{"id":5065,"date":"2017-12-31T16:32:19","date_gmt":"2017-12-31T16:32:19","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/cuny-hunter-collegealgebra\/?post_type=chapter&#038;p=5065"},"modified":"2023-11-08T13:20:36","modified_gmt":"2023-11-08T13:20:36","slug":"applications-of-quadratic-functions","status":"web-only","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/cuny-hunter-collegealgebra\/chapter\/applications-of-quadratic-functions\/","title":{"raw":"10.3 - Applications of Quadratic Functions","rendered":"10.3 &#8211; Applications of Quadratic Functions"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Objectives<\/h3>\r\n<ul>\r\n \t<li>(10.3.1) - Solve application problems involving quadratic functions\r\n<ul>\r\n \t<li>Objects in free fall<\/li>\r\n \t<li>Determining the width of a border<\/li>\r\n \t<li>Finding the maximum and minimum values of a quadratic function<\/li>\r\n<\/ul>\r\n<\/li>\r\n<\/ul>\r\n<\/div>\r\n<h1>(10.3.1) - Solve application problems involving quadratic functions<\/h1>\r\nQuadratic equations are widely used in science, business, and engineering. Quadratic equations are commonly used in situations where two things are multiplied together and they both depend on the same variable. For example, when working with area, if both dimensions are written in terms of the same variable, you use a quadratic equation. Because the quantity of a product sold often depends on the price, you sometimes use a quadratic equation to represent revenue as a product of the price and the quantity sold. Quadratic equations are also used when gravity is involved, such as the path of a ball or the shape of cables in a suspension bridge.\r\n<h3>Objects in free fall<\/h3>\r\nA very common and easy-to-understand application is the height of a ball thrown at the ground off a building. Because gravity will make the ball speed up as it falls, a quadratic equation can be used to estimate its height any time before it hits the ground.<i> Note: The equation isn't completely accurate, because friction from the air will slow the ball down a little. For our purposes, this is close enough.<\/i>\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example<\/h3>\r\nA ball is thrown off a building from 200 feet above the ground. Its starting velocity (also called <i>initial velocity<\/i>) is [latex]\u221210[\/latex] feet per second. (The negative value means it's heading toward the ground.)\r\n\r\nThe equation [latex]h=-16t^{2}-10t+200[\/latex]\u00a0can be used to model the height of the ball after [latex]t[\/latex] seconds. About how long does it take for the ball to hit the ground?\r\n\r\n[reveal-answer q=\"704677\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"704677\"]\r\n\r\nWhen the ball hits the ground, the height is 0. Substitute 0 for [latex]h[\/latex].\r\n<p style=\"text-align: center\">[latex]\\begin{array}{c}h=-16t^{2}-10t+200\\\\0=-16t^{2}-10t+200\\\\-16t^{2}-10t+200=0\\end{array}[\/latex]<\/p>\r\nThis equation is difficult to solve by factoring or by completing the square, so solve it by applying the Quadratic Formula, [latex] x=\\frac{-b\\pm \\sqrt{{{b}^{2}}-4ac}}{2a}[\/latex]. In this case, the variable is [latex]t[\/latex] rather than [latex]x[\/latex]. [latex]a=\u221216,b=\u221210[\/latex], and [latex]c=200[\/latex].\r\n<p style=\"text-align: center\">[latex]\\displaystyle t=\\frac{-(-10)\\pm \\sqrt{{{(-10)}^{2}}-4(-16)(200)}}{2(-16)}[\/latex]<\/p>\r\nSimplify. Be very careful with the signs.\r\n<p style=\"text-align: center\">[latex]\\large \\begin{array}{l}t=\\frac{10\\pm \\sqrt{100+12800}}{-32}\\\\\\,\\,=\\frac{10\\pm \\sqrt{12900}}{-32}\\end{array}[\/latex]<\/p>\r\nUse a calculator to find both roots.\r\n<p style=\"text-align: center\">[latex]t[\/latex] is approximately [latex]\u22123.86[\/latex] or [latex]3.24[\/latex].<\/p>\r\nConsider the roots logically. One solution, [latex]\u22123.86[\/latex], cannot be the time because it is a negative number. The other solution, [latex]3.24[\/latex] seconds, must be when the ball hits the ground.\r\n<h4>Answer<\/h4>\r\nThe ball hits the ground approximately [latex]3.24[\/latex] seconds after being thrown.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nIn the next video we show another example of how the quadratic equation can be used to find the time it takes for an object in free fall to hit the ground.\r\n\r\nhttps:\/\/youtu.be\/RcVeuJhcuL0\r\n\r\nHere are some more similar objects in free fall examples.\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Applying the Vertex and [latex]<em>x[\/latex]<\/em>-Intercepts of a Parabola<\/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\"]Solution[\/reveal-answer]\r\n[hidden-answer a=\"394530\"]\r\na. The ball reaches the maximum height at the vertex of the parabola.\r\n<p style=\"text-align: center\">[latex]\\large \\begin{array}{c} h=-\\frac{80}{2\\left(-16\\right)} \\text{ }=\\frac{80}{32}\\hfill \\\\ \\text{ }=\\frac{5}{2}\\hfill \\\\ \\text{ }=2.5\\hfill \\end{array}[\/latex]<\/p>\r\nThe ball reaches a maximum height after 2.5 seconds.