{"id":15306,"date":"2021-10-11T22:52:39","date_gmt":"2021-10-11T22:52:39","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/ccbcmd-math-1\/chapter\/introduction-minima-and-maxima-of-quadratic-functions\/"},"modified":"2021-12-29T19:08:00","modified_gmt":"2021-12-29T19:08:00","slug":"applications-of-quadratic-functions","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/ccbcmd-math-1\/chapter\/applications-of-quadratic-functions\/","title":{"raw":"Applications of Quadratic Functions","rendered":"Applications of Quadratic Functions"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul class=\"ul1\">\r\n \t<li class=\"li2\"><span class=\"s1\">Solve problems involving the roots and intercepts of a quadratic function.<\/span><\/li>\r\n \t<li class=\"li2\"><span class=\"s4\">Solve problems involving a quadratic function\u2019s minimum or maximum value.<\/span><\/li>\r\n<\/ul>\r\n<\/div>\r\nIn this section we will investigate quadratic functions further, including solving\u00a0problems involving revenue, area, and projectile motion. Working with quadratic functions can be less complex than working with higher degree polynomial functions, so they provide a good opportunity for a detailed study of function behavior.\r\n<p id=\"fs-id1165137431411\">There are many real-world scenarios that involve finding the maximum or minimum value of a quadratic function, such as applications involving area and revenue.<\/p>\r\n\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"975\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03010712\/CNX_Precalc_Figure_03_02_0092.jpg\" alt=\"Two graphs where the first graph shows the maximum value for f(x)=(x-2)^2+1 which occurs at (2, 1) and the second graph shows the minimum value for g(x)=-(x+3)^2+4 which occurs at (-3, 4).\" width=\"975\" height=\"558\" data-media-type=\"image\/jpg\" \/> <b>Figure 9<\/b>[\/caption]\r\n\r\n<div id=\"Example_03_02_05\" class=\"example\" data-type=\"example\">\r\n<div id=\"fs-id1165134378616\" class=\"exercise\" data-type=\"exercise\">\r\n<div id=\"fs-id1165134378618\" class=\"problem textbox shaded\" data-type=\"problem\">\r\n<h3 style=\"text-align: center;\" data-type=\"title\">Example<\/h3>\r\n<p id=\"fs-id1165137653457\">A backyard farmer wants to enclose a rectangular space for a new garden within her fenced backyard. She has purchased 80 feet of wire fencing to enclose three sides, and she will use a section of the backyard fence as the fourth side.<\/p>\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03010713\/CNX_Precalc_Figure_03_02_0102.jpg\" alt=\"Diagram of the garden and the backyard.\" width=\"487\" height=\"310\" data-media-type=\"image\/jpg\" \/>\r\n<ol id=\"fs-id1165135640934\" data-number-style=\"lower-alpha\">\r\n \t<li>Find a formula for the area enclosed by the fence if the sides of fencing perpendicular to the existing fence have length <em>L<\/em>.<\/li>\r\n \t<li>What dimensions should she make her garden to maximize the enclosed area?<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"89388\"]SOLUTION[\/reveal-answer]\r\n[hidden-answer a=\"89388\"]\r\n<p id=\"fs-id1165137836808\">Let\u2019s use a diagram above to record the given information. It is also helpful to introduce a temporary variable, <em>W<\/em>, to represent the width of the garden and the length of the fence section parallel to the backyard fence.<span id=\"fs-id1165135208803\" data-type=\"media\" data-alt=\"Diagram of the garden and the backyard.\">\r\n<\/span><\/p>\r\n\r\n<ol id=\"fs-id1165134363440\" data-number-style=\"lower-alpha\">\r\n \t<li>We know we have only 80 feet of fence available, and [latex]L+W+L=80[\/latex], or more simply, [latex]2L+W=80[\/latex]. This allows us to represent the width, <em>W<\/em>, in terms of <em>L<\/em>.\r\n<div id=\"eip-id1165135697866\" class=\"equation unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]W=80 - 2L[\/latex]<\/div>\r\n<p id=\"fs-id1165135435476\">Now we are ready to write an equation for the area the fence encloses. We know the area of a rectangle is length multiplied by width, so<\/p>\r\n\r\n<div id=\"eip-624\" class=\"equation unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{cases}\\text{ }A=LW=L\\left(80 - 2L\\right)\\hfill \\\\ A\\left(L\\right)=80L - 2{L}^{2}\\hfill \\end{cases}[\/latex]<\/div>\r\n<p id=\"fs-id1165135258914\">This formula represents the area of the fence in terms of the variable length <em>L<\/em>. The function, written in general form, is<\/p>\r\n\r\n<div id=\"eip-382\" class=\"equation unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]A\\left(L\\right)=-2{L}^{2}+80L[\/latex].<\/div><\/li>\r\n \t<li>The quadratic has a negative leading coefficient, so the graph will open downward, and the vertex will be the maximum value for the area. In finding the vertex, we must be careful because the equation is not written in standard polynomial form with decreasing powers. This is why we rewrote the function in general form above. Since <em>a<\/em>\u00a0is the coefficient of the squared term, [latex]a=-2,b=80[\/latex], and [latex]c=0[\/latex].<\/li>\r\n<\/ol>\r\n<p id=\"fs-id1165137772015\">To find the vertex:<\/p>\r\n\r\n<div id=\"eip-id1165135202446\" class=\"equation unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{cases}h=-\\frac{80}{2\\left(-2\\right)}\\hfill &amp; \\hfill &amp; \\hfill &amp; \\hfill &amp; k=A\\left(20\\right)\\hfill \\\\ \\text{ }=20\\hfill &amp; \\hfill &amp; \\text{and}\\hfill &amp; \\hfill &amp; \\text{ }=80\\left(20\\right)-2{\\left(20\\right)}^{2}\\hfill \\\\ \\hfill &amp; \\hfill &amp; \\hfill &amp; \\hfill &amp; \\text{ }=800\\hfill \\end{cases}[\/latex]<\/div>\r\n<p id=\"fs-id1165135174964\">The maximum value of the function is an area of 800 square feet, which occurs when [latex]L=20[\/latex] feet. When the shorter sides are 20 feet, there is 40 feet of fencing left for the longer side. To maximize the area, she should enclose the garden so the two shorter sides have length 20 feet and the longer side parallel to the existing fence has length 40 feet.<\/p>\r\n\r\n<h4 data-type=\"title\">Analysis of the Solution<\/h4>\r\nThis problem also could be solved by graphing the quadratic function. We can see where the maximum area occurs on a graph of the quadratic function in the figure below.