation involving only \(R\). To that end, we solve \(E1\) for \(t_{\text{down}}\) by dividing both sides 3 by the quantity \((6+R)\) to get \(t_{\text{down}} = \frac{5}{6+R}\). Similarly, we solve \(E2\) for \(t_{\text{up}}\) and get \(t_{\text{up}} = \frac{5}{6-R}\). Substituting these into \(E3\), we get: 4 \[\dfrac{5}{6+R} + \dfrac{5}{6 - R} = 3.\nonumber\] Clearing denominators, we get \(5(6-R) + 5(6+R) = 3(6+R)(6-R)\) which reduces to \(R^2 = 16\). We find \(R = \pm 4\), and since \(R\) represents the speed of the river, we choose \(R = 4\). On the day in question, the Meander River is flowing at a rate of \(4\) miles per hour.

One of the important lessons to learn from Example 4.3.2 is that speeds, and more generally, rates, are additive. As we see in our next example, the concept of rate and its associated principles can be applied to a wide variety of problems - not just ‘distance-rate-time’ scenarios.

## Example 4.3.3

Working alone, Taylor can weed the garden in 4 hours. If Carl helps, they can weed the garden in 3 hours. How long would it take for Carl to weed the garden on his own?

The key relationship between work and time which we use in this problem is: \[\text{amount of work done} = \text{rate of work} \cdot \text{time spent working}\nonumber\]

We are told that, working alone, Taylor can weed the garden in 4 hours. In Taylor’s case then: \[\begin{array}{rcl} \text{amount of work Taylor does} & = & \text{rate of Taylor working} \cdot \text{time Taylor spent working} \\ 1 \, \text{garden} & = & (\text{rate of Taylor working}) \cdot (4 \, \text{hours}) \\ \end{array}\nonumber\]

So we have that the rate Taylor works is \(\frac{1 \, \text{garden}}{ 4 \, \text{hours}} = \frac{1}{4} \frac{\text{garden}}{\text{hour}}\). We are also told that when working together, Taylor and Carl can weed the garden in just 3 hours. We have:

\[\begin{array}{rcl} \text{amount of work done together} & = & \text{rate of working together} \cdot \text{time spent working together} \\ 1 \, \text{garden} & = & (\text{rate of working together}) \cdot (3 \, \text{hours}) \\ \end{array}\nonumber\]

From this, we find that the rate of Taylor and Carl working together is \(\frac{1 \, \text{garden}}{3 \, \text{hours}} = \frac{1}{3} \frac{\text{garden}}{\text{hour}}\). We are asked to find out how long it would take for Carl to weed the garden on his own. Let us call this unknown \(t\), measured in hours to be consistent with the other times given to us in the problem. Then:

\[\begin{array}{rcl} \text{amount of work Carl does} & = & \text{rate of Carl working} \cdot \text{time Carl spent working} \\ 1 \, \text{garden} & = & (\text{rate of Carl working}) \cdot (t \, \text{hours}) \\ \end{array}\nonumber\]

In order to find \(t\), we need to find the rate of Carl working, so let’s call this quantity \(R\), with units \(\frac{\text{garden}}{\text{hour}}\). Using the fact that rates are additive, we have:

\[\begin{array}{rcl} \text{rate working together} & = & \text{rate of Taylor working} + \text{rate of Carl working} \\[4pt] \frac{1}{3} \frac{\text{garden}}{\text{hour}} & = & \frac{1}{4} \frac{\text{garden}}{\text{hour}} + R \frac{\text{garden}}{\text{hour}} \\ \end{array}\nonumber\]

so that \(R = \frac{1}{12} \frac{\text{garden}}{\text{hour}}\). Substituting this into our ‘work-rate-time’ equation for Carl, we get:

\[\begin{array}{rcl} 1 \, \text{garden} & = & (\text{rate of Carl working}) \cdot (t \, \text{hours}) \\[4pt] 1 \, \text{garden} & = & \left(\frac{1}{12} \frac{\text{garden}}{\text{hour}} \right) \cdot (t \, \text{hours}) \\ \end{array}\nonumber\]

Solving \(1 = \frac{1}{12} t\), we get \(t = 12\), so it takes Carl 12 hours to weed the garden on his own. 5

As is common with ‘word problems’ like Examples 4.3.2 and 4.3.3 , there is no short-cut to the answer. We encourage the reader to carefully think through and apply the basic principles of rate to each (potentially different!) situation. It is time well spent. We also encourage the tracking of units, especially in the early stages of the problem. Not only does this promote uniformity in the units, it also serves as a quick means to check if an equation makes sense. 6

Our next example deals with the average cost function, first introduced on page 82, as applied to PortaBoy Game systems from Example 2.1.5 in Section 2.1 .