\r\n\r\nb. To find the maximum height, find the <em>y\u00a0<\/em>coordinate of the vertex of the parabola.\r\n<p style=\"text-align: center\">[latex]\\large \\begin{array}{c}k=H\\left(-\\frac{b}{2a}\\right)\\hfill \\\\ \\text{ }=H\\left(2.5\\right)\\hfill \\\\ \\text{ }=-16{\\left(2.5\\right)}^{2}+80\\left(2.5\\right)+40\\hfill \\\\ \\text{ }=140\\hfill \\end{array}[\/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]\\large \\begin{array}{c} t=\\frac{-80\\pm \\sqrt{{80}^{2}-4\\left(-16\\right)\\left(40\\right)}}{2\\left(-16\\right)}\\hfill \\\\ \\text{ }=\\frac{-80\\pm \\sqrt{8960}}{-32}\\hfill \\end{array}[\/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]\\large \\begin{array}{c}t=\\frac{-80-\\sqrt{8960}}{-32}\\approx 5.458\\hfill &amp; \\text{or}\\hfill &amp; t=\\frac{-80+\\sqrt{8960}}{-32}\\approx -0.458\\hfill \\end{array}[\/latex]<\/p>\r\nThe second answer is outside the reasonable domain of our model, so we conclude the ball will hit the ground after about 5.458 seconds.\r\n\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2862\/2017\/12\/26165456\/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>EXAMPLE<\/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\"]Solution[\/reveal-answer]\r\n[hidden-answer a=\"174919\"]\r\n\r\na.\u00a03 seconds\r\n\r\nb.\u00a0256 feet\r\n\r\nc.\u00a07 seconds\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h3>Applications of quadratic functions: determining the width of a border<\/h3>\r\nThe area problem below does not look like it includes a Quadratic Formula of any type, and the problem seems to be something you have solved many times before by simply multiplying. But in order to solve it, you will need to use a quadratic equation.\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example<\/h3>\r\nBob made a quilt that is 4 ft [latex]\\times[\/latex] 5 ft. He has 10 sq. ft. of fabric he can use to add a border around the quilt. How wide should he make the border to use all the fabric? (The border must be the same width on all four sides.)\r\n\r\n[reveal-answer q=\"932211\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"932211\"]\r\n\r\nSketch the problem. Since you don\u2019t know the width of the border, you will let the variable [latex]x[\/latex]\u00a0represent the width.\r\n\r\nIn the diagram, the original quilt is indicated by the red rectangle. The border is the area between the red and blue lines.\r\n\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1468\/2016\/02\/04064559\/image052-2.gif\" alt=\"A blue rectangle. Within the blue rectangle are a pair of vertical parallel lines and a pair of horizontal parallel lines that create a smaller red rectangle. The lengths of this red rectangle are 4 feet and 5 feet. The line segments between the boundaries of the red rectangle and the bigger blue rectangle are all labeled x.\" width=\"321\" height=\"278\" \/>\r\n\r\nSince each side of the original 4 by 5 quilt has the border of width <i>x <\/i>added, the length of the quilt with the border will be [latex]5+2x[\/latex],\u00a0and the width will be\u00a0[latex]4+2x[\/latex].\r\n\r\n(Both dimensions are written in terms of the same variable, and you will multiply them to get an area! This is where you might start to think that a quadratic equation might be used to solve this problem.)\r\n\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1468\/2016\/02\/04064600\/image053-2.gif\" alt=\"A blue rectangle with one side a height of 4+2x and another side a length of 5+2x. Within the blue rectangle are a pair of vertical parallel lines and a pair of horizontal parallel lines that create a smaller red rectangle. The height of this red rectangle is 4 feet and the length is 5 feet. The line segments between the boundaries of the red rectangle and the bigger blue rectangle are all labeled x.\" width=\"330\" height=\"285\" \/>\r\n\r\nYou are only interested in the area of the border strips. Write an expression for the area of the border.\r\n<p style=\"text-align: center\">Area of border = Area of the blue rectangle minus the area of the red rectangle<\/p>\r\n<p style=\"text-align: center\">Area of border[latex]=\\left(4+2x\\right)\\left(5+2x\\right)\u2013\\left(4\\right)\\left(5\\right)[\/latex]<\/p>\r\nThere are 10 sq ft of fabric for the border, so set the area of border to be 10.\r\n<p style=\"text-align: center\">[latex]10=\\left(4+2x\\right)\\left(5+2x\\right)\u201320[\/latex]<\/p>\r\nMultiply [latex]\\left(4+2x\\right)\\left(5+2x\\right)[\/latex].\r\n<p style=\"text-align: center\">[latex]10=20+8x+10x+4x^{2}\u201320[\/latex]<\/p>\r\nSimplify.\r\n<p style=\"text-align: center\">[latex]10=18x+4x^{2}[\/latex]<\/p>\r\nSubtract 10 from both sides so that you have a quadratic equation in standard form and can apply the Quadratic Formula to find the roots of the equation.