{"id":16262,"date":"2019-10-02T20:23:08","date_gmt":"2019-10-02T20:23:08","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/wm-developmentalemporium\/chapter\/read-applications-of-adding-and-subtracting-polynomials\/"},"modified":"2020-10-22T09:30:23","modified_gmt":"2020-10-22T09:30:23","slug":"read-applications-of-adding-and-subtracting-polynomials","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/suny-rockland-developmentalemporium\/chapter\/read-applications-of-adding-and-subtracting-polynomials\/","title":{"raw":"12.4.a - Polynomials Involving Perimeter, Area, and Volume","rendered":"12.4.a &#8211; Polynomials Involving Perimeter, Area, and Volume"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Write polynomials involving perimeter,\u00a0area, and\u00a0volume<\/li>\r\n<\/ul>\r\n<\/div>\r\nIn this section we will explore ways that polynomials are used in applications of perimeter, area, and volume. First, we will see how a polynomial can be used to describe the perimeter of a rectangle.\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example<\/h3>\r\nA rectangular garden has one side with a length of [latex]x+7[\/latex] and another with a length [latex]2x + 3[\/latex]. Find the perimeter of the garden.<b> <\/b>\r\n\r\n<b><img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1468\/2016\/02\/04064121\/image004.jpg\" alt=\"Rectangle with height x+7 and length 2x+3.\" width=\"199\" height=\"112\" \/><\/b>\r\n\r\n[reveal-answer q=\"627193\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"627193\"]The perimeter of a rectangle is the sum of its side lengths.\r\n<p style=\"text-align: center\">[latex]\\left(x+7\\right)+\\left(2x+3\\right)+\\left(x+7\\right)+\\left(2x+3\\right)[\/latex]<\/p>\r\nRegroup by like terms.\r\n<p style=\"text-align: center\">[latex]\\left(x+2x+x+2x\\right)+\\left(7+3+7+3\\right)[\/latex]<\/p>\r\nAdd like terms.\r\n<p style=\"text-align: center\">[latex]6x+20[\/latex]<\/p>\r\n\r\n<h4>Answer<\/h4>\r\nThe perimeter is [latex]6x+20[\/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[ohm_question]48356[\/ohm_question]\r\n\r\n<\/div>\r\nIn the following video you are shown how to find the perimeter of a triangle whose sides are defined as polynomials.\r\n\r\nhttps:\/\/youtu.be\/BhRpZv0_0jE\r\n\r\nThe area of a circle can be found using the radius of the circle and the constant pi in the formula [latex]A=\\pi{r^2}[\/latex]. In the next example we will use this formula to find a polynomial that describes the area of an irregular shape.\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nFind a polynomial for the area of the shaded region of the figure.\r\n\r\n<img class=\"size-medium wp-image-4636 aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/117\/2016\/06\/06212731\/Screen-Shot-2016-06-06-at-2.18.00-PM-300x298.png\" alt=\"circle with middle extracted to form a ring shape. Inner radius labeled as r=3, outer radius labeled as R= r.\" width=\"300\" height=\"298\" \/>\r\n[reveal-answer q=\"667228\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"667228\"]\r\n\r\n<strong>Read and Understand:\u00a0<\/strong>We need to describe the area of the shaded region of this shape using polynomials. \u00a0We know the formula for the area of a circle is\u00a0[latex]A=\\pi{r^2}[\/latex]. \u00a0The figure we are working with is a circle with a smaller circle cut out.\r\n\r\n<strong>Define and Translate:\u00a0<\/strong>The larger circle has radius = r, and the smaller circle has radius= [latex]3[\/latex]. \u00a0If we find the area of the larger circle, then subtract the area of the smaller circle, the remaining area will be the shaded region. First define the area of the larger circle:\r\n<p style=\"text-align: center\">[latex]A_{1}=\\pi{r^2}[\/latex]<\/p>\r\n<p style=\"text-align: left\">Then define the area of the smaller circle.<\/p>\r\n<p style=\"text-align: center\">[latex]A_{2}=\\pi{3^2}=9\\pi[\/latex]<\/p>\r\n<p style=\"text-align: left\"><strong>Write and Solve:\u00a0<\/strong>The shaded region can be found by subtracting the smaller area from the larger area.