{"id":1416,"date":"2015-11-12T18:35:29","date_gmt":"2015-11-12T18:35:29","guid":{"rendered":"https:\/\/courses.candelalearning.com\/collegealgebra1xmaster\/?post_type=chapter&p=1416"},"modified":"2017-03-31T23:07:53","modified_gmt":"2017-03-31T23:07:53","slug":"solve-applied-problems-involving-rational-functions","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/ivytech-collegealgebra\/chapter\/solve-applied-problems-involving-rational-functions\/","title":{"raw":"Solve applied problems involving rational functions","rendered":"Solve applied problems involving rational functions"},"content":{"raw":"

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<\/strong> 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.<\/p>\r\n\r\n

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A General Note: Rational Function<\/h3>\r\n

A rational function<\/span> 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].<\/p>\r\n\r\n

[latex]f\\left(x\\right)=\\frac{P\\left(x\\right)}{Q\\left(x\\right)}=\\frac{{a}_{p}{x}^{p}+{a}_{p - 1}{x}^{p - 1}+...+{a}_{1}x+{a}_{0}}{{b}_{q}{x}^{q}+{b}_{q - 1}{x}^{q - 1}+...+{b}_{1}x+{b}_{0}},Q\\left(x\\right)\\ne 0[\/latex]<\/div>\r\n<\/div>\r\n
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Example 3: Solving an Applied Problem Involving a Rational Function<\/h3>\r\n

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?<\/p>\r\n\r\n<\/div>\r\n

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Solution<\/h3>\r\n

Let t<\/em>\u00a0be 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:<\/p>\r\n\r\n

[latex]\\begin{cases}\\text{water: }W\\left(t\\right)=100+10t\\text{ in gallons}\\\\ \\text{sugar: }S\\left(t\\right)=5+1t\\text{ in pounds}\\end{cases}[\/latex]<\/div>\r\n

The concentration, C<\/em>, will be the ratio of pounds of sugar to gallons of water<\/p>\r\n\r\n

[latex]C\\left(t\\right)=\\frac{5+t}{100+10t}[\/latex]<\/div>\r\n

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

[latex]\\begin{cases}C\\left(12\\right)=\\frac{5+12}{100+10\\left(12\\right)}\\hfill \\\\ \\text{ }=\\frac{17}{220}\\hfill \\end{cases}[\/latex]<\/div>\r\n

This means the concentration is 17 pounds of sugar to 220 gallons of water.<\/p>\r\n

At the beginning, the concentration is<\/p>\r\n\r\n

[latex]\\begin{cases}C\\left(0\\right)=\\frac{5+0}{100+10\\left(0\\right)}\\hfill \\\\ \\text{ }=\\frac{1}{20}\\hfill \\end{cases}[\/latex]<\/div>\r\n

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.<\/p>\r\n\r\n<\/div>\r\n

\r\n

Analysis of the Solution<\/h3>\r\n

To find the horizontal asymptote, divide the leading coefficient in the numerator by the leading coefficient in the denominator:<\/p>\r\n\r\n

[latex]\\frac{1}{10}=0.1[\/latex]<\/div>\r\n

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

\r\n

Try It 3<\/h3>\r\n

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.<\/p>\r\nSolution<\/a>\r\n\r\n<\/div>","rendered":"

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<\/strong> 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.<\/p>\n

\n

A General Note: Rational Function<\/h3>\n

A rational function<\/span> 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].<\/p>\n

[latex]f\\left(x\\right)=\\frac{P\\left(x\\right)}{Q\\left(x\\right)}=\\frac{{a}_{p}{x}^{p}+{a}_{p - 1}{x}^{p - 1}+...+{a}_{1}x+{a}_{0}}{{b}_{q}{x}^{q}+{b}_{q - 1}{x}^{q - 1}+...+{b}_{1}x+{b}_{0}},Q\\left(x\\right)\\ne 0[\/latex]<\/div>\n<\/div>\n
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\n
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Example 3: Solving an Applied Problem Involving a Rational Function<\/h3>\n

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?<\/p>\n<\/div>\n

\n

Solution<\/h3>\n

Let t<\/em>\u00a0be 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:<\/p>\n

[latex]\\begin{cases}\\text{water: }W\\left(t\\right)=100+10t\\text{ in gallons}\\\\ \\text{sugar: }S\\left(t\\right)=5+1t\\text{ in pounds}\\end{cases}[\/latex]<\/div>\n

The concentration, C<\/em>, will be the ratio of pounds of sugar to gallons of water<\/p>\n

[latex]C\\left(t\\right)=\\frac{5+t}{100+10t}[\/latex]<\/div>\n

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

[latex]\\begin{cases}C\\left(12\\right)=\\frac{5+12}{100+10\\left(12\\right)}\\hfill \\\\ \\text{ }=\\frac{17}{220}\\hfill \\end{cases}[\/latex]<\/div>\n

This means the concentration is 17 pounds of sugar to 220 gallons of water.<\/p>\n

At the beginning, the concentration is<\/p>\n

[latex]\\begin{cases}C\\left(0\\right)=\\frac{5+0}{100+10\\left(0\\right)}\\hfill \\\\ \\text{ }=\\frac{1}{20}\\hfill \\end{cases}[\/latex]<\/div>\n

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.<\/p>\n<\/div>\n

\n

Analysis of the Solution<\/h3>\n

To find the horizontal asymptote, divide the leading coefficient in the numerator by the leading coefficient in the denominator:<\/p>\n

[latex]\\frac{1}{10}=0.1[\/latex]<\/div>\n

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

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Try It 3<\/h3>\n

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.<\/p>\n

Solution<\/a><\/p>\n<\/div>\n\n\t\t\t

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