## Example 4.3.4

Given a cost function \(C(x)\), which returns the total cost of producing \(x\) items, recall that the average cost function, \(\overline{C}(x) = \frac{C(x)}{x}\) computes the cost per item when \(x\) items are produced. Suppose the cost \(C\), in dollars, to produce \(x\) PortaBoy game systems for a local retailer is \(C(x) = 80x + 150\), \(x \geq 0\).

- Find an expression for the average cost function \(\overline{C}(x)\).
- Solve \(\overline{C}(x) < 100\) and interpret.
- Determine the behavior of \(\overline{C}(x)\) as \(x \rightarrow \infty\) and interpret.
- From \(\overline{C}(x) = \frac{C(x)}{x}\), we obtain \(\overline{C}(x) = \frac{80x+150}{x}\). The domain of \(C\) is \(x \geq 0\), but since \(x=0\) causes problems for \(\overline{C}(x)\), we get our domain to be \(x>0\), or \((0, \infty)\).

\[\begin{array}{rclr} \dfrac{80x+150}{x} & < & 100 & \\[4pt] \dfrac{80x+150}{x} - 100 & < & 0 & \\[4pt] \dfrac{80x + 150 - 100x}{x} & < & 0 & \mbox{common denominator} \\[4pt] \dfrac{150 - 20x}{x} & < & 0 & \\ \end{array}\nonumber\]

If we take the left hand side to be a rational function \(r(x)\), we need to keep in mind that the applied domain of the problem is \(x > 0\). This means we consider only the positive half of the number line for our sign diagram. On \((0, \infty)\), \(r\) is defined everywhere so we need only look for zeros of \(r\). Setting \(r(x)=0\) gives \(150-20x =0\), so that \(x = \frac{15}{2}= 7.5\). The test intervals on our domain are \((0, 7.5)\) and \((7.5, \infty)\). We find \(r(x) < 0\) on \((7.5, \infty)\).

In the context of the problem, \(x\) represents the number of PortaBoy games systems produced and \(\overline{C}(x)\) is the average cost to produce each system. Solving \(\overline{C}(x) < 100\) means we are trying to find how many systems we need to produce so that the average cost is less than \(\$100\) per system. Our solution, \((7.5, \infty)\) tells us that we need to produce more than \(7.5\) systems to achieve this. Since it doesn’t make sense to produce half a system, our final answer is \([8, \infty)\).

- When we apply Theorem 4.2 to \(\overline{C}(x)\) we find that \(y=80\) is a horizontal asymptote to the graph of \(y=\overline{C}(x)\). To more precisely determine the behavior of \(\overline{C}(x)\) as \(x \rightarrow \infty\), we first use long division 7 and rewrite \(\overline{C}(x) = 80+\frac{150}{x}\). As \(x \rightarrow \infty\), \(\frac{150}{x} \rightarrow 0^{+}\), which means \(\overline{C}(x) \approx 80+\text { very small }(+)\). Thus the average cost per system is getting closer to \(\$ 80\) per system. If we set \(\overline{C}(x) = 80\), we get \(\frac{150}{x} = 0\), which is impossible, so we conclude that \(\overline{C}(x) > 80\) for all \(x > 0\). This means that the average cost per system is always greater than \(\$ 80\) per system, but the average cost is approaching this amount as more and more systems are produced. Looking back at Example 2.1.5 , we realize \(\$ 80\) is the variable cost per system \(-\) the cost per system above and beyond the fixed initial cost of \(\$150\). Another way to interpret our answer is that ‘infinitely’ many systems would need to be produced to effectively ‘zero out’ the fixed cost.

Our next example is another classic ‘box with no top’ problem.

## Example 4.3.5

A box with a square base and no top is to be constructed so that it has a volume of \(1000\) cubic centimeters. Let \(x\) denote the width of the box, in centimeters as seen below.