\r\n\r\n<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03010713\/CNX_Precalc_Figure_03_02_0112.jpg\" alt=\"Graph of the parabolic function A(L)=-2L^2+80L, which the x-axis is labeled Length (L) and the y-axis is labeled Area (A). The vertex is at (20, 800).\" width=\"487\" height=\"476\" data-media-type=\"image\/jpg\" \/>\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165133340409\" class=\"note precalculus howto textbox\" data-type=\"note\" data-has-label=\"true\" data-label=\"How To\">\r\n<h3 id=\"fs-id1165137803708\">How To: Given an application involving revenue, use a quadratic equation to find the maximum.<\/h3>\r\n<ol id=\"fs-id1165135436584\" data-number-style=\"arabic\">\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 <em data-effect=\"italics\">y<\/em>-value of the vertex.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div id=\"Example_03_02_06\" class=\"example\" data-type=\"example\">\r\n<div id=\"fs-id1165134278696\" class=\"exercise\" data-type=\"exercise\">\r\n<div id=\"fs-id1165137473136\" class=\"problem textbox shaded\" data-type=\"problem\">\r\n<h3 style=\"text-align: center;\" data-type=\"title\">Example<\/h3>\r\n<p id=\"fs-id1165137473142\">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. For example, a local newspaper currently has 84,000 subscribers at a quarterly charge of $30. Market research has suggested that if the owners raise the price to $32, they would lose 5,000 subscribers. Assuming that subscriptions are linearly related to the price, what price should the newspaper charge for a quarterly subscription to maximize their revenue?<\/p>\r\n[reveal-answer q=\"467644\"]SOLUTION[\/reveal-answer]\r\n[hidden-answer a=\"467644\"]\r\n<p id=\"fs-id1165135389888\">Revenue is the amount of money a company brings in. In this case, the revenue can be found by multiplying the price per subscription times the number of subscribers, or quantity. We can introduce variables, <em>p<\/em>\u00a0for price per subscription and <em>Q<\/em>\u00a0for quantity, giving us the equation [latex]\\text{Revenue}=pQ[\/latex].<\/p>\r\n<p id=\"fs-id1165134232972\">Because the number of subscribers changes with the price, we need to find a relationship between the variables. 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\r\n<div id=\"eip-id1165135246622\" class=\"equation unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{cases}m=\\frac{79,000 - 84,000}{32 - 30}\\hfill \\\\ \\text{ }=\\frac{-5,000}{2}\\hfill \\\\ \\text{ }=-2,500\\hfill \\end{cases}[\/latex]<\/div>\r\n<p id=\"fs-id1165135559520\">This tells us the paper will lose 2,500 subscribers for each dollar they raise the price. We can then solve for the <em data-effect=\"italics\">y<\/em>-intercept.<\/p>\r\n\r\n<div id=\"eip-id1165131968004\" class=\"equation unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{cases}\\text{ }Q=-2500p+b\\hfill &amp; \\text{Substitute in the point }Q=84,000\\text{ and }p=30\\hfill \\\\ 84,000=-2500\\left(30\\right)+b\\hfill &amp; \\text{Solve for }b\\hfill \\\\ \\text{ }b=159,000\\hfill &amp; \\hfill \\end{cases}[\/latex]<\/div>\r\n<p id=\"fs-id1165137933138\">This gives us the linear equation [latex]Q=-2,500p+159,000[\/latex] relating cost and subscribers. We now return to our revenue equation.<\/p>\r\n\r\n<div id=\"eip-id1165132337192\" class=\"equation unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{cases}\\text{Revenue}=pQ\\hfill \\\\ \\text{Revenue}=p\\left(-2,500p+159,000\\right)\\hfill \\\\ \\text{Revenue}=-2,500{p}^{2}+159,000p\\hfill \\end{cases}[\/latex]<\/div>\r\n<p id=\"fs-id1165135502033\">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>\r\n\r\n<div id=\"eip-id1165135170999\" class=\"equation unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{cases}h=-\\frac{159,000}{2\\left(-2,500\\right)}\\hfill \\\\ \\text{ }=31.8\\hfill \\end{cases}[\/latex]<\/div>\r\n<p id=\"fs-id1165137647087\">The model tells us that the maximum revenue will occur if the newspaper charges $31.80 for a subscription. To find what the maximum revenue is, we evaluate the revenue function.<\/p>\r\n\r\n<div id=\"eip-id1165134323486\" class=\"equation unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{cases}\\text{maximum revenue}=-2,500{\\left(31.8\\right)}^{2}+159,000\\left(31.8\\right)\\hfill \\\\ \\text{ }=2,528,100\\hfill \\end{cases}[\/latex]<\/div>\r\n<div data-type=\"equation\" data-label=\"\">\r\n<h4 data-type=\"title\">Analysis of the Solution<\/h4>\r\nThis could also be solved by graphing the quadratic. We can see the maximum revenue on a graph of the quadratic function.\r\n\r\n<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03010713\/CNX_Precalc_Figure_03_02_0122.jpg\" alt=\"Graph of the parabolic function which the x-axis is labeled Price (p) and the y-axis is labeled Revenue ($). The vertex is at (31.80, 258100).\" width=\"487\" height=\"327\" data-media-type=\"image\/jpg\" \/>\r\n\r\n<\/div>\r\n[\/hidden-answer]<span style=\"text-align: center; font-size: 0.9em;\">\u00a0<\/span>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<h2><\/h2>\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nA ball is thrown upward from the top of a 40 foot high building at a speed of 80 feet per second. The ball\u2019s height above ground can be modeled by the equation [latex]H\\left(t\\right)=-16{t}^{2}+80t+40[\/latex].\r\n\r\na. When does the ball reach the maximum height?\r\n\r\nb. What is the maximum height of the ball?\r\n\r\nc. When does the ball hit the ground?\r\n\r\n[reveal-answer q=\"394530\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"394530\"]\r\n\r\na. The ball reaches the maximum height at the vertex of the parabola.\r\n<p style=\"text-align: center;\">[latex]\\begin{align}h&amp;=-\\dfrac{80}{2\\left(-16\\right)} \\\\[1mm]&amp;=\\dfrac{80}{32} \\\\[1mm] &amp;=\\dfrac{5}{2} \\\\[1mm] &amp;=2.5 \\end{align}[\/latex]<\/p>\r\nThe ball reaches a maximum height after 2.5 seconds.\r\n\r\nb. To find the maximum height, find the [latex]y[\/latex]<em>\u00a0<\/em>coordinate of the vertex of the parabola.