\r\n<p style=\"text-align: center\">[latex]\\begin{array}{c}0=18x+4x^{2}-10\\\\\\\\\\text{or}\\\\\\\\4x^{2}-10\\\\\\\\2\\left(2x^{2}+9x-5\\right)=0\\end{array}[\/latex]<\/p>\r\nFactor out the greatest common factor, 2, so that you can work with the simpler equivalent equation, [latex]2x^{2}+9x\u20135=0[\/latex].\r\n<p style=\"text-align: center\">[latex]\\large \\begin{array}{r}2\\left(2x^{2}+9x-5\\right)=0\\\\\\\\\\frac{2\\left(2x^{2}+9x-5\\right)}{2}=\\frac{0}{2}\\\\\\\\2x^{2}+9x-5=0\\end{array}[\/latex]<\/p>\r\nUse the Quadratic Formula. In this case, [latex]a=2,b=9[\/latex], and [latex]c=\u22125[\/latex].\r\n<p style=\"text-align: center\">[latex]\\large \\begin{array}{l}x=\\frac{-b\\pm \\sqrt{{{b}^{2}}-4ac}}{2a}\\\\\\\\x=\\frac{-9\\pm \\sqrt{{{9}^{2}}-4(2)(-5)}}{2(2)}\\end{array}[\/latex]<\/p>\r\nSimplify.\r\n<p style=\"text-align: center\">[latex]\\displaystyle x=\\frac{-9\\pm \\sqrt{121}}{4}=\\frac{-9\\pm 11}{4}[\/latex]<\/p>\r\nFind the solutions, making sure that the [latex]\\pm[\/latex] is evaluated for both values.\r\n<p style=\"text-align: center\">[latex]\\large \\begin{array}{c}x=\\frac{-9+11}{4}=\\frac{2}{4}=\\frac{1}{2}=0.5\\\\\\\\\\text{or}\\\\\\\\x=\\frac{-9-11}{4}=\\frac{-20}{4}=-5\\end{array}[\/latex]<\/p>\r\nIgnore the solution [latex]x=\u22125[\/latex], since the width could not be negative.\r\n<h4>Answer<\/h4>\r\nThe width of the border should be 0.5 ft.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nHere is a video which gives another example of using the quadratic formula for a geometry problem involving the border around a quilt.\r\n\r\nhttps:\/\/youtu.be\/Zxe-SdwutxA\r\n<h3>Finding the maximum and minimum values of a quadratic function<\/h3>\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\/2862\/2017\/12\/26165500\/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\" \/><a href=\"https:\/\/s3-us-west-2.amazonaws.com\/oerfiles\/College+Algebra\/calculator.html\" target=\"_blank\" rel=\"noopener\"><img class=\"alignnone size-full wp-image-3370\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2862\/2017\/12\/26165501\/calculator.png\" alt=\"\" width=\"251\" height=\"46\" \/><\/a>\r\n<div class=\"textbox exercises\">\r\n<h3>ExAMPLE<\/h3>\r\nFind two numbers [latex]x[\/latex] and [latex]y[\/latex] whose difference is 100 and whose product is a minimum.\r\n\r\n[reveal-answer q=\"42108\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"42108\"]\r\n\r\nWe are trying to find the minimum of the product [latex]P=xy[\/latex] of two numbers, such that their difference is 100: [latex]y-x=100[\/latex]. First, we rewrite one variable in terms of the other:\r\n<p style=\"text-align: center\">[latex]y-x=100 \\rightarrow y=100+x[\/latex]<\/p>\r\nNext, we plug in the above relationship between the variables into the first equation:\r\n<p style=\"text-align: center\">[latex]P=xy=x(100+x) = 100x+x^2 = x^2+100x[\/latex]<\/p>\r\nAs a result, we get a quadratic function [latex]P(x)=x^2+100x[\/latex]. The graph of this quadratic function opens upwards, and its vertex is the minimum, So if we find the vertex of this parabola, we will find the minimum product. The vertex is:\r\n<p style=\"text-align: center\">[latex]\\displaystyle \\displaystyle \\left(-\\frac{b}{2a}, P\\left(-\\frac{b}{2a}\\right)\\right) = \\left(-\\frac{(100)}{2(1)}, P\\left(-\\frac{(100)}{2(1)}\\right)\\right) = (-50,-2,500)[\/latex]<\/p>\r\nThus the minimum of the parabola occurs at [latex]x=-50[\/latex], and is [latex]-2,500[\/latex]. So one of the numbers is [latex]x=-50[\/latex], the other we obtain by plugging in [latex]x=-50[\/latex] in to [latex]y-x=100[\/latex]:\r\n<p style=\"text-align: center\">[latex]\\begin{array}{cc}y-(-50)&amp;=&amp;100 \\\\ y+50 &amp;=&amp; 100 \\\\ y &amp;=&amp; 50\\end{array}[\/latex]<\/p>\r\n<strong>Answer<\/strong>\r\n\r\n[latex]x=-50[\/latex] and [latex]y=50[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\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\"]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\/2862\/2017\/12\/26165503\/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, <em>W<\/em>, to represent the width of the garden and the length of the fence section parallel to the backyard fence.\r\n\r\n1)\u00a0 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<p style=\"text-align: center\">[latex]W=80 - 2L[\/latex]<\/p>\r\nNow 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\r\n<p style=\"text-align: center\">[latex]\\begin{array}{l}\\text{ }A&amp;=&amp;LW=L\\left(80 - 2L\\right)\\hfill \\\\ A\\left(L\\right)&amp;=&amp;80L - 2{L}^{2}\\hfill \\end{array}[\/latex]<\/p>\r\nThis formula represents the area of the fence in terms of the variable length [latex]L[\/latex]. The function, written in general form, is\r\n<p style=\"text-align: center\">[latex]A\\left(L\\right)=-2{L}^{2}+80L[\/latex].<\/p>\r\n2) 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].