<\/p>\r\n<p style=\"text-align: center\">[latex]\\begin{array}{c}A_{1}-A_{2}\\\\=\\pi{r^2}-9\\pi\\end{array}[\/latex]<\/p>\r\n<p style=\"text-align: left\">The area of the shaded region is [latex]\\pi{r^2}-9\\pi[\/latex]<\/p>\r\n\r\n<h4 style=\"text-align: left\">Answer<\/h4>\r\n[latex]\\pi{r^2}-9\\pi[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nIn the video that follows, you will be shown an example of determining the area of a rectangle whose sides are defined as polynomials.\r\nhttps:\/\/youtu.be\/5Q2htATOIik\r\n\r\n&nbsp;\r\n\r\n[caption id=\"attachment_4638\" align=\"alignleft\" width=\"208\"]<img class=\" wp-image-4638\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/117\/2016\/06\/06215403\/Screen-Shot-2016-06-06-at-2.53.40-PM-300x236.png\" alt=\"pi\" width=\"208\" height=\"164\" \/> Pi[\/caption]\r\n<h3>A note about pi.<\/h3>\r\nIt is easy to confuse pi as a variable because we use a greek letter to represent it. \u00a0We use a greek letter instead of a number because nobody has been able to find an end to the number of digits of pi. \u00a0To be precise and thorough, we use the greek letter as a way to say: \"we are including all the digits of pi without having to write them\". The expression for the area of the shaded region in the example above included both the variable r, which represented an unknown radius and the number pi. \u00a0If we needed to use this expression to build a physical object or instruct a machine to cut specific dimensions, we would round pi to an appropriate number of decimal places.\r\n\r\n&nbsp;\r\n\r\nIn the next example, we will write the area for a rectangle in two different ways, one as the product of two binomials and the other as the sum of four rectangles. Because we are describing the same shape two different ways, we should end up with the same expression no matter what way we define the area.\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nWrite two different polynomials that describe the area of of the figure. For one expression, think of the rectangle as one large figure, and for the other expression, think of the rectangle as the sum of [latex]4[\/latex] different rectangles.\r\n\r\n<img class=\"alignnone size-medium wp-image-4640 aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/117\/2016\/06\/06225159\/Screen-Shot-2016-06-06-at-3.51.10-PM-300x234.png\" alt=\"Rectangle with side length y+9 and y+7\" width=\"300\" height=\"234\" \/>\r\n[reveal-answer q=\"15093\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"15093\"]\r\n\r\nFirst, we will define the polynomial that describes the area of the rectangle as one figure.\r\n\r\n<strong>Read and Understand:\u00a0<\/strong>We are tasked with writing an\u00a0expressions for the area of the figure above. The area of a rectangle is given as [latex]A=lw[\/latex]. \u00a0We need to consider the whole figure in our dimensions.\r\n\r\n<strong>Define and Translate:<\/strong>\u00a0Use the formula for area:\u00a0\u00a0[latex]A=lw[\/latex], the sides of the figure are the sum of the defined sides. \u00a0[latex]\\begin{array}{c}l=\\left(y+7\\right)\\\\w=\\left(y+9\\right)\\end{array}[\/latex]\r\n\r\nYou could define\u00a0[latex]\\begin{array}{c}w=\\left(y+7\\right)\\\\l=\\left(y+9\\right)\\end{array}[\/latex] because it doesn't matter in which order you multiply.\r\n\r\n&nbsp;\r\n\r\n<strong>Write and Solve:\u00a0<\/strong>\r\n\r\n[latex]\\begin{array}{c}A=lw\\\\l=\\left(y+7\\right)\\\\w=\\left(y+9\\right)\\end{array}[\/latex]\r\n\r\nWe can use either method we learned to multiply binomials to simplify this expression, we will use a table.\r\n<table class=\" aligncenter\" style=\"width: 20%\">\r\n<tbody>\r\n<tr>\r\n<td><\/td>\r\n<td>[latex]y[\/latex]<\/td>\r\n<td>[latex]+7[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]y[\/latex]<\/td>\r\n<td>[latex]y^2[\/latex]<\/td>\r\n<td>[latex]7y[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]+9[\/latex]<\/td>\r\n<td>[latex]9y[\/latex]<\/td>\r\n<td>[latex]63[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<p style=\"text-align: left\">Sum the terms from the table, and simplify:<\/p>\r\n<p style=\"text-align: center\">[latex]\\begin{array}{c}\\left(y+7\\right)\\left(y+9\\right)\\\\=y^2+7y+9y+63\\\\=y^2+16y+63\\end{array}[\/latex]<\/p>\r\n\r\n<h4 style=\"text-align: left\">Answer 1<\/h4>\r\n[latex]A=y^2+16y+63[\/latex]\r\n\r\nNow we will find an expression for the area of the whole figure as comprised by the areas of the four rectangles added together.