\r\n<p style=\"text-align: center;\">[latex]\\begin{align}k&amp;=H\\left(2.5\\right) \\\\[1mm] &amp;=-16{\\left(2.5\\right)}^{2}+80\\left(2.5\\right)+40 \\\\[1mm] &amp;=140\\hfill \\end{align}[\/latex]<\/p>\r\nThe ball reaches a maximum height of 140 feet.\r\n\r\nc. To find when the ball hits the ground, we need to determine when the height is zero, [latex]H\\left(t\\right)=0[\/latex].\r\n\r\nWe use the quadratic formula.\r\n<p style=\"text-align: center;\">[latex]\\begin{align} t&amp;=\\dfrac{-80\\pm \\sqrt{{80}^{2}-4\\left(-16\\right)\\left(40\\right)}}{2\\left(-16\\right)} \\\\[1mm] &amp;=\\dfrac{-80\\pm \\sqrt{8960}}{-32} \\end{align}[\/latex]<\/p>\r\nBecause the square root does not simplify nicely, we can use a calculator to approximate the values of the solutions.\r\n<p style=\"text-align: center;\">[latex]t=\\dfrac{-80+\\sqrt{8960}}{-32}\\approx 5.458\\hspace{3mm}[\/latex] or [latex]\\hspace{3mm}t=\\dfrac{-80-\\sqrt{8960}}{-32}\\approx -0.458[\/latex]<\/p>\r\nSince the domain starts at [latex]t=0[\/latex] when the ball is thrown, the second answer is outside the reasonable domain of our model. The ball will hit the ground after about 5.46 seconds.\r\n\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/02170404\/CNX_Precalc_Figure_03_02_0162.jpg\" alt=\"Graph of a negative parabola where x goes from -1 to 6.\" width=\"487\" height=\"254\" \/>\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nA rock is thrown upward from the top of a 112-foot high cliff overlooking the ocean at a speed of 96 feet per second. The rock\u2019s height above ocean can be modeled by the equation [latex]H\\left(t\\right)=-16{t}^{2}+96t+112[\/latex].\r\n\r\na. When does the rock reach the maximum height?\r\n\r\nb. What is the maximum height of the rock?\r\n\r\nc. When does the rock hit the ocean?\r\n\r\n[reveal-answer q=\"174919\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"174919\"]\r\n\r\na.\u00a03 seconds \u00a0b.\u00a0256 feet \u00a0c.\u00a07 seconds\r\n\r\n[\/hidden-answer]\r\n<iframe id=\"mom2\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=15809&amp;theme=oea&amp;iframe_resize_id=mom2\" width=\"100%\" height=\"375\"><\/iframe>\r\n\r\n<\/div>\r\n<h2>Applying<\/h2>\r\nThere are many real-world scenarios that involve finding the maximum or minimum value of a quadratic function, such as applications involving area and revenue.\r\n\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/02170347\/CNX_Precalc_Figure_03_02_0092.jpg\" alt=\"Two graphs where the first graph shows the maximum value for f(x)=(x-2)^2+1 which occurs at (2, 1) and the second graph shows the minimum value for g(x)=-(x+3)^2+4 which occurs at (-3, 4).\" width=\"975\" height=\"558\" \/>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Finding the Maximum Value of a Quadratic Function<\/h3>\r\nA backyard farmer wants to enclose a rectangular space for a new garden within her fenced backyard. She has purchased 80 feet of wire fencing to enclose three sides, and she will use a section of the backyard fence as the fourth side.\r\n<ol>\r\n \t<li>Find a formula for the area enclosed by the fence if the sides of fencing perpendicular to the existing fence have length [latex]L[\/latex].<\/li>\r\n \t<li>What dimensions should she make her garden to maximize the enclosed area?<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"704029\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"704029\"]\r\n\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/02170349\/CNX_Precalc_Figure_03_02_0102.jpg\" alt=\"Diagram of the garden and the backyard.\" width=\"487\" height=\"310\" \/>\r\n\r\nLet\u2019s use a diagram such as the one above\u00a0to record the given information. It is also helpful to introduce a temporary variable, [latex]W[\/latex], to represent the width of the garden and the length of the fence section parallel to the backyard fence.\r\n<ol>\r\n \t<li>We know we have only 80 feet of fence available, and [latex]L+W+L=80[\/latex], or more simply, [latex]2L+W=80[\/latex]. This allows us to represent the width, [latex]W[\/latex], in terms of [latex]L[\/latex].\r\n[latex]W=80 - 2L[\/latex]Now we are ready to write an equation for the area the fence encloses. We know the area of a rectangle is length multiplied by width, so [latex]A=LW=L\\left(80 - 2L\\right)=80L - 2{L}^{2}[\/latex]. This formula represents the area of the fence in terms of the variable length [latex]L[\/latex]. The function, written in general form, is[latex]A\\left(L\\right)=-2{L}^{2}+80L[\/latex].<\/li>\r\n \t<li>The quadratic has a negative leading coefficient, so the graph will open downward, and the vertex will be the maximum value for the area. In finding the vertex, we must be careful because the equation is not written in standard polynomial form with decreasing powers. This is why we rewrote the function in general form above. Since [latex]a[\/latex]\u00a0is the coefficient of the squared term, [latex]a=-2,b=80[\/latex], and [latex]c=0[\/latex].<\/li>\r\n<\/ol>\r\nTo find the vertex:\r\n<p style=\"text-align: center;\">[latex]h=-\\dfrac{80}{2\\left(-2\\right)}=20[\/latex]<\/p>\r\n<p style=\"text-align: center;\">and<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{align}k&amp;=A\\left(20\\right) \\\\&amp;=80\\left(20\\right)-2{\\left(20\\right)}^{2}\\\\&amp;=800 \\end{align}[\/latex]<\/p>\r\nThe maximum value of the function is an area of 800 square feet, which occurs when [latex]L=20[\/latex] feet. When the shorter sides are 20 feet, there is 40 feet of fencing left for the longer side. To maximize the area, she should enclose the garden so the two shorter sides have length 20 feet and the longer side parallel to the existing fence has length 40 feet.\r\n<h4>Analysis of the Solution<\/h4>\r\nThis problem also could be solved by graphing the quadratic function. We can see where the maximum area occurs on a graph of the quadratic function below.\r\n\r\n<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/02170352\/CNX_Precalc_Figure_03_02_0112.jpg\" alt=\"Graph of the parabolic function A(L)=-2L^2+80L, which the x-axis is labeled Length (L) and the y-axis is labeled Area (A). The vertex is at (20, 800).