\r\n\r\nTo find the vertex:\r\n<p style=\"text-align: center\">[latex]\\large \\begin{array}{l}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{array}[\/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\/2862\/2017\/12\/26165506\/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\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\"]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]\\large \\begin{array}{c}m=\\frac{79,000 - 84,000}{32 - 30}\\hfill \\\\ \\text{ }=\\frac{-5,000}{2}\\hfill \\\\ \\text{ }=-2,500\\hfill \\end{array}[\/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 <em>y<\/em>-intercept.\r\n<p style=\"text-align: center\">[latex]\\begin{array}{c}\\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{array}[\/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{array}{c}\\text{Revenue}=pQ\\hfill \\\\ \\text{Revenue}=p\\left(-2,500p+159,000\\right)\\hfill \\\\ \\text{Revenue}=-2,500{p}^{2}+159,000p\\hfill \\end{array}[\/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]\\large \\begin{array}{c}h=-\\frac{159,000}{2\\left(-2,500\\right)}\\hfill \\\\ \\text{ }=31.8\\hfill \\end{array}[\/latex]<\/p>\r\nThe model tells us that the maximum revenue will occur if the newspaper charges $31.80 for a subscription. To find what the maximum revenue is, we evaluate the revenue function.\r\n<p style=\"text-align: center\">[latex]\\begin{array}{c}\\text{maximum revenue}&amp;=&amp;-2,500{\\left(31.8\\right)}^{2}+159,000\\left(31.8\\right)\\hfill \\\\ \\text{ }&amp;=&amp;2,528,100\\hfill \\end{array}[\/latex]<\/p>\r\n\r\n<h4>Analysis of the Solution<\/h4>\r\nThis could also be solved by graphing the quadratic. We can see the maximum revenue on a graph of the quadratic function.\r\n\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2862\/2017\/12\/26165508\/CNX_Precalc_Figure_03_02_0122.jpg\" alt=\"Graph of the parabolic function which the x-axis is labeled Price (p) and the y-axis is labeled Revenue ($). The vertex is at (31.80, 258100).\" width=\"487\" height=\"327\" \/>\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nA coordinate grid has been superimposed over the quadratic path of a basketball in the picture below. Find an equation for the path of the ball. Does the shooter make the basket?\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2862\/2017\/12\/26165447\/CNX_Precalc_Figure_03_02_0082.jpg\" alt=\"Stop motioned picture of a boy throwing a basketball into a hoop to show the parabolic curve it makes.\" width=\"487\" height=\"424\" \/> (credit: modification of work by Dan Meyer)[\/caption]\r\n\r\n[reveal-answer q=\"283194\"]Solution[\/reveal-answer]\r\n[hidden-answer a=\"283194\"]\r\n\r\nThe path passes through the origin and has vertex at [latex]\\left(-4,\\text{ }7\\right)[\/latex], so [latex]\\left(h\\right)x=-\\frac{7}{16}{\\left(x+4\\right)}^{2}+7[\/latex]. To make the shot, [latex]h\\left(-7.5\\right)[\/latex] would need to be about 4 but [latex]h\\left(-7.5\\right)\\approx 1.64[\/latex]; he doesn\u2019t make it.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Objectives<\/h3>\n<ul>\n<li>(10.3.1) &#8211; Solve application problems involving quadratic functions\n<ul>\n<li>Objects in free fall<\/li>\n<li>Determining the width of a border<\/li>\n<li>Finding the maximum and minimum values of a quadratic function<\/li>\n<\/ul>\n<\/li>\n<\/ul>\n<\/div>\n<h1>(10.3.1) &#8211; Solve application problems involving quadratic functions<\/h1>\n<p>Quadratic equations are widely used in science, business, and engineering. Quadratic equations are commonly used in situations where two things are multiplied together and they both depend on the same variable. For example, when working with area, if both dimensions are written in terms of the same variable, you use a quadratic equation. Because the quantity of a product sold often depends on the price, you sometimes use a quadratic equation to represent revenue as a product of the price and the quantity sold. Quadratic equations are also used when gravity is involved, such as the path of a ball or the shape of cables in a suspension bridge.<\/p>\n<h3>Objects in free fall<\/h3>\n<p>A very common and easy-to-understand application is the height of a ball thrown at the ground off a building. Because gravity will make the ball speed up as it falls, a quadratic equation can be used to estimate its height any time before it hits the ground.<i> Note: The equation isn&#8217;t completely accurate, because friction from the air will slow the ball down a little. For our purposes, this is close enough.<\/i><\/p>\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>A ball is thrown off a building from 200 feet above the ground. Its starting velocity (also called <i>initial velocity<\/i>) is [latex]\u221210[\/latex] feet per second. (The negative value means it&#8217;s heading toward the ground.)<\/p>\n<p>The equation [latex]h=-16t^{2}-10t+200[\/latex]\u00a0can be used to model the height of the ball after [latex]t[\/latex] seconds. About how long does it take for the ball to hit the ground?