\r\n\r\n<strong>Read and Understand:\u00a0<\/strong>\u00a0The area of a rectangle is given as [latex]A=lw[\/latex]. \u00a0We need to first define the areas of each rectangle, then sum them all together to get the area of the whole figure. It helps to label the four rectangle in the figure so you can keep the dimensions organized.\r\n\r\n<img class=\"alignnone size-medium wp-image-4641 aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/117\/2016\/06\/07004311\/Screen-Shot-2016-06-06-at-5.42.17-PM-300x229.png\" alt=\"Rectangle made from four rectangles labeled 1, 2, 3, 4.\" width=\"300\" height=\"229\" \/>\r\n\r\n<strong>Define and Translate:<\/strong>\u00a0Use the formula for area:\u00a0\u00a0[latex]A=lw[\/latex] for each rectangle:\r\n<p style=\"text-align: center\">[latex]A_{1}=7\\cdot{y}[\/latex]<\/p>\r\n<p style=\"text-align: center\">[latex]A_{2}=7\\cdot{9}=63[\/latex]<\/p>\r\n<p style=\"text-align: center\">[latex]A_{3}=y\\cdot{y}=y^2[\/latex]<\/p>\r\n<p style=\"text-align: center\">[latex]A_{4}=y\\cdot{9}[\/latex]<\/p>\r\n<strong>Write and Solve:\u00a0<\/strong>\r\n<p style=\"text-align: center\">[latex]\\begin{array}{c}A=A_{1}+A_{2}+A_{3}+A_{4}\\\\=7y+63+y^2+9y\\\\\\text{ reorganize and simplify }\\\\=y^2+16y+63\\end{array}[\/latex]<\/p>\r\n\r\n<h4 style=\"text-align: left\">Answer 2<\/h4>\r\n[latex]A=y^2+16y+63[\/latex]\r\n\r\nHopefully, it isn't surprising that both expressions simplify to the same thing.\r\n<p style=\"text-align: center\">[\/hidden-answer]<\/p>\r\n\r\n<\/div>\r\nThe last example we will provide in this section is one for volume. \u00a0The volume of regular solids such as spheres, cylinders, cones and rectangular prisms are known. \u00a0We will find an expression for the volume of a cylinder, which is defined as [latex]V=\\pi{r^2}h[\/latex].\r\n\r\n&nbsp;\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nDefine a polynomial that describes the volume of the cylinder shown in the figure:\r\n\r\n<img class=\"alignnone size-medium wp-image-4643\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/117\/2016\/06\/07010208\/Screen-Shot-2016-06-06-at-6.01.33-PM-232x300.png\" alt=\"Cylinder with height = 7 and radius = (t-2)\" width=\"232\" height=\"300\" \/>\r\n[reveal-answer q=\"137608\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"137608\"]\r\n\r\n<strong>Read and Understand:\u00a0<\/strong> We are tasked with writing an\u00a0expressions for the volume\u00a0of the cylinder in the figure\u00a0above. The volume of a cylinder\u00a0is given as[latex]V=\\pi{r^2}h[\/latex], where [latex]\\pi[\/latex] is a constant, and r is the radius and h is the height of the cylinder.\r\n\r\n<strong>Define and Translate:<\/strong>\u00a0Use the formula for volume:[latex]V=\\pi{r^2}h[\/latex], \u00a0we need to define r and h.\r\n\r\n[latex]\\begin{array}{c}r=\\left(t-2\\right)\\\\h=7\\end{array}[\/latex]\r\n\r\n<strong>Write and Solve:\u00a0<\/strong>Substitute r and h into the formula for volume.\r\n<p style=\"text-align: center\">[latex]\\begin{array}{c}V=\\pi{r^2}h\\\\=\\pi\\left(t-2\\right)^2\\cdot{7}\\\\=7\\pi\\left({t^2}-2t-2t+4\\right)\\\\=7\\pi\\left({t^2}-4t+4\\right)\\end{array}[\/latex]<\/p>\r\n<p style=\"text-align: left\">Note that we usually write other constants that are multiplied by [latex]\\pi[\/latex] in front of it. \u00a0We can now distribute [latex]7\\pi[\/latex] to each term in the polynomial.<\/p>\r\n<p style=\"text-align: center\">[latex]\\begin{array}{c}7\\pi\\left({t^2}-4t+4\\right)\\\\=\\left(7\\pi\\right){t^2}-\\left(7\\pi\\right)\\cdot{4t}+\\left(7\\pi\\right)\\cdot{4}\\\\=7\\pi{t^2}-28\\pi{t}+28\\pi\\end{array}[\/latex].<\/p>\r\n<p style=\"text-align: left\">Note again how we left [latex]\\pi[\/latex] as a greek letter. \u00a0If we needed to use this calculation for measurement of materials, we would round pi, or a computer would round for us.