{"id":1864,"date":"2024-06-14T21:30:40","date_gmt":"2024-06-14T21:30:40","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/?post_type=chapter&#038;p=1864"},"modified":"2025-08-14T01:13:12","modified_gmt":"2025-08-14T01:13:12","slug":"analysis-of-quadratic-functions-learn-it-3","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/chapter\/analysis-of-quadratic-functions-learn-it-3\/","title":{"raw":"Analysis of Quadratic Functions: Learn It 3","rendered":"Analysis of Quadratic Functions: Learn It 3"},"content":{"raw":"<div class=\"page\" title=\"Page 483\">\r\n<div class=\"layoutArea\">\r\n<div class=\"column\">\r\n<h2>Finding Maximum Revenue<\/h2>\r\nQuadratic functions aren't just abstract math concepts\u2014they're super useful in real-life situations, especially in business! One cool way we can use quadratic functions is to figure out how to maximize revenue.\r\n\r\nThink of it this way: if you\u2019re running a business and selling a product, you want to find the best price to sell that product so you make the most money. <strong>Revenue<\/strong> (the money you bring in) is calculated by multiplying the price per unit by the number of units sold. But here's the catch\u2014the number of units sold usually changes with the price, creating a quadratic relationship.\r\n\r\nLet's see how to use quadratic functions to find that sweet spot\u2014the price that brings in the most revenue! By looking at the graph of the quadratic function, especially the vertex, we can figure out the optimal price to charge.\r\n\r\n<section class=\"textbox questionHelp\" aria-label=\"Question Help\"><strong>How To: Given an application involving revenue, use a quadratic equation to find the maximum.<\/strong>\r\n<ol>\r\n \t<li>Write a quadratic equation for revenue.<\/li>\r\n \t<li>Find the vertex of the quadratic equation.<\/li>\r\n \t<li>Determine the [latex]y[\/latex]-value of the vertex.<\/li>\r\n<\/ol>\r\n<\/section><section class=\"textbox example\">\r\n<div class=\"page\" title=\"Page 483\">\r\n<div class=\"layoutArea\">\r\n<div class=\"column\">\r\n\r\nThe unit price of an item affects its supply and demand. That is, if the unit price goes up, the demand for the item will usually decrease.\r\n\r\nA local newspaper currently has [latex]84,000[\/latex] subscribers at a quarterly charge of [latex]\\$30[\/latex]. Market research has suggested that if the owners raise the price to [latex]\\$32[\/latex], they would lose [latex]5,000[\/latex] subscribers.\r\n\r\nAssuming that subscriptions are linearly related to the price, what price should the newspaper charge for a quarterly subscription to maximize their revenue?\r\n\r\n[reveal-answer q=\"963099\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"963099\"]\r\n\r\n<strong>Revenue<\/strong> (the money you bring in) is calculated by multiplying the price per unit by the number of units sold.\r\n<p style=\"padding-left: 40px;\">In this scenario, the revenue can be found by multiplying the price per subscription times the number of subscribers, or quantity.<\/p>\r\nLet's introduce variables: Let [latex]p[\/latex] for price per subscription and [latex]Q[\/latex] for quantity, giving us the equation [latex]\\text{Revenue}=pQ[\/latex].\r\n\r\nSubscriptions are linearly related to the price. Let's find the linear function represent the subscription.\r\n<ul>\r\n \t<li>Slope<\/li>\r\n<\/ul>\r\n<p style=\"padding-left: 40px;\">We know that currently [latex]p=30[\/latex] and [latex]Q=84,000[\/latex]. We also know that if the price rises to $32, the newspaper would lose 5,000 subscribers, giving a second pair of values, [latex]p=32[\/latex] and [latex]Q=79,000[\/latex]. From this we can find a linear equation relating the two quantities. The slope will be<\/p>\r\n<p style=\"text-align: center; padding-left: 40px;\">[latex]\\begin{align}m&amp;=\\dfrac{79,000 - 84,000}{32 - 30} \\\\ &amp;=\\dfrac{-5,000}{2} \\\\ &amp;=-2,500 \\end{align}[\/latex]<\/p>\r\n<p style=\"padding-left: 40px;\">This tells us the paper will lose 2,500 subscribers for each dollar they raise the price.