- Express the height \(h\) in centimeters as a function of the width \(x\) and state the applied domain.
- Solve \(h(x) \geq x\) and interpret.
- Find and interpret the behavior of \(h(x)\) as \(x \rightarrow 0^{+}\) and as \(x \rightarrow \infty\).
- Express the surface area \(S\) of the box as a function of \(x\) and state the applied domain.
- Use a calculator to approximate (to two decimal places) the dimensions of the box which minimize the surface area.
- We are told that the volume of the box is \(1000\) cubic centimeters and that \(x\) represents the width, in centimeters. From geometry, we know \(\mbox{Volume} = \mbox{width} \times \mbox{height} \times \mbox{depth}\). Since the base of the box is a square, the width and the depth are both \(x\) centimeters. Using \(h\) for the height, we have \(1000 = x^2h\), so that \(h = \frac{1000}{x^2}\). Using function notation, 8 \(h(x) = \frac{1000}{x^2}\) As for the applied domain, in order for there to be a box at all, \(x > 0\), and since every such choice of \(x\) will return a positive number for the height \(h\) we have no other restrictions and conclude our domain is \((0, \infty)\).

\[\begin{array}{rclr} h(x) & \geq & x & \\[4pt] \dfrac{1000}{x^2} & \geq & x & \\[4pt] \dfrac{1000}{x^2} - x & \geq & 0 \\[4pt] \dfrac{1000-x^3}{x^2} & \geq & 0 & \mbox{common denominator} \\[10pt] \end{array}\nonumber\]

We consider the left hand side of the inequality as our rational function \(r(x)\). We see \(r\) is undefined at \(x=0\), but, as in the previous example, the applied domain of the problem is \(x > 0\), so we are considering only the behavior of \(r\) on \((0, \infty)\). The sole zero of \(r\) comes when \(1000-x^3 = 0\), which is \(x=10\). Choosing test values in the intervals \((0,10)\) and \((10, \infty)\) gives the following diagram.

We see \(r(x) > 0\) on \((0,10)\), and since \(r(x) = 0\) at \(x=10\), our solution is \((0,10]\). In the context of the problem, \(h\) represents the height of the box while \(x\) represents the width (and depth) of the box. Solving \(h(x) \geq x\) is tantamount to finding the values of \(x\) which result in a box where the height is at least as big as the width (and, in this case, depth.) Our answer tells us the width of the box can be at most \(10\) centimeters for this to happen.

- As \(x \rightarrow 0^{+}\), \(h(x) = \frac{1000}{x^2} \rightarrow \infty\). This means that the smaller the width \(x\) (and, in this case, depth), the larger the height \(h\) has to be in order to maintain a volume of \(1000\) cubic centimeters. As \(x \rightarrow \infty\), we find \(h(x) \rightarrow 0^{+}\), which means that in order to maintain a volume of \(1000\) cubic centimeters, the width and depth must get bigger as the height becomes smaller.
- Since the box has no top, the surface area can be found by adding the area of each of the sides to the area of the base. The base is a square of dimensions \(x\) by \(x\), and each side has dimensions \(x\) by \(h\). We get the surface area, \(S = x^2+4xh\). To get \(S\) as a function of \(x\), we substitute \(h = \frac{1000}{x^2}\) to obtain \(S = x^2+4x \left( \frac{1000}{x^2}\right)\). Hence, as a function of \(x\), \(S(x) = x^2 + \frac{4000}{x}\). The domain of \(S\) is the same as \(h\), namely \((0, \infty)\), for the same reasons as above.
- A first attempt at the graph of \(y=S(x)\) on the calculator may lead to frustration. Chances are good that the first window chosen to view the graph will suggest \(y=S(x)\) has the \(x\)-axis as a horizontal asymptote. From the formula \(S(x) = x^2 + \frac{4000}{x}\), however, we get \(S(x) \approx x^2\) as \(x \rightarrow \infty\), so \(S(x) \rightarrow \infty\). Readjusting the window, we find \(S\) does possess a relative minimum at \(x \approx 12.60\). As far as we can tell, 9 this is the only relative extremum, so it is the absolute minimum as well. This means that the width and depth of the box should each measure approximately \(12.60\) centimeters. To determine the height, we find \(h(12.60) \approx 6.30\), so the height of the box should be approximately \(6.30\) centimeters.

## 4.3.1 Variation

In many instances in the sciences, rational functions are encountered as a result of fundamental natural laws which are typically a result of assuming certain basic relationships between variables. These basic relationships are summarized in the definition below.