<\/p>\r\n\r\n<h4 style=\"text-align: left\">Answer<\/h4>\r\n[latex]V=\\pi{r^2}h=7\\pi{t^2}-28\\pi{t}+28\\pi[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nIn this last video, we present another example of finding the volume of a cylinder whose dimensions include polynomials.\r\n\r\nhttps:\/\/youtu.be\/g-g_nSsfGs4\r\n\r\n&nbsp;\r\n\r\nIn this section we defined polynomials that represent perimeter, area and volume of well-known shapes. \u00a0We also introduced some convention about how to use and write [latex]\\pi[\/latex] when it is combined with other constants and variables. The next application will introduce you to cost and revenue polynomials.\u00a0 Next we will see that cost and revenue equations can be polynomials.","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Write polynomials involving perimeter,\u00a0area, and\u00a0volume<\/li>\n<\/ul>\n<\/div>\n<p>In this section we will explore ways that polynomials are used in applications of perimeter, area, and volume. First, we will see how a polynomial can be used to describe the perimeter of a rectangle.<\/p>\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>A rectangular garden has one side with a length of [latex]x+7[\/latex] and another with a length [latex]2x + 3[\/latex]. Find the perimeter of the garden.<b> <\/b><\/p>\n<p><b><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\/04064121\/image004.jpg\" alt=\"Rectangle with height x+7 and length 2x+3.\" width=\"199\" height=\"112\" \/><\/b><\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q627193\">Show Solution<\/span><\/p>\n<div id=\"q627193\" class=\"hidden-answer\" style=\"display: none\">The perimeter of a rectangle is the sum of its side lengths.<\/p>\n<p style=\"text-align: center\">[latex]\\left(x+7\\right)+\\left(2x+3\\right)+\\left(x+7\\right)+\\left(2x+3\\right)[\/latex]<\/p>\n<p>Regroup by like terms.<\/p>\n<p style=\"text-align: center\">[latex]\\left(x+2x+x+2x\\right)+\\left(7+3+7+3\\right)[\/latex]<\/p>\n<p>Add like terms.<\/p>\n<p style=\"text-align: center\">[latex]6x+20[\/latex]<\/p>\n<h4>Answer<\/h4>\n<p>The perimeter is [latex]6x+20[\/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=\"ohm48356\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=48356&theme=oea&iframe_resize_id=ohm48356&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<p>In the following video you are shown how to find the perimeter of a triangle whose sides are defined as polynomials.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-1\" title=\"Ex: Polynomial Addition Application - Perimeter\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/BhRpZv0_0jE?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<p>The area of a circle can be found using the radius of the circle and the constant pi in the formula [latex]A=\\pi{r^2}[\/latex]. In the next example we will use this formula to find a polynomial that describes the area of an irregular shape.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Find a polynomial for the area of the shaded region of the figure.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"size-medium wp-image-4636 aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/117\/2016\/06\/06212731\/Screen-Shot-2016-06-06-at-2.18.00-PM-300x298.png\" alt=\"circle with middle extracted to form a ring shape. Inner radius labeled as r=3, outer radius labeled as R= r.\" width=\"300\" height=\"298\" \/><\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q667228\">Show Solution<\/span><\/p>\n<div id=\"q667228\" class=\"hidden-answer\" style=\"display: none\">\n<p><strong>Read and Understand:\u00a0<\/strong>We need to describe the area of the shaded region of this shape using polynomials. \u00a0We know the formula for the area of a circle is\u00a0[latex]A=\\pi{r^2}[\/latex]. \u00a0The figure we are working with is a circle with a smaller circle cut out.