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q704677\">Show Solution<\/span><\/p>\n<div id=\"q704677\" class=\"hidden-answer\" style=\"display: none\">\n<p>When the ball hits the ground, the height is 0. Substitute 0 for [latex]h[\/latex].<\/p>\n<p style=\"text-align: center\">[latex]\\begin{array}{c}h=-16t^{2}-10t+200\\\\0=-16t^{2}-10t+200\\\\-16t^{2}-10t+200=0\\end{array}[\/latex]<\/p>\n<p>This equation is difficult to solve by factoring or by completing the square, so solve it by applying the Quadratic Formula, [latex]x=\\frac{-b\\pm \\sqrt{{{b}^{2}}-4ac}}{2a}[\/latex]. In this case, the variable is [latex]t[\/latex] rather than [latex]x[\/latex]. [latex]a=\u221216,b=\u221210[\/latex], and [latex]c=200[\/latex].<\/p>\n<p style=\"text-align: center\">[latex]\\displaystyle t=\\frac{-(-10)\\pm \\sqrt{{{(-10)}^{2}}-4(-16)(200)}}{2(-16)}[\/latex]<\/p>\n<p>Simplify. Be very careful with the signs.<\/p>\n<p style=\"text-align: center\">[latex]\\large \\begin{array}{l}t=\\frac{10\\pm \\sqrt{100+12800}}{-32}\\\\\\,\\,=\\frac{10\\pm \\sqrt{12900}}{-32}\\end{array}[\/latex]<\/p>\n<p>Use a calculator to find both roots.<\/p>\n<p style=\"text-align: center\">[latex]t[\/latex] is approximately [latex]\u22123.86[\/latex] or [latex]3.24[\/latex].<\/p>\n<p>Consider the roots logically. One solution, [latex]\u22123.86[\/latex], cannot be the time because it is a negative number. The other solution, [latex]3.24[\/latex] seconds, must be when the ball hits the ground.<\/p>\n<h4>Answer<\/h4>\n<p>The ball hits the ground approximately [latex]3.24[\/latex] seconds after being thrown.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>In the next video we show another example of how the quadratic equation can be used to find the time it takes for an object in free fall to hit the ground.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-1\" title=\"Quadratic Formula Application - Time for an Object to Hit the Ground\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/RcVeuJhcuL0?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<p>Here are some more similar objects in free fall examples.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example: Applying the Vertex and [latex]<em>x[\/latex]<\/em>-Intercepts of a Parabola<\/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\">Solution<\/span><\/p>\n<div id=\"q394530\" class=\"hidden-answer\" style=\"display: none\">\na. The ball reaches the maximum height at the vertex of the parabola.<\/p>\n<p style=\"text-align: center\">[latex]\\large \\begin{array}{c} h=-\\frac{80}{2\\left(-16\\right)} \\text{ }=\\frac{80}{32}\\hfill \\\\ \\text{ }=\\frac{5}{2}\\hfill \\\\ \\text{ }=2.5\\hfill \\end{array}[\/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 <em>y\u00a0<\/em>coordinate of the vertex of the parabola.<\/p>\n<p style=\"text-align: center\">[latex]\\large \\begin{array}{c}k=H\\left(-\\frac{b}{2a}\\right)\\hfill \\\\ \\text{ }=H\\left(2.5\\right)\\hfill \\\\ \\text{ }=-16{\\left(2.5\\right)}^{2}+80\\left(2.5\\right)+40\\hfill \\\\ \\text{ }=140\\hfill \\end{array}[\/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]\\large \\begin{array}{c} t=\\frac{-80\\pm \\sqrt{{80}^{2}-4\\left(-16\\right)\\left(40\\right)}}{2\\left(-16\\right)}\\hfill \\\\ \\text{ }=\\frac{-80\\pm \\sqrt{8960}}{-32}\\hfill \\end{array}[\/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]\\large \\begin{array}{c}t=\\frac{-80-\\sqrt{8960}}{-32}\\approx 5.458\\hfill & \\text{or}\\hfill & t=\\frac{-80+\\sqrt{8960}}{-32}\\approx -0.458\\hfill \\end{array}[\/latex]<\/p>\n<p>The second answer is outside the reasonable domain of our model, so we conclude the ball will hit the ground after about 5.458 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\/2862\/2017\/12\/26165456\/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>EXAMPLE<\/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\">Solution<\/span><\/p>\n<div id=\"q174919\" class=\"hidden-answer\" style=\"display: none\">\n<p>a.\u00a03 seconds<\/p>\n<p>b.\u00a0256 feet<\/p>\n<p>c.\u00a07 seconds<\/p>\n<\/div>\n<\/div>\n<\/div>\n<h3>Applications of quadratic functions: determining the width of a border<\/h3>\n<p>The area problem below does not look like it includes a Quadratic Formula of any type, and the problem seems to be something you have solved many times before by simply multiplying. But in order to solve it, you will need to use a quadratic equation.<\/p>\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>Bob made a quilt that is 4 ft [latex]\\times[\/latex] 5 ft. He has 10 sq. ft. of fabric he can use to add a border around the quilt. How wide should he make the border to use all the fabric? (The border must be the same width on all four sides.)<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q932211\">Show Solution<\/span><\/p>\n<div id=\"q932211\" class=\"hidden-answer\" style=\"display: none\">\n<p>Sketch the problem. Since you don\u2019t know the width of the border, you will let the variable [latex]x[\/latex]\u00a0represent the width.