## Definition 4.5

Suppose \(x\), \(y\) and \(z\) are variable quantities. We say

- \(y\) varies directly with (or is directly proportional to ) \(x\) if there is a constant \(k\) such that \(y=kx\).
- \(y\) varies inversely with (or is inversely proportional to ) \(x\) if there is a constant \(k\) such that \(y=\frac{k}{x}\).
- \(z\) varies jointly with (or is jointly proportional to ) \(x\) and \(y\) if there is a constant \(k\) such that \(z = kxy\).

The constant \(k\) in the above definitions is called the constant of proportionality .

## Example 4.3.6

Translate the following into mathematical equations using Definition 4.5 .

- Hooke’s Law : The force \(F\) exerted on a spring is directly proportional the extension \(x\) of the spring.
- Boyle’s Law : At a constant temperature, the pressure \(P\) of an ideal gas is inversely proportional to its volume \(V\).
- The volume \(V\) of a right circular cone varies jointly with the height \(h\) of the cone and the square of the radius \(r\) of the base.
- Ohm’s Law : The current \(I\) through a conductor between two points is directly proportional to the voltage \(V\) between the two points and inversely proportional to the resistance \(R\) between the two points.
- Newton’s Law of Universal Gravitation : Suppose two objects, one of mass \(m\) and one of mass \(M\), are positioned so that the distance between their centers of mass is \(r\). The gravitational force \(F\) exerted on the two objects varies directly with the product of the two masses and inversely with the square of the distance between their centers of mass.
- Applying the definition of direct variation, we get \(F = k x\) for some constant \(k\).
- Since \(P\) and \(V\) are inversely proportional, we write \(P = \frac{k}{V}\).
- There is a bit of ambiguity here. It’s clear that the volume and the height of the cone are represented by the quantities \(V\) and \(h\), respectively, but does \(r\) represent the radius of the base or the square of the radius of the base? It is the former. Usually, if an algebraic operation is specified (like squaring), it is meant to be expressed in the formula. We apply Definition 4.5 to get \(V = k h r^{2}\).
- Even though the problem doesn’t use the phrase ‘varies jointly’, it is implied by the fact that the current \(I\) is related to two different quantities. Since \(I\) varies directly with \(V\) but inversely with \(R\), we write \(I = \frac{k V}{R}\).
- We write the product of the masses \(mM\) and the square of the distance as \(r^2\). We have that \(F\) varies directly with \(mM\) and inversely with \(r^2\), so \(F = \frac{kmM}{r^2}\).

In many of the formulas in the previous example, more than two varying quantities are related. In practice, however, usually all but two quantities are held constant in an experiment and the data collected is used to relate just two of the variables. Comparing just two varying quantities allows us to view the relationship between them as functional, as the next example illustrates.

## Example 4.3.7

According to this website the actual data relating the volume \(V\) of a gas and its pressure \(P\) used by Boyle and his assistant in 1662 to verify the gas law that bears his name is given below.

\[\begin{array}{|c||c|c|c|c|c|c|c|c|c|c|c|c|c|} \hline V & 48 & 46 & 44 & 42 & 40 & 38 & 36 & 34 & 32 & 30 & 28 & 26 & 24 \\ \hline P & 29.13 & 30.56 & 31.94 & 33.5 & 35.31 & 37 & 39.31 & 41.63 & 44.19 & 47.06 & 50.31 & 54.31 & 58.81 \\ \hline \end{array}\nonumber\]

\[\begin{array}{|c||c||c|c|c|c|c|c|c|c|c|c|c|c|} \hline V & 23 & 22 & 21 & 20 & 19 & 18 & 17 & 16 & 15 & 14 & 13 & 12 \\ \hline P & 61.31 & 64.06 & 67.06 & 70.69 & 74.13 & 77.88 & 82.75 & 87.88 & 93.06 & 100.44 & 107.81 & 117.56 \\ \hline \end{array}\nonumber\]

- Use your calculator to generate a scatter diagram for these data using \(V\) as the independent variable and \(P\) as the dependent variable. Does it appear from the graph that \(P\) is inversely proportional to \(V\)? Explain.
- Assuming that \(P\) and \(V\) do vary inversely, use the data to approximate the constant of proportionality.
- Use your calculator to determine a ‘Power Regression’ for this data 10 and use it verify your results in 1 and 2.
- If \(P\) really does vary inversely with \(V\), then \(P = \frac{k}{V}\) for some constant \(k\). From the data plot, the points do seem to lie along a curve like \(y = \frac{k}{x}\).