<\/p>\n<p><strong>Define and Translate:\u00a0<\/strong>The larger circle has radius = r, and the smaller circle has radius= [latex]3[\/latex]. \u00a0If we find the area of the larger circle, then subtract the area of the smaller circle, the remaining area will be the shaded region. First define the area of the larger circle:<\/p>\n<p style=\"text-align: center\">[latex]A_{1}=\\pi{r^2}[\/latex]<\/p>\n<p style=\"text-align: left\">Then define the area of the smaller circle.<\/p>\n<p style=\"text-align: center\">[latex]A_{2}=\\pi{3^2}=9\\pi[\/latex]<\/p>\n<p style=\"text-align: left\"><strong>Write and Solve:\u00a0<\/strong>The shaded region can be found by subtracting the smaller area from the larger area.<\/p>\n<p style=\"text-align: center\">[latex]\\begin{array}{c}A_{1}-A_{2}\\\\=\\pi{r^2}-9\\pi\\end{array}[\/latex]<\/p>\n<p style=\"text-align: left\">The area of the shaded region is [latex]\\pi{r^2}-9\\pi[\/latex]<\/p>\n<h4 style=\"text-align: left\">Answer<\/h4>\n<p>[latex]\\pi{r^2}-9\\pi[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>In the video that follows, you will be shown an example of determining the area of a rectangle whose sides are defined as polynomials.<br \/>\n<iframe loading=\"lazy\" id=\"oembed-2\" title=\"Ex:  Find the Area of a Rectangle Using a Polynomial\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/5Q2htATOIik?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<p>&nbsp;<\/p>\n<div id=\"attachment_4638\" style=\"width: 218px\" class=\"wp-caption alignleft\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-4638\" class=\"wp-image-4638\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/117\/2016\/06\/06215403\/Screen-Shot-2016-06-06-at-2.53.40-PM-300x236.png\" alt=\"pi\" width=\"208\" height=\"164\" \/><\/p>\n<p id=\"caption-attachment-4638\" class=\"wp-caption-text\">Pi<\/p>\n<\/div>\n<h3>A note about pi.<\/h3>\n<p>It is easy to confuse pi as a variable because we use a greek letter to represent it. \u00a0We use a greek letter instead of a number because nobody has been able to find an end to the number of digits of pi. \u00a0To be precise and thorough, we use the greek letter as a way to say: &#8220;we are including all the digits of pi without having to write them&#8221;. The expression for the area of the shaded region in the example above included both the variable r, which represented an unknown radius and the number pi. \u00a0If we needed to use this expression to build a physical object or instruct a machine to cut specific dimensions, we would round pi to an appropriate number of decimal places.<\/p>\n<p>&nbsp;<\/p>\n<p>In the next example, we will write the area for a rectangle in two different ways, one as the product of two binomials and the other as the sum of four rectangles. Because we are describing the same shape two different ways, we should end up with the same expression no matter what way we define the area.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Write two different polynomials that describe the area of of the figure. For one expression, think of the rectangle as one large figure, and for the other expression, think of the rectangle as the sum of [latex]4[\/latex] different rectangles.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-medium wp-image-4640 aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/117\/2016\/06\/06225159\/Screen-Shot-2016-06-06-at-3.51.10-PM-300x234.png\" alt=\"Rectangle with side length y+9 and y+7\" width=\"300\" height=\"234\" \/><\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q15093\">Show Solution<\/span><\/p>\n<div id=\"q15093\" class=\"hidden-answer\" style=\"display: none\">\n<p>First, we will define the polynomial that describes the area of the rectangle as one figure.<\/p>\n<p><strong>Read and Understand:\u00a0<\/strong>We are tasked with writing an\u00a0expressions for the area of the figure above. The area of a rectangle is given as [latex]A=lw[\/latex]. \u00a0We need to consider the whole figure in our dimensions.<\/p>\n<p><strong>Define and Translate:<\/strong>\u00a0Use the formula for area:\u00a0\u00a0[latex]A=lw[\/latex], the sides of the figure are the sum of the defined sides. \u00a0[latex]\\begin{array}{c}l=\\left(y+7\\right)\\\\w=\\left(y+9\\right)\\end{array}[\/latex]<\/p>\n<p>You could define\u00a0[latex]\\begin{array}{c}w=\\left(y+7\\right)\\\\l=\\left(y+9\\right)\\end{array}[\/latex] because it doesn&#8217;t matter in which order you multiply.