<\/p>\n<p>In the diagram, the original quilt is indicated by the red rectangle. The border is the area between the red and blue lines.<\/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\/1468\/2016\/02\/04064559\/image052-2.gif\" alt=\"A blue rectangle. Within the blue rectangle are a pair of vertical parallel lines and a pair of horizontal parallel lines that create a smaller red rectangle. The lengths of this red rectangle are 4 feet and 5 feet. The line segments between the boundaries of the red rectangle and the bigger blue rectangle are all labeled x.\" width=\"321\" height=\"278\" \/><\/p>\n<p>Since each side of the original 4 by 5 quilt has the border of width <i>x <\/i>added, the length of the quilt with the border will be [latex]5+2x[\/latex],\u00a0and the width will be\u00a0[latex]4+2x[\/latex].<\/p>\n<p>(Both dimensions are written in terms of the same variable, and you will multiply them to get an area! This is where you might start to think that a quadratic equation might be used to solve this problem.)<\/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\/1468\/2016\/02\/04064600\/image053-2.gif\" alt=\"A blue rectangle with one side a height of 4+2x and another side a length of 5+2x. Within the blue rectangle are a pair of vertical parallel lines and a pair of horizontal parallel lines that create a smaller red rectangle. The height of this red rectangle is 4 feet and the length is 5 feet. The line segments between the boundaries of the red rectangle and the bigger blue rectangle are all labeled x.\" width=\"330\" height=\"285\" \/><\/p>\n<p>You are only interested in the area of the border strips. Write an expression for the area of the border.<\/p>\n<p style=\"text-align: center\">Area of border = Area of the blue rectangle minus the area of the red rectangle<\/p>\n<p style=\"text-align: center\">Area of border[latex]=\\left(4+2x\\right)\\left(5+2x\\right)\u2013\\left(4\\right)\\left(5\\right)[\/latex]<\/p>\n<p>There are 10 sq ft of fabric for the border, so set the area of border to be 10.<\/p>\n<p style=\"text-align: center\">[latex]10=\\left(4+2x\\right)\\left(5+2x\\right)\u201320[\/latex]<\/p>\n<p>Multiply [latex]\\left(4+2x\\right)\\left(5+2x\\right)[\/latex].<\/p>\n<p style=\"text-align: center\">[latex]10=20+8x+10x+4x^{2}\u201320[\/latex]<\/p>\n<p>Simplify.<\/p>\n<p style=\"text-align: center\">[latex]10=18x+4x^{2}[\/latex]<\/p>\n<p>Subtract 10 from both sides so that you have a quadratic equation in standard form and can apply the Quadratic Formula to find the roots of the equation.<\/p>\n<p style=\"text-align: center\">[latex]\\begin{array}{c}0=18x+4x^{2}-10\\\\\\\\\\text{or}\\\\\\\\4x^{2}-10\\\\\\\\2\\left(2x^{2}+9x-5\\right)=0\\end{array}[\/latex]<\/p>\n<p>Factor out the greatest common factor, 2, so that you can work with the simpler equivalent equation, [latex]2x^{2}+9x\u20135=0[\/latex].<\/p>\n<p style=\"text-align: center\">[latex]\\large \\begin{array}{r}2\\left(2x^{2}+9x-5\\right)=0\\\\\\\\\\frac{2\\left(2x^{2}+9x-5\\right)}{2}=\\frac{0}{2}\\\\\\\\2x^{2}+9x-5=0\\end{array}[\/latex]<\/p>\n<p>Use the Quadratic Formula. In this case, [latex]a=2,b=9[\/latex], and [latex]c=\u22125[\/latex].<\/p>\n<p style=\"text-align: center\">[latex]\\large \\begin{array}{l}x=\\frac{-b\\pm \\sqrt{{{b}^{2}}-4ac}}{2a}\\\\\\\\x=\\frac{-9\\pm \\sqrt{{{9}^{2}}-4(2)(-5)}}{2(2)}\\end{array}[\/latex]<\/p>\n<p>Simplify.<\/p>\n<p style=\"text-align: center\">[latex]\\displaystyle x=\\frac{-9\\pm \\sqrt{121}}{4}=\\frac{-9\\pm 11}{4}[\/latex]<\/p>\n<p>Find the solutions, making sure that the [latex]\\pm[\/latex] is evaluated for both values.<\/p>\n<p style=\"text-align: center\">[latex]\\large \\begin{array}{c}x=\\frac{-9+11}{4}=\\frac{2}{4}=\\frac{1}{2}=0.5\\\\\\\\\\text{or}\\\\\\\\x=\\frac{-9-11}{4}=\\frac{-20}{4}=-5\\end{array}[\/latex]<\/p>\n<p>Ignore the solution [latex]x=\u22125[\/latex], since the width could not be negative.<\/p>\n<h4>Answer<\/h4>\n<p>The width of the border should be 0.5 ft.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>Here is a video which gives another example of using the quadratic formula for a geometry problem involving the border around a quilt.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-2\" title=\"Quadratic Formula Application - Determine the Width of a Border\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/Zxe-SdwutxA?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<h3>Finding the maximum and minimum values of a quadratic function<\/h3>\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\/2862\/2017\/12\/26165500\/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\" \/><a href=\"https:\/\/s3-us-west-2.amazonaws.com\/oerfiles\/College+Algebra\/calculator.html\" target=\"_blank\" rel=\"noopener\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-3370\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2862\/2017\/12\/26165501\/calculator.png\" alt=\"\" width=\"251\" height=\"46\" \/><\/a><\/p>\n<div class=\"textbox exercises\">\n<h3>ExAMPLE<\/h3>\n<p>Find two numbers [latex]x[\/latex] and [latex]y[\/latex] whose difference is 100 and whose product is a minimum.