\" width=\"487\" height=\"476\" \/>\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nThe problem we solved above is called a constrained optimization problem. We can optimize our desired outcome given a constraint, which in this case was a limited amount of fencing materials. Try it yourself in the next problem.\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n<iframe id=\"mom2\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=2451&amp;theme=oea&amp;iframe_resize_id=mom2\" width=\"100%\" height=\"350\"><\/iframe>\r\n\r\n<\/div>\r\n&nbsp;\r\n<div class=\"textbox\">\r\n<h3>How To: Given an application involving revenue, use a quadratic equation to find the maximum.<\/h3>\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<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Finding Maximum Revenue<\/h3>\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. For example, a local newspaper currently has 84,000 subscribers at a quarterly charge of $30. Market research has suggested that if the owners raise the price to $32, they would lose 5,000 subscribers. Assuming 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=\"230015\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"230015\"]\r\n\r\nRevenue is the amount of money a company brings in. In this case, the revenue can be found by multiplying the price per subscription times the number of subscribers, or quantity. We can introduce variables, [latex]p[\/latex]\u00a0for price per subscription and [latex]Q[\/latex]\u00a0for quantity, giving us the equation [latex]\\text{Revenue}=pQ[\/latex].\r\n\r\nBecause the number of subscribers changes with the price, we need to find a relationship between the variables. 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\r\n<p style=\"text-align: center;\">[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\nThis tells us the paper will lose 2,500 subscribers for each dollar they raise the price. We can then solve for the [latex]y[\/latex]-intercept.\r\n<p style=\"text-align: center;\">[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. We now return to our 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\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<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 $31.80 for a subscription. To 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\r\n<h4>Analysis of the Solution<\/h4>\r\nThis could also be solved by graphing the quadratic. We can see the maximum revenue on a graph of the quadratic function.\r\n\r\n<img class=\"aligncenter\" 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=\"487\" height=\"327\" \/>\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nIn 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<div class=\"textbox key-takeaways\">\r\n<h3>try it<\/h3>\r\nPreviously,\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<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n<iframe id=\"mom5\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=15552&amp;theme=oea&amp;iframe_resize_id=mom5\" width=\"100%\" height=\"350\"><\/iframe>\r\n\r\n<\/div>\r\n<h2>Key Equations<\/h2>\r\nThe quadratic formula [latex]x=\\frac{-b\\pm \\sqrt{{b}^{2}-4ac}}{2a}[\/latex]\r\n\r\nThe discriminant is defined as [latex]b^2-4ac[\/latex]\r\n<h2>Key Concepts<\/h2>\r\n<ul>\r\n \t<li>The zeros, or [latex]x[\/latex]-intercepts, are the points at which the parabola crosses the [latex]x[\/latex]-axis. The [latex]y[\/latex]-intercept is the point at which the parabola crosses the [latex]y[\/latex]<em>-<\/em>axis.<\/li>\r\n \t<li>The vertex can be found from an equation representing a quadratic function.<\/li>\r\n \t<li>A quadratic function\u2019s minimum or maximum value is given by the [latex]y[\/latex]-value of the vertex.<\/li>\r\n \t<li>The minium or maximum value of a quadratic function can be used to determine the range of the function and to solve many kinds of real-world problems, including problems involving area and revenue.<\/li>\r\n \t<li>Some quadratic equations must be solved by using the quadratic formula.<\/li>\r\n \t<li>The vertex and the intercepts can be identified and interpreted to solve real-world problems.<\/li>\r\n \t<li>Some quadratic functions have complex roots.<\/li>\r\n<\/ul>\r\n<h2>Glossary<\/h2>\r\n<dl id=\"fs-id1165135623614\" class=\"definition\">\r\n \t<dt><strong>discriminant<\/strong><\/dt>\r\n \t<dd>the value under the radical in the quadratic formula, [latex]b^2-4ac[\/latex], which tells whether the quadratic has real or complex roots<\/dd>\r\n \t<dt><strong>vertex<\/strong><\/dt>\r\n \t<dd id=\"fs-id1165135623619\">the point at which a parabola changes direction, corresponding to the minimum or maximum value of the quadratic function<\/dd>\r\n<\/dl>\r\n<dl id=\"fs-id1165135623624\" class=\"definition\">\r\n \t<dt><strong>vertex form of a quadratic function<\/strong><\/dt>\r\n \t<dd id=\"fs-id1165135623630\">another name for the standard form of a quadratic function<\/dd>\r\n<\/dl>\r\n<dl id=\"fs-id1165135623634\" class=\"definition\">\r\n \t<dt><strong>zeros<\/strong><\/dt>\r\n \t<dd id=\"fs-id1165135623639\">in a given function, the values of [latex]x[\/latex] at which [latex]y=0[\/latex], also called roots<\/dd>\r\n<\/dl>","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<ul class=\"ul1\">\n<li class=\"li2\"><span class=\"s1\">Solve problems involving the roots and intercepts of a quadratic function.<\/span><\/li>\n<li class=\"li2\"><span class=\"s4\">Solve problems involving a quadratic function\u2019s minimum or maximum value.<\/span><\/li>\n<\/ul>\n<\/div>\n<p>In this section we will investigate quadratic functions further, including solving\u00a0problems involving revenue, area, and projectile motion. Working with quadratic functions can be less complex than working with higher degree polynomial functions, so they provide a good opportunity for a detailed study of function behavior.<\/p>\n<p id=\"fs-id1165137431411\">There are many real-world scenarios that involve finding the maximum or minimum value of a quadratic function, such as applications involving area and revenue.