<\/p>\r\n\r\n<ul>\r\n \t<li>[latex]y[\/latex]-intercept<\/li>\r\n<\/ul>\r\n<p style=\"padding-left: 40px;\">[latex]\\begin{align}&amp;Q=-2500p+b &amp;&amp;\\text{Substitute in the point }Q=84,000\\text{ and }p=30 \\\\ &amp;84,000=-2500\\left(30\\right)+b &amp;&amp;\\text{Solve for }b \\\\ &amp;b=159,000 \\end{align}[\/latex]<\/p>\r\nThis gives us the linear equation [latex]Q=-2,500p+159,000[\/latex] relating cost and subscribers.\r\n\r\nBack to the revenue equation.\r\n<p style=\"text-align: center;\">[latex]\\begin{align}&amp;\\text{Revenue}=pQ \\\\ &amp;\\text{Revenue}=p\\left(-2,500p+159,000\\right) \\\\ &amp;\\text{Revenue}=-2,500{p}^{2}+159,000p \\end{align}[\/latex]<\/p>\r\n<strong>Revenue Function: [latex]R(p) = -2,500{p}^{2}+159,000p [\/latex]<\/strong>\r\n\r\nWe now have a quadratic function for revenue as a function of the subscription charge. To find the price that will maximize revenue for the newspaper, we can find the vertex.\r\n\r\n[caption id=\"\" align=\"alignright\" width=\"356\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/02170354\/CNX_Precalc_Figure_03_02_0122.jpg\" alt=\"Graph of the parabolic function which the x-axis is labeled Price (p) and the y-axis is labeled Revenue ($). The vertex is at (31.80, 258100).\" width=\"356\" height=\"239\" \/> Graph of the parabolic function with the vertex labeled[\/caption]\r\n<p style=\"text-align: center;\">[latex]\\begin{align}h&amp;=-\\dfrac{159,000}{2\\left(-2,500\\right)} \\\\ &amp;=31.8 \\end{align}[\/latex]<\/p>\r\nThe model tells us that the maximum revenue will occur if the newspaper charges [latex]\\$31.80[\/latex] for a subscription.\r\n\r\nTo find what the maximum revenue is, we evaluate the revenue function.\r\n<p style=\"text-align: center;\">[latex]\\begin{align}\\text{maximum revenue}&amp;=-2,500{\\left(31.8\\right)}^{2}+159,000\\left(31.8\\right) \\\\ &amp;=\\$2,528,100\\hfill \\end{align}[\/latex]<\/p>\r\n<strong>The price should the newspaper charge for a quarterly subscription is [latex]\\$31.80[\/latex] to obtain the maximum revenue of [latex]\\$2,528,100[\/latex].<\/strong>\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/section>In the example above, we knew the number of subscribers to a newspaper and used that information to find the optimal price for each subscription. What if the price of subscriptions is affected by competition?\r\n\r\n<section class=\"textbox example\" aria-label=\"Example\">Previously,\u00a0we found\u00a0a quadratic function that modeled revenue as a function of price.\r\n<p style=\"text-align: center;\">[latex]\\text{Revenue}-2,500{p}^{2}+159,000p[\/latex]<\/p>\r\n<p style=\"text-align: left;\">We found that selling the paper at [latex]\\$31.80[\/latex] per subscription would maximize revenue. \u00a0What if your closest competitor sells their paper for [latex]\\$25.00[\/latex] per subscription? What is the maximum revenue you can make you sell your paper for the same?<\/p>\r\n[reveal-answer q=\"410084\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"410084\"]\r\n\r\nEvaluating the function for [latex]p=25[\/latex] gives [latex]\\$2,412,500[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/section><section class=\"textbox tryIt\">[ohm2_question hide_question_numbers=1]24489[\/ohm2_question]<\/section><\/div>\r\n<\/div>\r\n<\/div>","rendered":"<div class=\"page\" title=\"Page 483\">\n<div class=\"layoutArea\">\n<div class=\"column\">\n<h2>Finding Maximum Revenue<\/h2>\n<p>Quadratic functions aren&#8217;t just abstract math concepts\u2014they&#8217;re super useful in real-life situations, especially in business! One cool way we can use quadratic functions is to figure out how to maximize revenue.