## 4.3.2 Exercises

In Exercises 1 - 6, solve the rational equation. Be sure to check for extraneous solutions.

- \(\dfrac{x}{5x + 4} = 3\)
- \(\dfrac{3x - 1}{x^{2} + 1} = 1\)
- \(\dfrac{1}{x + 3} + \dfrac{1}{x - 3} = \dfrac{x^{2} - 3}{x^{2} - 9}\)
- \(\dfrac{2x + 17}{x + 1} = x + 5\)
- \(\dfrac{x^{2} - 2x + 1}{x^{3} + x^{2} - 2x} = 1\)
- \(\dfrac{-x^{3} + 4x}{x^{2} - 9} = 4x\)

In Exercises 7 - 20, solve the rational inequality. Express your answer using interval notation.

- \(\dfrac{1}{x + 2} \geq 0\)
- \(\dfrac{x - 3}{x + 2} \leq 0\)
- \(\dfrac{x}{x^{2} - 1} > 0\)
- \(\dfrac{4x}{x^2+4} \geq 0\)
- \(\dfrac{x^2-x-12}{x^2+x-6} > 0\)
- \(\dfrac{3x^2-5x-2}{x^2-9} < 0\)
- \(\dfrac{x^3+2x^2+x}{x^2-x-2} \geq 0\)
- \(\dfrac{x^{2} + 5x + 6}{x^{2} - 1} > 0\)
- \(\dfrac{3x - 1}{x^{2} + 1} \leq 1\)
- \(\dfrac{2x + 17}{x + 1} > x + 5\)
- \(\dfrac{-x^{3} + 4x}{x^{2} - 9} \geq 4x\)
- \(\dfrac{1}{x^{2} + 1} < 0\)
- \(\dfrac{x^4-4x^3+x^2-2x-15}{x^3-4x^2} \geq x\)
- \(\dfrac{5x^3-12x^2+9x+10}{x^2-1}\geq 3x-1\)
- Carl and Mike start a 3 mile race at the same time. If Mike ran the race at 6 miles per hour and finishes the race 10 minutes before Carl, how fast does Carl run?
- One day, Donnie observes that the wind is blowing at 6 miles per hour. A unladen swallow nesting near Donnie’s house flies three quarters of a mile down the road (in the direction of the wind), turns around, and returns exactly 4 minutes later. What is the airspeed of the unladen swallow? (Here, ‘airspeed’ is the speed that the swallow can fly in still air.)
- In order to remove water from a flooded basement, two pumps, each rated at 40 gallons per minute, are used. After half an hour, the one pump burns out, and the second pump finishes removing the water half an hour later. How many gallons of water were removed from the basement?
- A faucet can fill a sink in 5 minutes while a drain will empty the same sink in 8 minutes. If the faucet is turned on and the drain is left open, how long will it take to fill the sink?
- Working together, Daniel and Donnie can clean the llama pen in 45 minutes. On his own, Daniel can clean the pen in an hour. How long does it take Donnie to clean the llama pen on his own?
- In Exercise 32, the function \(C(x) = .03x^{3} - 4.5x^{2} + 225x + 250\), for \(x \geq 0\) was used to model the cost (in dollars) to produce \(x\) PortaBoy game systems. Using this cost function, find the number of PortaBoys which should be produced to minimize the average cost \(\overline{C}\). Round your answer to the nearest number of systems.
- Suppose we are in the same situation as Example 4.3.5 . If the volume of the box is to be \(500\) cubic centimeters, use your calculator to find the dimensions of the box which minimize the surface area. What is the minimum surface area? Round your answers to two decimal places.
- The box for the new Sasquatch-themed cereal, ‘Crypt-Os’, is to have a volume of \(140\) cubic inches. For aesthetic reasons, the height of the box needs to be \(1.62\) times the width of the base of the box. 13 Find the dimensions of the box which will minimize the surface area of the box. What is the minimum surface area? Round your answers to two decimal places.
- Sally is Skippy’s neighbor from Exercise 19 in Section 2.3 . Sally also wants to plant a vegetable garden along the side of her home. She doesn’t have any fencing, but wants to keep the size of the garden to 100 square feet. What are the dimensions of the garden which will minimize the amount of fencing she needs to buy? What is the minimum amount of fencing she needs to buy? Round your answers to the nearest foot. (Note: Since one side of the garden will border the house, Sally doesn’t need fencing along that side.)
- Find an expression for the volume \(V\) of the can in terms of the height \(h\) and the base radius \(r\).
- Find an expression for the surface area \(S\) of the can in terms of the height \(h\) and the base radius \(r\). (Hint: The top and bottom of the can are circles of radius \(r\) and the side of the can is really just a rectangle that has been bent into a cylinder.)
- Using the fact that \(V = 33.6\), write \(S\) as a function of \(r\) and state its applied domain.
- Use your graphing calculator to find the dimensions of the can which has minimal surface area.
- A right cylindrical drum is to hold 7.35 cubic feet of liquid. Find the dimensions (radius of the base and height) of the drum which would minimize the surface area. What is the minimum surface area? Round your answers to two decimal places.
- In Exercise 71 in Section 1.4 , the population of Sasquatch in Portage County was modeled by the function \(P(t) = \frac{150t}{t + 15}\), where \(t = 0\) represents the year 1803. When were there fewer than 100 Sasquatch in Portage County?