<\/p>\n<div style=\"width: 985px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03010712\/CNX_Precalc_Figure_03_02_0092.jpg\" alt=\"Two graphs where the first graph shows the maximum value for f(x)=(x-2)^2+1 which occurs at (2, 1) and the second graph shows the minimum value for g(x)=-(x+3)^2+4 which occurs at (-3, 4).\" width=\"975\" height=\"558\" data-media-type=\"image\/jpg\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 9<\/b><\/p>\n<\/div>\n<div id=\"Example_03_02_05\" class=\"example\" data-type=\"example\">\n<div id=\"fs-id1165134378616\" class=\"exercise\" data-type=\"exercise\">\n<div id=\"fs-id1165134378618\" class=\"problem textbox shaded\" data-type=\"problem\">\n<h3 style=\"text-align: center;\" data-type=\"title\">Example<\/h3>\n<p id=\"fs-id1165137653457\">A backyard farmer wants to enclose a rectangular space for a new garden within her fenced backyard. She has purchased 80 feet of wire fencing to enclose three sides, and she will use a section of the backyard fence as the fourth side.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03010713\/CNX_Precalc_Figure_03_02_0102.jpg\" alt=\"Diagram of the garden and the backyard.\" width=\"487\" height=\"310\" data-media-type=\"image\/jpg\" \/><\/p>\n<ol id=\"fs-id1165135640934\" data-number-style=\"lower-alpha\">\n<li>Find a formula for the area enclosed by the fence if the sides of fencing perpendicular to the existing fence have length <em>L<\/em>.<\/li>\n<li>What dimensions should she make her garden to maximize the enclosed area?<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q89388\">SOLUTION<\/span><\/p>\n<div id=\"q89388\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165137836808\">Let\u2019s use a diagram above to record the given information. It is also helpful to introduce a temporary variable, <em>W<\/em>, to represent the width of the garden and the length of the fence section parallel to the backyard fence.<span id=\"fs-id1165135208803\" data-type=\"media\" data-alt=\"Diagram of the garden and the backyard.\"><br \/>\n<\/span><\/p>\n<ol id=\"fs-id1165134363440\" data-number-style=\"lower-alpha\">\n<li>We know we have only 80 feet of fence available, and [latex]L+W+L=80[\/latex], or more simply, [latex]2L+W=80[\/latex]. This allows us to represent the width, <em>W<\/em>, in terms of <em>L<\/em>.\n<div id=\"eip-id1165135697866\" class=\"equation unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]W=80 - 2L[\/latex]<\/div>\n<p id=\"fs-id1165135435476\">Now we are ready to write an equation for the area the fence encloses. We know the area of a rectangle is length multiplied by width, so<\/p>\n<div id=\"eip-624\" class=\"equation unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{cases}\\text{ }A=LW=L\\left(80 - 2L\\right)\\hfill \\\\ A\\left(L\\right)=80L - 2{L}^{2}\\hfill \\end{cases}[\/latex]<\/div>\n<p id=\"fs-id1165135258914\">This formula represents the area of the fence in terms of the variable length <em>L<\/em>. The function, written in general form, is<\/p>\n<div id=\"eip-382\" class=\"equation unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]A\\left(L\\right)=-2{L}^{2}+80L[\/latex].<\/div>\n<\/li>\n<li>The quadratic has a negative leading coefficient, so the graph will open downward, and the vertex will be the maximum value for the area. In finding the vertex, we must be careful because the equation is not written in standard polynomial form with decreasing powers. This is why we rewrote the function in general form above. Since <em>a<\/em>\u00a0is the coefficient of the squared term, [latex]a=-2,b=80[\/latex], and [latex]c=0[\/latex].<\/li>\n<\/ol>\n<p id=\"fs-id1165137772015\">To find the vertex:<\/p>\n<div id=\"eip-id1165135202446\" class=\"equation unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{cases}h=-\\frac{80}{2\\left(-2\\right)}\\hfill & \\hfill & \\hfill & \\hfill & k=A\\left(20\\right)\\hfill \\\\ \\text{ }=20\\hfill & \\hfill & \\text{and}\\hfill & \\hfill & \\text{ }=80\\left(20\\right)-2{\\left(20\\right)}^{2}\\hfill \\\\ \\hfill & \\hfill & \\hfill & \\hfill & \\text{ }=800\\hfill \\end{cases}[\/latex]<\/div>\n<p id=\"fs-id1165135174964\">The maximum value of the function is an area of 800 square feet, which occurs when [latex]L=20[\/latex] feet. When the shorter sides are 20 feet, there is 40 feet of fencing left for the longer side. To maximize the area, she should enclose the garden so the two shorter sides have length 20 feet and the longer side parallel to the existing fence has length 40 feet.<\/p>\n<h4 data-type=\"title\">Analysis of the Solution<\/h4>\n<p>This problem also could be solved by graphing the quadratic function. We can see where the maximum area occurs on a graph of the quadratic function in the figure below.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03010713\/CNX_Precalc_Figure_03_02_0112.jpg\" alt=\"Graph of the parabolic function A(L)=-2L^2+80L, which the x-axis is labeled Length (L) and the y-axis is labeled Area (A). The vertex is at (20, 800).\" width=\"487\" height=\"476\" data-media-type=\"image\/jpg\" \/><\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165133340409\" class=\"note precalculus howto textbox\" data-type=\"note\" data-has-label=\"true\" data-label=\"How To\">\n<h3 id=\"fs-id1165137803708\">How To: Given an application involving revenue, use a quadratic equation to find the maximum.<\/h3>\n<ol id=\"fs-id1165135436584\" data-number-style=\"arabic\">\n<li>Write a quadratic equation for revenue.<\/li>\n<li>Find the vertex of the quadratic equation.<\/li>\n<li>Determine the <em data-effect=\"italics\">y<\/em>-value of the vertex.<\/li>\n<\/ol>\n<\/div>\n<div id=\"Example_03_02_06\" class=\"example\" data-type=\"example\">\n<div id=\"fs-id1165134278696\" class=\"exercise\" data-type=\"exercise\">\n<div id=\"fs-id1165137473136\" class=\"problem textbox shaded\" data-type=\"problem\">\n<h3 style=\"text-align: center;\" data-type=\"title\">Example<\/h3>\n<p id=\"fs-id1165137473142\">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. For example, a local newspaper currently has 84,000 subscribers at a quarterly charge of $30. Market research has suggested that if the owners raise the price to $32, they would lose 5,000 subscribers. 