<\/p>\n<p>Think of it this way: if you\u2019re running a business and selling a product, you want to find the best price to sell that product so you make the most money. <strong>Revenue<\/strong> (the money you bring in) is calculated by multiplying the price per unit by the number of units sold. But here&#8217;s the catch\u2014the number of units sold usually changes with the price, creating a quadratic relationship.<\/p>\n<p>Let&#8217;s see how to use quadratic functions to find that sweet spot\u2014the price that brings in the most revenue! By looking at the graph of the quadratic function, especially the vertex, we can figure out the optimal price to charge.<\/p>\n<section class=\"textbox questionHelp\" aria-label=\"Question Help\"><strong>How To: Given an application involving revenue, use a quadratic equation to find the maximum.<\/strong><\/p>\n<ol>\n<li>Write a quadratic equation for revenue.<\/li>\n<li>Find the vertex of the quadratic equation.<\/li>\n<li>Determine the [latex]y[\/latex]-value of the vertex.<\/li>\n<\/ol>\n<\/section>\n<section class=\"textbox example\">\n<div class=\"page\" title=\"Page 483\">\n<div class=\"layoutArea\">\n<div class=\"column\">\n<p>The unit price of an item affects its supply and demand. That is, if the unit price goes up, the demand for the item will usually decrease.<\/p>\n<p>A local newspaper currently has [latex]84,000[\/latex] subscribers at a quarterly charge of [latex]\\$30[\/latex]. Market research has suggested that if the owners raise the price to [latex]\\$32[\/latex], they would lose [latex]5,000[\/latex] subscribers.<\/p>\n<p>Assuming that subscriptions are linearly related to the price, what price should the newspaper charge for a quarterly subscription to maximize their revenue?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q963099\">Show Answer<\/button><\/p>\n<div id=\"q963099\" class=\"hidden-answer\" style=\"display: none\">\n<p><strong>Revenue<\/strong> (the money you bring in) is calculated by multiplying the price per unit by the number of units sold.<\/p>\n<p style=\"padding-left: 40px;\">In this scenario, the revenue can be found by multiplying the price per subscription times the number of subscribers, or quantity.<\/p>\n<p>Let&#8217;s introduce variables: Let [latex]p[\/latex] for price per subscription and [latex]Q[\/latex] for quantity, giving us the equation [latex]\\text{Revenue}=pQ[\/latex].<\/p>\n<p>Subscriptions are linearly related to the price. Let&#8217;s find the linear function represent the subscription.<\/p>\n<ul>\n<li>Slope<\/li>\n<\/ul>\n<p style=\"padding-left: 40px;\">We know that currently [latex]p=30[\/latex] and [latex]Q=84,000[\/latex]. We also know that if the price rises to $32, the newspaper would lose 5,000 subscribers, giving a second pair of values, [latex]p=32[\/latex] and [latex]Q=79,000[\/latex]. From this we can find a linear equation relating the two quantities. The slope will be<\/p>\n<p style=\"text-align: center; padding-left: 40px;\">[latex]\\begin{align}m&=\\dfrac{79,000 - 84,000}{32 - 30} \\\\ &=\\dfrac{-5,000}{2} \\\\ &=-2,500 \\end{align}[\/latex]<\/p>\n<p style=\"padding-left: 40px;\">This tells us the paper will lose 2,500 subscribers for each dollar they raise the price.<\/p>\n<ul>\n<li>[latex]y[\/latex]-intercept<\/li>\n<\/ul>\n<p style=\"padding-left: 40px;\">[latex]\\begin{align}&Q=-2500p+b &&\\text{Substitute in the point }Q=84,000\\text{ and }p=30 \\\\ &84,000=-2500\\left(30\\right)+b &&\\text{Solve for }b \\\\ &b=159,000 \\end{align}[\/latex]<\/p>\n<p>This gives us the linear equation [latex]Q=-2,500p+159,000[\/latex] relating cost and subscribers.<\/p>\n<p>Back to the revenue equation.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}&\\text{Revenue}=pQ \\\\ &\\text{Revenue}=p\\left(-2,500p+159,000\\right) \\\\ &\\text{Revenue}=-2,500{p}^{2}+159,000p \\end{align}[\/latex]<\/p>\n<p><strong>Revenue Function: [latex]R(p) = -2,500{p}^{2}+159,000p[\/latex]<\/strong><\/p>\n<p>We now have a quadratic function for revenue as a function of the subscription charge. To find the price that will maximize revenue for the newspaper, we can find the vertex.