In Exercises 33 - 38, translate the following into mathematical equations.

- At a constant pressure, the temperature \(T\) of an ideal gas is directly proportional to its volume \(V\). (This is Charles’s Law )
- The frequency of a wave \(f\) is inversely proportional to the wavelength of the wave \(\lambda\).
- The density \(d\) of a material is directly proportional to the mass of the object \(m\) and inversely proportional to its volume \(V\).
- The square of the orbital period of a planet \(P\) is directly proportional to the cube of the semi-major axis of its orbit \(a\). (This is Kepler’s Third Law of Planetary Motion )
- The drag of an object traveling through a fluid \(D\) varies jointly with the density of the fluid \(\rho\) and the square of the velocity of the object \(\nu\).
- Suppose two electric point charges, one with charge \(q\) and one with charge \(Q\), are positioned \(r\) units apart. The electrostatic force \(F\) exerted on the charges varies directly with the product of the two charges and inversely with the square of the distance between the charges. (This is Coulomb’s Law )
- According to this webpage , the frequency \(f\) of a vibrating string is given by \(f = \dfrac{1}{2L} \sqrt{\dfrac{T}{\mu}}\) where \(T\) is the tension, \(\mu\) is the linear mass 15 of the string and \(L\) is the length of the vibrating part of the string. Express this relationship using the language of variation.
- Express this relationship as a mathematical equation.
- If a person who was \(5\) feet, \(10\) inches tall weighed 235 pounds had a Body Mass Index of 33.7, what is the value of the constant of proportionality?
- Rewrite the mathematical equation found in part 40a to include the value of the constant found in part 40b and then find your Body Mass Index.
- We know that the circumference of a circle varies directly with its radius with \(2\pi\) as the constant of proportionality. (That is, we know \(C = 2\pi r.\)) With the help of your classmates, compile a list of other basic geometric relationships which can be seen as variations.