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\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q467644\">SOLUTION<\/span><\/p>\n<div id=\"q467644\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165135389888\">Revenue is the amount of money a company brings in. In this case, the revenue can be found by multiplying the price per subscription times the number of subscribers, or quantity. We can introduce variables, <em>p<\/em>\u00a0for price per subscription and <em>Q<\/em>\u00a0for quantity, giving us the equation [latex]\\text{Revenue}=pQ[\/latex].<\/p>\n<p id=\"fs-id1165134232972\">Because the number of subscribers changes with the price, we need to find a relationship between the variables. 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<div id=\"eip-id1165135246622\" class=\"equation unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{cases}m=\\frac{79,000 - 84,000}{32 - 30}\\hfill \\\\ \\text{ }=\\frac{-5,000}{2}\\hfill \\\\ \\text{ }=-2,500\\hfill \\end{cases}[\/latex]<\/div>\n<p id=\"fs-id1165135559520\">This tells us the paper will lose 2,500 subscribers for each dollar they raise the price. We can then solve for the <em data-effect=\"italics\">y<\/em>-intercept.<\/p>\n<div id=\"eip-id1165131968004\" class=\"equation unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{cases}\\text{ }Q=-2500p+b\\hfill & \\text{Substitute in the point }Q=84,000\\text{ and }p=30\\hfill \\\\ 84,000=-2500\\left(30\\right)+b\\hfill & \\text{Solve for }b\\hfill \\\\ \\text{ }b=159,000\\hfill & \\hfill \\end{cases}[\/latex]<\/div>\n<p id=\"fs-id1165137933138\">This gives us the linear equation [latex]Q=-2,500p+159,000[\/latex] relating cost and subscribers. We now return to our revenue equation.<\/p>\n<div id=\"eip-id1165132337192\" class=\"equation unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{cases}\\text{Revenue}=pQ\\hfill \\\\ \\text{Revenue}=p\\left(-2,500p+159,000\\right)\\hfill \\\\ \\text{Revenue}=-2,500{p}^{2}+159,000p\\hfill \\end{cases}[\/latex]<\/div>\n<p id=\"fs-id1165135502033\">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<div id=\"eip-id1165135170999\" class=\"equation unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{cases}h=-\\frac{159,000}{2\\left(-2,500\\right)}\\hfill \\\\ \\text{ }=31.8\\hfill \\end{cases}[\/latex]<\/div>\n<p id=\"fs-id1165137647087\">The model tells us that the maximum revenue will occur if the newspaper charges $31.80 for a subscription. To find what the maximum revenue is, we evaluate the revenue function.<\/p>\n<div id=\"eip-id1165134323486\" class=\"equation unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{cases}\\text{maximum revenue}=-2,500{\\left(31.8\\right)}^{2}+159,000\\left(31.8\\right)\\hfill \\\\ \\text{ }=2,528,100\\hfill \\end{cases}[\/latex]<\/div>\n<div data-type=\"equation\" data-label=\"\">\n<h4 data-type=\"title\">Analysis of the Solution<\/h4>\n<p>This could also be solved by graphing the quadratic. We can see the maximum revenue on a graph of the quadratic function.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03010713\/CNX_Precalc_Figure_03_02_0122.jpg\" alt=\"Graph of the parabolic function which the x-axis is labeled Price (p) and the y-axis is labeled Revenue ($). The vertex is at (31.80, 258100).\" width=\"487\" height=\"327\" data-media-type=\"image\/jpg\" \/><\/p>\n<\/div>\n<\/div>\n<\/div>\n<p><span style=\"text-align: center; font-size: 0.9em;\">\u00a0<\/span><\/p>\n<\/div>\n<\/div>\n<\/div>\n<h2><\/h2>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>A ball is thrown upward from the top of a 40 foot high building at a speed of 80 feet per second. The ball\u2019s height above ground can be modeled by the equation [latex]H\\left(t\\right)=-16{t}^{2}+80t+40[\/latex].<\/p>\n<p>a. When does the ball reach the maximum height?<\/p>\n<p>b. What is the maximum height of the ball?<\/p>\n<p>c. When does the ball hit the ground?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q394530\">Show Solution<\/span><\/p>\n<div id=\"q394530\" class=\"hidden-answer\" style=\"display: none\">\n<p>a. The ball reaches the maximum height at the vertex of the parabola.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}h&=-\\dfrac{80}{2\\left(-16\\right)} \\\\[1mm]&=\\dfrac{80}{32} \\\\[1mm] &=\\dfrac{5}{2} \\\\[1mm] &=2.5 \\end{align}[\/latex]<\/p>\n<p>The ball reaches a maximum height after 2.5 seconds.<\/p>\n<p>b. To find the maximum height, find the [latex]y[\/latex]<em>\u00a0<\/em>coordinate of the vertex of the parabola.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}k&=H\\left(2.5\\right) \\\\[1mm] &=-16{\\left(2.5\\right)}^{2}+80\\left(2.5\\right)+40 \\\\[1mm] &=140\\hfill \\end{align}[\/latex]<\/p>\n<p>The ball reaches a maximum height of 140 feet.<\/p>\n<p>c. To find when the ball hits the ground, we need to determine when the height is zero, [latex]H\\left(t\\right)=0[\/latex].<\/p>\n<p>We use the quadratic formula.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align} t&=\\dfrac{-80\\pm \\sqrt{{80}^{2}-4\\left(-16\\right)\\left(40\\right)}}{2\\left(-16\\right)} \\\\[1mm] &=\\dfrac{-80\\pm \\sqrt{8960}}{-32} \\end{align}[\/latex]<\/p>\n<p>Because the square root does not simplify nicely, we can use a calculator to approximate the values of the solutions.<\/p>\n<p style=\"text-align: center;\">[latex]t=\\dfrac{-80+\\sqrt{8960}}{-32}\\approx 5.458\\hspace{3mm}[\/latex] or [latex]\\hspace{3mm}t=\\dfrac{-80-\\sqrt{8960}}{-32}\\approx -0.458[\/latex]<\/p>\n<p>Since the domain starts at [latex]t=0[\/latex] when the ball is thrown, the second answer is outside the reasonable domain of our model. The ball will hit the ground after about 5.46 seconds.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/02170404\/CNX_Precalc_Figure_03_02_0162.jpg\" alt=\"Graph of a negative parabola where x goes from -1 to 6.\" width=\"487\" height=\"254\" \/><\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>A rock is thrown upward from the top of a 112-foot high cliff overlooking the ocean at a speed of 96 feet per second. The rock\u2019s height above ocean can be modeled by the equation [latex]H\\left(t\\right)=-16{t}^{2}+96t+112[\/latex].