<\/p>\n<p>&nbsp;<\/p>\n<p><strong>Write and Solve:\u00a0<\/strong><\/p>\n<p>[latex]\\begin{array}{c}A=lw\\\\l=\\left(y+7\\right)\\\\w=\\left(y+9\\right)\\end{array}[\/latex]<\/p>\n<p>We can use either method we learned to multiply binomials to simplify this expression, we will use a table.<\/p>\n<table class=\"aligncenter\" style=\"width: 20%\">\n<tbody>\n<tr>\n<td><\/td>\n<td>[latex]y[\/latex]<\/td>\n<td>[latex]+7[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]y[\/latex]<\/td>\n<td>[latex]y^2[\/latex]<\/td>\n<td>[latex]7y[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]+9[\/latex]<\/td>\n<td>[latex]9y[\/latex]<\/td>\n<td>[latex]63[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p style=\"text-align: left\">Sum the terms from the table, and simplify:<\/p>\n<p style=\"text-align: center\">[latex]\\begin{array}{c}\\left(y+7\\right)\\left(y+9\\right)\\\\=y^2+7y+9y+63\\\\=y^2+16y+63\\end{array}[\/latex]<\/p>\n<h4 style=\"text-align: left\">Answer 1<\/h4>\n<p>[latex]A=y^2+16y+63[\/latex]<\/p>\n<p>Now we will find an expression for the area of the whole figure as comprised by the areas of the four rectangles added together.<\/p>\n<p><strong>Read and Understand:\u00a0<\/strong>\u00a0The area of a rectangle is given as [latex]A=lw[\/latex]. \u00a0We need to first define the areas of each rectangle, then sum them all together to get the area of the whole figure. It helps to label the four rectangle in the figure so you can keep the dimensions organized.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-medium wp-image-4641 aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/117\/2016\/06\/07004311\/Screen-Shot-2016-06-06-at-5.42.17-PM-300x229.png\" alt=\"Rectangle made from four rectangles labeled 1, 2, 3, 4.\" width=\"300\" height=\"229\" \/><\/p>\n<p><strong>Define and Translate:<\/strong>\u00a0Use the formula for area:\u00a0\u00a0[latex]A=lw[\/latex] for each rectangle:<\/p>\n<p style=\"text-align: center\">[latex]A_{1}=7\\cdot{y}[\/latex]<\/p>\n<p style=\"text-align: center\">[latex]A_{2}=7\\cdot{9}=63[\/latex]<\/p>\n<p style=\"text-align: center\">[latex]A_{3}=y\\cdot{y}=y^2[\/latex]<\/p>\n<p style=\"text-align: center\">[latex]A_{4}=y\\cdot{9}[\/latex]<\/p>\n<p><strong>Write and Solve:\u00a0<\/strong><\/p>\n<p style=\"text-align: center\">[latex]\\begin{array}{c}A=A_{1}+A_{2}+A_{3}+A_{4}\\\\=7y+63+y^2+9y\\\\\\text{ reorganize and simplify }\\\\=y^2+16y+63\\end{array}[\/latex]<\/p>\n<h4 style=\"text-align: left\">Answer 2<\/h4>\n<p>[latex]A=y^2+16y+63[\/latex]<\/p>\n<p>Hopefully, it isn&#8217;t surprising that both expressions simplify to the same thing.<\/p>\n<p style=\"text-align: center\"><\/div>\n<\/div>\n<\/div>\n<p>The last example we will provide in this section is one for volume. \u00a0The volume of regular solids such as spheres, cylinders, cones and rectangular prisms are known. \u00a0We will find an expression for the volume of a cylinder, which is defined as [latex]V=\\pi{r^2}h[\/latex].<\/p>\n<p>&nbsp;<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Define a polynomial that describes the volume of the cylinder shown in the figure:<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-medium wp-image-4643\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/117\/2016\/06\/07010208\/Screen-Shot-2016-06-06-at-6.01.33-PM-232x300.png\" alt=\"Cylinder with height = 7 and radius = (t-2)\" width=\"232\" height=\"300\" \/><\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q137608\">Show Solution<\/span><\/p>\n<div id=\"q137608\" class=\"hidden-answer\" style=\"display: none\">\n<p><strong>Read and Understand:\u00a0<\/strong> We are tasked with writing an\u00a0expressions for the volume\u00a0of the cylinder in the figure\u00a0above. The volume of a cylinder\u00a0is given as[latex]V=\\pi{r^2}h[\/latex], where [latex]\\pi[\/latex] is a constant, and r is the radius and h is the height of the cylinder.<\/p>\n<p><strong>Define and Translate:<\/strong>\u00a0Use the formula for volume:[latex]V=\\pi{r^2}h[\/latex], \u00a0we need to define r and h.