## 4.3.3 Answers

- \(x = -\frac{6}{7}\)
- \(x = 1, \; x = 2\)
- \(x = -6, \; x = 2\)
- No solution
- \(x = 0, \; x = \pm 2\sqrt{2}\)
- \((-2, \infty)\)
- \((-2, 3]\)
- \((-1, 0) \cup (1, \infty)\)
- \([0, \infty)\)
- \((-\infty, -3) \cup (-3,2) \cup (4, \infty)\)
- \(\left(-3, -\frac{1}{3} \right) \cup (2,3)\)
- \((-1,0] \cup (2, \infty)\)
- \((-\infty, -3) \cup (-2, -1) \cup (1, \infty)\)
- \((-\infty, 1] \cup [2, \infty)\)
- \((-\infty, -6) \cup (-1, 2)\)
- \((-\infty, -3) \cup \left[-2\sqrt{2}, 0\right] \cup \left[2\sqrt{2}, 3\right)\)
- \([-3,0) \cup (0,4) \cup [5, \infty)\)
- \(\left(-1,-\frac{1}{2}\right] \cup (1, \infty)\)
- 4.5 miles per hour
- 24 miles per hour
- 3600 gallons
- \(\frac{40}{3} \approx 13.33\) minutes
- The absolute minimum of \(y=\overline{C}(x)\) occurs at \(\approx (75.73, 59.57)\). Since \(x\) represents the number of game systems, we check \(\overline{C}(75) \approx 59.58\) and \(\overline{C}(76) \approx 59.57\). Hence, to minimize the average cost, \(76\) systems should be produced at an average cost of \(\$59.57\) per system.
- The width (and depth) should be \(10.00\) centimeters, the height should be \(5.00\) centimeters. The minimum surface area is \(300.00\) square centimeters.
- The width of the base of the box should be \(\approx 4.12\) inches, the height of the box should be \(\approx 6.67\) inches, and the depth of the base of the box should be \(\approx 5.09\) inches; minimum surface area \(\approx 164.91\) square inches.
- The dimensions are \(\approx 7\) feet by \(\approx 14\) feet; minimum amount of fencing required \(\approx 28\) feet.
- \(V = \pi r^{2}h\)
- \(S = 2 \pi r^{2} + 2\pi r h\)
- \(S(r) = 2\pi r^{2} + \frac{67.2}{r}, \;\) Domain \(r > 0\)
- \(r \approx 1.749\,\)in. and \(h \approx 3.498\,\)in.
- The radius of the drum should be \(\approx 1.05\) feet and the height of the drum should be \(\approx 2.12\) feet. The minimum surface area of the drum is \(\approx 20.93\) cubic feet.
- \(P(t) < 100\) on \((-15, 30)\), and the portion of this which lies in the applied domain is \([0,30)\). Since \(t=0\) corresponds to the year 1803, from 1803 through the end of 1832, there were fewer than 100 Sasquatch in Portage County.
- \(T = k V\)
- 16 \(f = \dfrac{k}{\lambda}\)
- \(d = \dfrac{k m}{V}\)
- \(P^2 = k a^3\)
- 17 \(D = k \rho \nu^2\)
- 18 \(F = \dfrac{kqQ}{r^2}\)
- Rewriting \(f = \dfrac{1}{2L} \sqrt{\dfrac{T}{\mu}}\) as \(f = \dfrac{\frac{1}{2} \sqrt{T}}{L \sqrt{\mu}}\) we see that the frequency \(f\) varies directly with the square root of the tension and varies inversely with the length and the square root of the linear mass.
- \(B = \dfrac{kW}{h^{2}}\)
- 19 \(k = 702.68\)
- \(B = \dfrac{702.68W}{h^{2}}\)

1 There is no asymptote at \(x = 1\) since the graph is well behaved near \(x = 1\). According to Theorem 4.1 , there must be a hole there.

2 This is called a system of equations. No doubt, you’ve had experience with these things before, and we will study systems in greater detail in Chapter 8 .

3 While we usually discourage dividing both sides of an equation by a variable expression, we know \((6+R) \neq 0\) since otherwise we couldn’t possibly multiply it by \(t_{\text {down }}\) and get 5.

4 The reader is encouraged to verify that the units in this equation are the same on both sides. To get you started, the units on the ‘3’ is ‘hours.’

5 Carl would much rather spend his time writing open-source Mathematics texts than gardening anyway.

6 In other words, make sure you don’t try to add apples to oranges!

7 In this case, long division amounts to term-by-term division.

8 That is, \(h(x)\) means ‘\(h\) of \(x\)’, not ‘\(h\) times \(x\)’ here.

9 without Calculus, that is...

10 We will talk more about this in the coming chapters.

11 You can use tell the calculator to do this arithmetic on the lists and save yourself some time.

12 We will revisit this example once we have developed logarithms in Chapter 6 to see how we can actually ‘linearize’ this data and do a linear regression to obtain the same result.

13 1.62 is a crude approximation of the so-called ‘Golden Ratio’ \(\phi=\frac{1+\sqrt{5}}{2}\).

14 According to www.dictionary.com , there are different values given for this conversion. We will stick with 33.6in 3 for this problem.

15 Also known as the linear density. It is simply a measure of mass per unit length.

16 The character λ is the lower case Greek letter ‘lambda.’

17 The characters \(\rho\) and \(\nu\) are the lower case Greek letters ‘rho’ and ‘nu,’ respectively.

18 Note the similarity to this formula and Newton’s Law of Universal Gravitation as discussed in Example 5.

19 The CDC uses 703.

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This module is written to help you solve real-life problems involving rational functions, equations, and inequalities. You will be introduced to different types of word problems and situations. These problems can be transformed into rational functions, equations, and inequalities.

518 Share 21K views 1 year ago GRADE 11 FIRST QUARTER Solving Real-Life Problems Involving Rational Functions, Equations, and Inequalities A Garden Plot Vincent is a farmer. He loves to plant...

This video lesson is about solving real-life problems involving rational functions, rational equation, and rational inequality.