<\/p>\n<p>a. When does the rock reach the maximum height?<\/p>\n<p>b. What is the maximum height of the rock?<\/p>\n<p>c. When does the rock hit the ocean?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q174919\">Show Solution<\/span><\/p>\n<div id=\"q174919\" class=\"hidden-answer\" style=\"display: none\">\n<p>a.\u00a03 seconds \u00a0b.\u00a0256 feet \u00a0c.\u00a07 seconds<\/p>\n<\/div>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"mom2\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=15809&amp;theme=oea&amp;iframe_resize_id=mom2\" width=\"100%\" height=\"375\"><\/iframe><\/p>\n<\/div>\n<h2>Applying<\/h2>\n<p>There are many real-world scenarios that involve finding the maximum or minimum value of a quadratic function, such as applications involving area and revenue.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/02170347\/CNX_Precalc_Figure_03_02_0092.jpg\" alt=\"Two graphs where the first graph shows the maximum value for f(x)=(x-2)^2+1 which occurs at (2, 1) and the second graph shows the minimum value for g(x)=-(x+3)^2+4 which occurs at (-3, 4).\" width=\"975\" height=\"558\" \/><\/p>\n<div class=\"textbox exercises\">\n<h3>Example: Finding the Maximum Value of a Quadratic Function<\/h3>\n<p>A backyard farmer wants to enclose a rectangular space for a new garden within her fenced backyard. She has purchased 80 feet of wire fencing to enclose three sides, and she will use a section of the backyard fence as the fourth side.<\/p>\n<ol>\n<li>Find a formula for the area enclosed by the fence if the sides of fencing perpendicular to the existing fence have length [latex]L[\/latex].<\/li>\n<li>What dimensions should she make her garden to maximize the enclosed area?<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q704029\">Show Solution<\/span><\/p>\n<div id=\"q704029\" class=\"hidden-answer\" style=\"display: none\">\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/02170349\/CNX_Precalc_Figure_03_02_0102.jpg\" alt=\"Diagram of the garden and the backyard.\" width=\"487\" height=\"310\" \/><\/p>\n<p>Let\u2019s use a diagram such as the one above\u00a0to record the given information. It is also helpful to introduce a temporary variable, [latex]W[\/latex], to represent the width of the garden and the length of the fence section parallel to the backyard fence.<\/p>\n<ol>\n<li>We know we have only 80 feet of fence available, and [latex]L+W+L=80[\/latex], or more simply, [latex]2L+W=80[\/latex]. This allows us to represent the width, [latex]W[\/latex], in terms of [latex]L[\/latex].<br \/>\n[latex]W=80 - 2L[\/latex]Now we are ready to write an equation for the area the fence encloses. We know the area of a rectangle is length multiplied by width, so [latex]A=LW=L\\left(80 - 2L\\right)=80L - 2{L}^{2}[\/latex]. This formula represents the area of the fence in terms of the variable length [latex]L[\/latex]. The function, written in general form, is[latex]A\\left(L\\right)=-2{L}^{2}+80L[\/latex].<\/li>\n<li>The quadratic has a negative leading coefficient, so the graph will open downward, and the vertex will be the maximum value for the area. In finding the vertex, we must be careful because the equation is not written in standard polynomial form with decreasing powers. This is why we rewrote the function in general form above. Since [latex]a[\/latex]\u00a0is the coefficient of the squared term, [latex]a=-2,b=80[\/latex], and [latex]c=0[\/latex].<\/li>\n<\/ol>\n<p>To find the vertex:<\/p>\n<p style=\"text-align: center;\">[latex]h=-\\dfrac{80}{2\\left(-2\\right)}=20[\/latex]<\/p>\n<p style=\"text-align: center;\">and<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}k&=A\\left(20\\right) \\\\&=80\\left(20\\right)-2{\\left(20\\right)}^{2}\\\\&=800 \\end{align}[\/latex]<\/p>\n<p>The maximum value of the function is an area of 800 square feet, which occurs when [latex]L=20[\/latex] feet. When the shorter sides are 20 feet, there is 40 feet of fencing left for the longer side. To maximize the area, she should enclose the garden so the two shorter sides have length 20 feet and the longer side parallel to the existing fence has length 40 feet.<\/p>\n<h4>Analysis of the Solution<\/h4>\n<p>This problem also could be solved by graphing the quadratic function. We can see where the maximum area occurs on a graph of the quadratic function below.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/02170352\/CNX_Precalc_Figure_03_02_0112.jpg\" alt=\"Graph of the parabolic function A(L)=-2L^2+80L, which the x-axis is labeled Length (L) and the y-axis is labeled Area (A). The vertex is at (20, 800).\" width=\"487\" height=\"476\" \/><\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>The problem we solved above is called a constrained optimization problem. We can optimize our desired outcome given a constraint, which in this case was a limited amount of fencing materials. Try it yourself in the next problem.<\/p>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"mom2\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=2451&amp;theme=oea&amp;iframe_resize_id=mom2\" width=\"100%\" height=\"350\"><\/iframe><\/p>\n<\/div>\n<p>&nbsp;<\/p>\n<div class=\"textbox\">\n<h3>How To: Given an application involving revenue, use a quadratic equation to find the maximum.<\/h3>\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<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Finding Maximum Revenue<\/h3>\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. For example, a local newspaper currently has 84,000 subscribers at a quarterly charge of $30. Market research has suggested that if the owners raise the price to $32, they would lose 5,000 subscribers. 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\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q230015\">Show Solution<\/span><\/p>\n<div id=\"q230015\" class=\"hidden-answer\" style=\"display: none\">\n<p>Revenue is the amount of money a company brings in. In this case, the revenue can be found by multiplying the price per subscription times the number of subscribers, or quantity. We can introduce variables, [latex]p[\/latex]\u00a0for price per subscription and [latex]Q[\/latex]\u00a0for quantity, giving us the equation [latex]\\text{Revenue}=pQ[\/latex].