<\/p>\n<figure style=\"width: 356px\" class=\"wp-caption alignright\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/02170354\/CNX_Precalc_Figure_03_02_0122.jpg\" alt=\"Graph of the parabolic function which the x-axis is labeled Price (p) and the y-axis is labeled Revenue ($). The vertex is at (31.80, 258100).\" width=\"356\" height=\"239\" \/><figcaption class=\"wp-caption-text\">Graph of the parabolic function with the vertex labeled<\/figcaption><\/figure>\n<p style=\"text-align: center;\">[latex]\\begin{align}h&=-\\dfrac{159,000}{2\\left(-2,500\\right)} \\\\ &=31.8 \\end{align}[\/latex]<\/p>\n<p>The model tells us that the maximum revenue will occur if the newspaper charges [latex]\\$31.80[\/latex] for a subscription.<\/p>\n<p>To find what the maximum revenue is, we evaluate the revenue function.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}\\text{maximum revenue}&=-2,500{\\left(31.8\\right)}^{2}+159,000\\left(31.8\\right) \\\\ &=\\$2,528,100\\hfill \\end{align}[\/latex]<\/p>\n<p><strong>The price should the newspaper charge for a quarterly subscription is [latex]\\$31.80[\/latex] to obtain the maximum revenue of [latex]\\$2,528,100[\/latex].<\/strong><\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/section>\n<p>In the example above, we knew the number of subscribers to a newspaper and used that information to find the optimal price for each subscription. What if the price of subscriptions is affected by competition?<\/p>\n<section class=\"textbox example\" aria-label=\"Example\">Previously,\u00a0we found\u00a0a quadratic function that modeled revenue as a function of price.<\/p>\n<p style=\"text-align: center;\">[latex]\\text{Revenue}-2,500{p}^{2}+159,000p[\/latex]<\/p>\n<p style=\"text-align: left;\">We found that selling the paper at [latex]\\$31.80[\/latex] per subscription would maximize revenue. \u00a0What if your closest competitor sells their paper for [latex]\\$25.00[\/latex] per subscription? What is the maximum revenue you can make you sell your paper for the same?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q410084\">Show Solution<\/button><\/p>\n<div id=\"q410084\" class=\"hidden-answer\" style=\"display: none\">\n<p>Evaluating the function for [latex]p=25[\/latex] gives [latex]\\$2,412,500[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox tryIt\"><iframe loading=\"lazy\" id=\"ohm24489\" class=\"resizable\" src=\"https:\/\/ohm.one.lumenlearning.com\/multiembedq.php?id=24489&theme=lumen&iframe_resize_id=ohm24489&source=tnh\" width=\"100%\" height=\"150\"><\/iframe><\/section>\n<\/div>\n<\/div>\n<\/div>\n","protected":false},"author":12,"menu_order":20,"template":"","meta":{"_candela_citation":"[]","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"part":185,"module-header":"learn_it","content_attributions":[],"internal_book_links":[],"video_content":null,"cc_video_embed_content":{"cc_scripts":"","media_targets":[]},"try_it_collection":null,"_links":{"self":[{"href":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/1864"}],"collection":[{"href":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/wp-json\/wp\/v2\/users\/12"}],"version-history":[{"count":5,"href":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/1864\/revisions"}],"predecessor-version":[{"id":7733,"href":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/1864\/revisions\/7733"}],"part":[{"href":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/wp-json\/pressbooks\/v2\/parts\/185"}],"metadata":[{"href":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/1864\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/wp-json\/wp\/v2\/media?parent=1864"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/wp-json\/pressbooks\/v2\/chapter-type?post=1864"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/wp-json\/wp\/v2\/contributor?post=1864"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/wp-json\/wp\/v2\/license?post=1864"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}