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q42108\">Show Answer<\/span><\/p>\n<div id=\"q42108\" class=\"hidden-answer\" style=\"display: none\">\n<p>We are trying to find the minimum of the product [latex]P=xy[\/latex] of two numbers, such that their difference is 100: [latex]y-x=100[\/latex]. First, we rewrite one variable in terms of the other:<\/p>\n<p style=\"text-align: center\">[latex]y-x=100 \\rightarrow y=100+x[\/latex]<\/p>\n<p>Next, we plug in the above relationship between the variables into the first equation:<\/p>\n<p style=\"text-align: center\">[latex]P=xy=x(100+x) = 100x+x^2 = x^2+100x[\/latex]<\/p>\n<p>As a result, we get a quadratic function [latex]P(x)=x^2+100x[\/latex]. The graph of this quadratic function opens upwards, and its vertex is the minimum, So if we find the vertex of this parabola, we will find the minimum product. The vertex is:<\/p>\n<p style=\"text-align: center\">[latex]\\displaystyle \\displaystyle \\left(-\\frac{b}{2a}, P\\left(-\\frac{b}{2a}\\right)\\right) = \\left(-\\frac{(100)}{2(1)}, P\\left(-\\frac{(100)}{2(1)}\\right)\\right) = (-50,-2,500)[\/latex]<\/p>\n<p>Thus the minimum of the parabola occurs at [latex]x=-50[\/latex], and is [latex]-2,500[\/latex]. So one of the numbers is [latex]x=-50[\/latex], the other we obtain by plugging in [latex]x=-50[\/latex] in to [latex]y-x=100[\/latex]:<\/p>\n<p style=\"text-align: center\">[latex]\\begin{array}{cc}y-(-50)&=&100 \\\\ y+50 &=& 100 \\\\ y &=& 50\\end{array}[\/latex]<\/p>\n<p><strong>Answer<\/strong><\/p>\n<p>[latex]x=-50[\/latex] and [latex]y=50[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\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\">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\/2862\/2017\/12\/26165503\/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, <em>W<\/em>, to represent the width of the garden and the length of the fence section parallel to the backyard fence.<\/p>\n<p>1)\u00a0 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].<\/p>\n<p style=\"text-align: center\">[latex]W=80 - 2L[\/latex]<\/p>\n<p>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<p style=\"text-align: center\">[latex]\\begin{array}{l}\\text{ }A&=&LW=L\\left(80 - 2L\\right)\\hfill \\\\ A\\left(L\\right)&=&80L - 2{L}^{2}\\hfill \\end{array}[\/latex]<\/p>\n<p>This formula represents the area of the fence in terms of the variable length [latex]L[\/latex]. The function, written in general form, is<\/p>\n<p style=\"text-align: center\">[latex]A\\left(L\\right)=-2{L}^{2}+80L[\/latex].<\/p>\n<p>2) 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].<\/p>\n<p>To find the vertex:<\/p>\n<p style=\"text-align: center\">[latex]\\large \\begin{array}{l}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{array}[\/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\/2862\/2017\/12\/26165506\/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<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\">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]\\large \\begin{array}{c}m=\\frac{79,000 - 84,000}{32 - 30}\\hfill \\\\ \\text{ }=\\frac{-5,000}{2}\\hfill \\\\ \\text{ }=-2,500\\hfill \\end{array}[\/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 <em>y<\/em>-intercept.<\/p>\n<p style=\"text-align: center\">[latex]\\begin{array}{c}\\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{array}[\/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{array}{c}\\text{Revenue}=pQ\\hfill \\\\ \\text{Revenue}=p\\left(-2,500p+159,000\\right)\\hfill \\\\ \\text{Revenue}=-2,500{p}^{2}+159,000p\\hfill \\end{array}[\/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]\\large \\begin{array}{c}h=-\\frac{159,000}{2\\left(-2,500\\right)}\\hfill \\\\ \\text{ }=31.8\\hfill \\end{array}[\/latex]<\/p>\n<p>The model tells us that the maximum revenue will occur if the newspaper charges $31.80 for a subscription. To find what the maximum revenue is, we evaluate the revenue function.<\/p>\n<p style=\"text-align: center\">[latex]\\begin{array}{c}\\text{maximum revenue}&=&-2,500{\\left(31.8\\right)}^{2}+159,000\\left(31.8\\right)\\hfill \\\\ \\text{ }&=&2,528,100\\hfill \\end{array}[\/latex]<\/p>\n<h4>Analysis of the Solution<\/h4>\n<p>This could also be solved by graphing the quadratic. We can see the maximum revenue on a graph of the quadratic function.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2862\/2017\/12\/26165508\/CNX_Precalc_Figure_03_02_0122.jpg\" alt=\"Graph of the parabolic function which the x-axis is labeled Price (p) and the y-axis is labeled Revenue ($). The vertex is at (31.80, 258100).