<\/p>\n<p>Because the number of subscribers changes with the price, we need to find a relationship between the variables. 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;\">[latex]\\begin{align}m&=\\dfrac{79,000 - 84,000}{32 - 30} \\\\ &=\\dfrac{-5,000}{2} \\\\ &=-2,500 \\end{align}[\/latex]<\/p>\n<p>This tells us the paper will lose 2,500 subscribers for each dollar they raise the price. We can then solve for the [latex]y[\/latex]-intercept.<\/p>\n<p style=\"text-align: center;\">[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. We now return to our 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>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<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 $31.80 for a subscription. 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<h4>Analysis of the Solution<\/h4>\n<p>This could also be solved by graphing the quadratic. We can see the maximum revenue on a graph of the quadratic function.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" 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=\"487\" height=\"327\" \/><\/p>\n<\/div>\n<\/div>\n<\/div>\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<div class=\"textbox key-takeaways\">\n<h3>try it<\/h3>\n<p>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\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q410084\">Show Solution<\/span><\/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<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"mom5\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=15552&amp;theme=oea&amp;iframe_resize_id=mom5\" width=\"100%\" height=\"350\"><\/iframe><\/p>\n<\/div>\n<h2>Key Equations<\/h2>\n<p>The quadratic formula [latex]x=\\frac{-b\\pm \\sqrt{{b}^{2}-4ac}}{2a}[\/latex]<\/p>\n<p>The discriminant is defined as [latex]b^2-4ac[\/latex]<\/p>\n<h2>Key Concepts<\/h2>\n<ul>\n<li>The zeros, or [latex]x[\/latex]-intercepts, are the points at which the parabola crosses the [latex]x[\/latex]-axis. The [latex]y[\/latex]-intercept is the point at which the parabola crosses the [latex]y[\/latex]<em>&#8211;<\/em>axis.<\/li>\n<li>The vertex can be found from an equation representing a quadratic function.<\/li>\n<li>A quadratic function\u2019s minimum or maximum value is given by the [latex]y[\/latex]-value of the vertex.<\/li>\n<li>The minium or maximum value of a quadratic function can be used to determine the range of the function and to solve many kinds of real-world problems, including problems involving area and revenue.<\/li>\n<li>Some quadratic equations must be solved by using the quadratic formula.<\/li>\n<li>The vertex and the intercepts can be identified and interpreted to solve real-world problems.<\/li>\n<li>Some quadratic functions have complex roots.<\/li>\n<\/ul>\n<h2>Glossary<\/h2>\n<dl id=\"fs-id1165135623614\" class=\"definition\">\n<dt><strong>discriminant<\/strong><\/dt>\n<dd>the value under the radical in the quadratic formula, [latex]b^2-4ac[\/latex], which tells whether the quadratic has real or complex roots<\/dd>\n<dt><strong>vertex<\/strong><\/dt>\n<dd id=\"fs-id1165135623619\">the point at which a parabola changes direction, corresponding to the minimum or maximum value of the quadratic function<\/dd>\n<\/dl>\n<dl id=\"fs-id1165135623624\" class=\"definition\">\n<dt><strong>vertex form of a quadratic function<\/strong><\/dt>\n<dd id=\"fs-id1165135623630\">another name for the standard form of a quadratic function<\/dd>\n<\/dl>\n<dl id=\"fs-id1165135623634\" class=\"definition\">\n<dt><strong>zeros<\/strong><\/dt>\n<dd id=\"fs-id1165135623639\">in a given function, the values of [latex]x[\/latex] at which [latex]y=0[\/latex], also called roots<\/dd>\n<\/dl>\n\n\t\t\t <section class=\"citations-section\" role=\"contentinfo\">\n\t\t\t <h3>Candela Citations<\/h3>\n\t\t\t\t\t <div>\n\t\t\t\t\t\t <div id=\"citation-list-15306\">\n\t\t\t\t\t\t\t <div class=\"licensing\"><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Original<\/div><ul class=\"citation-list\"><li>Revision and Adaptation. <strong>Provided by<\/strong>: Lumen Learning. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Question ID 121416, 121401. <strong>Provided by<\/strong>: Lumen Learning. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em>. <strong>License Terms<\/strong>: IMathAS Community License CC-BY + GPL<\/li><\/ul><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Shared previously<\/div><ul class=\"citation-list\"><li>College Algebra. <strong>Authored by<\/strong>: Abramson, Jay et al.. <strong>Provided by<\/strong>: OpenStax. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"http:\/\/cnx.org\/contents\/9b08c294-057f-4201-9f48-5d6ad992740d@5.2\">http:\/\/cnx.org\/contents\/9b08c294-057f-4201-9f48-5d6ad992740d@5.2<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em>. <strong>License Terms<\/strong>: Download for free at http:\/\/cnx.org\/contents\/9b08c294-057f-4201-9f48-5d6ad992740d@5.2<\/li><li>Question ID 15809. <strong>Authored by<\/strong>: Sousa, James, mb Lippman, David. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em>. <strong>License Terms<\/strong>: IMathAS Community License CC-BY + GPL<\/li><li>Question ID 35145. <strong>Authored by<\/strong>: Jim Smart. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em>. <strong>License Terms<\/strong>: IMathAS Community License CC-BY + GPL<\/li><li>Question ID 2451. <strong>Authored by<\/strong>: Anderson, Tophe. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em>. <strong>License Terms<\/strong>: IMathAS Community License CC-BY + GPL<\/li><li>Question ID 15552. <strong>Authored by<\/strong>: David Lippman. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em>. <strong>License Terms<\/strong>: IMathAS Community License CC-BY + GPL<\/li><\/ul><\/div>\n\t\t\t\t\t\t <\/div>\n\t\t\t\t\t <\/div>\n\t\t\t <\/section>","protected":false},"author":167848,"menu_order":22,"template":"","meta":{"_candela_citation":"[{\"type\":\"original\",\"description\":\"Revision and Adaptation\",\"author\":\"\",\"organization\":\"Lumen 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