Analysis of the Solution To find the horizontal asymptote, divide the leading coefficient in the numerator by the leading coefficient in the denominator: \displaystyle \frac {1} {10}=0.1 101 = 0.1 Notice the horizontal asymptote is \displaystyle y=\text { }0.1 y = 0.1.

Finding the Common Denominator When we add or subtract two or more rationals, we need to find the least common denominator (LCD), just like when we add or subtract regular fractions. If the denominators are the same, we can just add the numerators across, leaving the denominators as they are. We then must be sure we can't do any further factoring:

In this section, we solve equations and inequalities involving rational functions and explore associated application problems. Our first example showcases the critical difference in procedure between solving a rational equation and a rational inequality. Example 4.3.1. Solve x3 − 2x + 1 x − 1 = 1 2x − 1. Solve x3 − 2x + 1 x − 1 ≥ 1 ...

In this lesson, you will learn how to SOLVE PROBLEMS INVOLVING RATIONAL FUNCTIONS, EQUATIONS, AND INEQUALITIES.00:00- Introduction00:20- Example 1 (Motion Pr...

SOLVING REAL - LIFE PROBLEMS INVOLVING RATIONAL FUNCTIONS, EQUATIONS, AND INEQUALITIESAt the end of this video lesson, you are expected to:1. Solve real - li...

This topic covers: - Simplifying rational expressions - Multiplying, dividing, adding, & subtracting rational expressions - Rational equations - Graphing rational functions (including horizontal & vertical asymptotes) - Modeling with rational functions - Rational inequalities - Partial fraction expansion

A rational inequality is an inequality that contains a rational expression. Inequalities such as 3 2x > 1, 2x x − 3 < 4, 2x − 3 x − 6 ≥ x, and 1 4 − 2 x2 ≤ 3 x are rational inequalities as they each contain a rational expression. When we solve a rational inequality, we will use many of the techniques we used solving linear inequalities.

Autoplay 50K views Solving a Rational Equation Let's begin by looking at solving an equation with rational functions in it. I'll give you a simple equation for the sake of keeping...

This video tutorial is about problems involving rational functions, equations, and inequalities witj several examplesThanks for watching

Solving Real-Life Problems Involving Rational Functions, Equations, and InequalitiesMix mix mix! How many liters of pure alcohol must be added to 30 liters o...

Rational Functions Word Problems - Work, Tank And Pipe. Here are a few examples of work problems that are solved with rational equations. Examples: Sam can paint a house in 5 hours. Gary can do it in 4 hours. How long will it take the two working together? Joy can file 100 claims in 5 hours. Stephen can file 100 claims in 8 hours.

A good way to remember this first part of solving the problem is to think about what identifies a rational equation. Yes, fractions, and fractions have denominators. Step 1 is to find the common ...

1. solve real-life problems involving rational functions, equations, and inequalities; 2. carefully analyze and understand word problems before solving them; and 3. create real-life word problems about rational functions, equations and inequalities. genmath_q1_mod10_SolvingRealLifeProblemsInvolvingRational_v2 Share with your friends!

Section 4.8 : Rational Functions. Back to Problem List. 1. Sketch the graph of the following function. Clearly identify all intercepts and asymptotes. f (x) = −4 x−2 f ( x) = − 4 x − 2. Show All Steps Hide All Steps. Start Solution.

For the learner: Welcome to General Mathematics Grade 11 Alternative Delivery Mode (ADM) Module on Solving Real-Life Problems Involving Rational Functions, Equations and Inequalities. The hand is one of the most symbolized parts of the human body. It is often used to depict skill, action and purpose.

This module was designed to help learners gain understanding about rational functions. It is composed of two lessons. The first lesson tackles about representing real life situations using rational functions, and the second lesson will delve about distinguishing rational function, rational equation and rational inequality.

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.

This module was designed and written with you in mind. It is here to help you solve real-life problems involving exponential functions, equations, and inequalities. Most of the time, students like you ask why you need to study Mathematics. Even though you know the answer, still you keep on asking this question because perhaps you

It is here to help you solve real-life problems involving exponential functions, equations, and ... 1. recall how to solve exponential functions, equations, and inequalities; and. 2. solve real-life problems involving exponential functions, equations, and inequalities.

4.3: Rational Inequalities and Applications. In this section, we solve equations and inequalities involving rational functions and explore associated application problems. Our first example showcases the critical difference in procedure between solving a rational equation and a rational inequality.