<\/p>\n<p>[latex]\\begin{array}{c}r=\\left(t-2\\right)\\\\h=7\\end{array}[\/latex]<\/p>\n<p><strong>Write and Solve:\u00a0<\/strong>Substitute r and h into the formula for volume.<\/p>\n<p style=\"text-align: center\">[latex]\\begin{array}{c}V=\\pi{r^2}h\\\\=\\pi\\left(t-2\\right)^2\\cdot{7}\\\\=7\\pi\\left({t^2}-2t-2t+4\\right)\\\\=7\\pi\\left({t^2}-4t+4\\right)\\end{array}[\/latex]<\/p>\n<p style=\"text-align: left\">Note that we usually write other constants that are multiplied by [latex]\\pi[\/latex] in front of it. \u00a0We can now distribute [latex]7\\pi[\/latex] to each term in the polynomial.<\/p>\n<p style=\"text-align: center\">[latex]\\begin{array}{c}7\\pi\\left({t^2}-4t+4\\right)\\\\=\\left(7\\pi\\right){t^2}-\\left(7\\pi\\right)\\cdot{4t}+\\left(7\\pi\\right)\\cdot{4}\\\\=7\\pi{t^2}-28\\pi{t}+28\\pi\\end{array}[\/latex].<\/p>\n<p style=\"text-align: left\">Note again how we left [latex]\\pi[\/latex] as a greek letter. \u00a0If we needed to use this calculation for measurement of materials, we would round pi, or a computer would round for us.<\/p>\n<h4 style=\"text-align: left\">Answer<\/h4>\n<p>[latex]V=\\pi{r^2}h=7\\pi{t^2}-28\\pi{t}+28\\pi[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>In this last video, we present another example of finding the volume of a cylinder whose dimensions include polynomials.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-3\" title=\"Polynomial Multiplication Application - Volume of a Cylinder\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/g-g_nSsfGs4?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<p>&nbsp;<\/p>\n<p>In this section we defined polynomials that represent perimeter, area and volume of well-known shapes. \u00a0We also introduced some convention about how to use and write [latex]\\pi[\/latex] when it is combined with other constants and variables. The next application will introduce you to cost and revenue polynomials.\u00a0 Next we will see that cost and revenue equations can be polynomials.<\/p>\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-16262\">\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>Screenshot PI. <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>Polynomial Multiplication Application - Volume of a Cylinder. <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com) for Lumen Learning. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/g-g_nSsfGs4\">https:\/\/youtu.be\/g-g_nSsfGs4<\/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>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 class=\"license-attribution-dropdown-subheading\">CC licensed content, Shared previously<\/div><ul class=\"citation-list\"><li>Ex: Polynomial Addition Application - Perimeter. <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com). <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/BhRpZv0_0jE\">https:\/\/youtu.be\/BhRpZv0_0jE<\/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>Ex: Find the Area of a Rectangle Using a Polynomial. <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com) . <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/5Q2htATOIik\">https:\/\/youtu.be\/5Q2htATOIik<\/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>Unit 11: Exponents and Polynomials, from Developmental Math: An Open Program. <strong>Provided by<\/strong>: Monterey Institute of Technology and Education. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"http:\/\/nrocnetwork.org\/resources\/downloads\/nroc-math-open-textbook-units-1-12-pdf-and-word-formats\/\">http:\/\/nrocnetwork.org\/resources\/downloads\/nroc-math-open-textbook-units-1-12-pdf-and-word-formats\/<\/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>\n\t\t\t\t\t\t <\/div>\n\t\t\t\t\t <\/div>\n\t\t\t <\/section>","protected":false},"author":169554,"menu_order":19,"template":"","meta":{"_candela_citation":"[{\"type\":\"original\",\"description\":\"Screenshot PI\",\"author\":\"\",\"organization\":\"Lumen Learning\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"Ex: Polynomial Addition Application - 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