\" width=\"487\" height=\"327\" \/><\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>A coordinate grid has been superimposed over the quadratic path of a basketball in the picture below. Find an equation for the path of the ball. Does the shooter make the basket?<\/p>\n<div style=\"width: 497px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2862\/2017\/12\/26165447\/CNX_Precalc_Figure_03_02_0082.jpg\" alt=\"Stop motioned picture of a boy throwing a basketball into a hoop to show the parabolic curve it makes.\" width=\"487\" height=\"424\" \/><\/p>\n<p class=\"wp-caption-text\">(credit: modification of work by Dan Meyer)<\/p>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q283194\">Solution<\/span><\/p>\n<div id=\"q283194\" class=\"hidden-answer\" style=\"display: none\">\n<p>The path passes through the origin and has vertex at [latex]\\left(-4,\\text{ }7\\right)[\/latex], so [latex]\\left(h\\right)x=-\\frac{7}{16}{\\left(x+4\\right)}^{2}+7[\/latex]. To make the shot, [latex]h\\left(-7.5\\right)[\/latex] would need to be about 4 but [latex]h\\left(-7.5\\right)\\approx 1.64[\/latex]; he doesn\u2019t make it.<\/p>\n<\/div>\n<\/div>\n<\/div>\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-5065\">\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>Quadratic Formula Application - Time for an Object to Hit the Ground. <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com) for Lumen Learning. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/RcVeuJhcuL0\">https:\/\/youtu.be\/RcVeuJhcuL0<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Quadratic Formula Application - Determine the Width of a Border. <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com) for Lumen Learning. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/Zxe-SdwutxA\">https:\/\/youtu.be\/Zxe-SdwutxA<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/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@3.278:1\/Preface\">http:\/\/cnx.org\/contents\/9b08c294-057f-4201-9f48-5d6ad992740d@3.278:1\/Preface<\/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@3.278:1\/Preface<\/li><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><\/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":60342,"menu_order":4,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc\",\"description\":\"College Algebra\",\"author\":\"Abramson, Jay, et al.\",\"organization\":\"OpenStax\",\"url\":\"http:\/\/cnx.org\/contents\/9b08c294-057f-4201-9f48-5d6ad992740d@3.278:1\/Preface\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"Download for free at : http:\/\/cnx.org\/contents\/9b08c294-057f-4201-9f48-5d6ad992740d@3.278:1\/Preface\"},{\"type\":\"cc\",\"description\":\"Revision and Adaptation\",\"author\":\"\",\"organization\":\"Lumen Learning\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"original\",\"description\":\"Quadratic Formula Application - Time for an Object to Hit the Ground\",\"author\":\"James Sousa (Mathispower4u.com) for Lumen Learning\",\"organization\":\"\",\"url\":\"https:\/\/youtu.be\/RcVeuJhcuL0\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"original\",\"description\":\"Quadratic Formula Application - Determine the Width of a Border\",\"author\":\"James Sousa (Mathispower4u.com) for Lumen Learning\",\"organization\":\"\",\"url\":\"https:\/\/youtu.be\/Zxe-SdwutxA\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"}]","CANDELA_OUTCOMES_GUID":"","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-5065","chapter","type-chapter","status-web-only","hentry"],"part":1787,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/cuny-hunter-collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/5065","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/cuny-hunter-collegealgebra\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/cuny-hunter-collegealgebra\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/cuny-hunter-collegealgebra\/wp-json\/wp\/v2\/users\/60342"}],"version-history":[{"count":21,"href":"https:\/\/courses.lumenlearning.com\/cuny-hunter-collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/5065\/revisions"}],"predecessor-version":[{"id":5741,"href":"https:\/\/courses.lumenlearning.com\/cuny-hunter-collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/5065\/revisions\/5741"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/cuny-hunter-collegealgebra\/wp-json\/pressbooks\/v2\/parts\/1787"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/cuny-hunter-collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/5065\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/cuny-hunter-collegealgebra\/wp-json\/wp\/v2\/media?parent=5065"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/cuny-hunter-collegealgebra\/wp-json\/pressbooks\/v2\/chapter-type?post=5065"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/cuny-hunter-collegealgebra\/wp-json\/wp\/v2\/contributor?post=5065"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/cuny-hunter-collegealgebra\/wp-json\/wp\/v2\/license?post=5065"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}