{"id":1362,"date":"2024-05-10T17:30:27","date_gmt":"2024-05-10T17:30:27","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/?post_type=chapter&#038;p=1362"},"modified":"2025-08-13T15:48:44","modified_gmt":"2025-08-13T15:48:44","slug":"applications-and-inequalities-learn-it-2","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/chapter\/applications-and-inequalities-learn-it-2\/","title":{"raw":"Applications of Non-Linear Equations: Learn It 2","rendered":"Applications of Non-Linear Equations: Learn It 2"},"content":{"raw":"<h2>Areas and Volumes<\/h2>\r\n<section class=\"textbox recall\">\r\n<ul>\r\n \t<li>Perimeter of a rectangle: [latex]P = 2L+2W[\/latex]<\/li>\r\n \t<li>Area of a square: [latex]A = s^2[\/latex]<\/li>\r\n \t<li>Area of a rectangle: [latex]A = lw[\/latex]<\/li>\r\n \t<li>Area of a triangle: [latex]A = \\dfrac{1}{2}bh[\/latex]<\/li>\r\n \t<li>Area of a cicle: [latex]A = \\pi r^2[\/latex]<\/li>\r\n \t<li>Volume of a box: [latex]V = LWH[\/latex]<\/li>\r\n \t<li>Volume of a sphere: [latex]V = \\dfrac{4}{3} \\pi r^3[\/latex]<\/li>\r\n<\/ul>\r\nMaahi is building a little free library (a small house-shaped book repository), whose front is in the shape of a square topped with a triangle. There will be a rectangular door through which people can take and donate books. Maahi wants to find the area of the front of the library so that they can purchase the correct amount of paint. Note, Maahi is not painting the door.\r\n\r\nUsing the measurements of the front of the house, shown below, <span style=\"font-family: 'Public Sans', -apple-system, BlinkMacSystemFont, 'Segoe UI', Roboto, Oxygen-Sans, Ubuntu, Cantarell, 'Helvetica Neue', sans-serif;\">create an expression that represents the area of the front of the library.<\/span>\r\n\r\n[caption id=\"attachment_6795\" align=\"aligncenter\" width=\"271\"]<img class=\"wp-image-6795 size-full\" src=\"https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/42\/2024\/05\/17154248\/large-AreaDiagram1.png\" alt=\"A simple house-shaped diagram drawn in purple. The house has a rectangular base with a triangular roof on top. Inside the rectangular portion is a smaller rectangle that could represent a door. The diagram includes several measurements: the height of the door is labeled as &quot;x&quot;, the space above the door to the roof is labeled as &quot;1 foot&quot;, the height of the entire rectangular portion is marked as &quot;2x&quot;, and a horizontal measurement in the roof area is labeled as &quot;3\/2 feet&quot;.\" width=\"271\" height=\"234\" \/> House diagram with labels[\/caption]\r\n\r\n[reveal-answer q=\"7453\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"7453\"]\r\n\r\nTo calculate the area of the front of the little free library that Maahi needs to paint, we need to consider the areas of both the square and the triangular part of the front, minus the area of the door.\r\n\r\nFirst, find the area of the square in square feet.\r\n\r\n<center>[latex]\\text{Area}_{\\text{square}} = (2x)^2 = 4x^2[\/latex]<\/center>Then, find the area of the triangle in square feet.\r\n\r\n<center>[latex]\\text{Area}_{\\text{triangle}} = \\frac{1}{2} \\times \\text{base} \\times \\text{height} = \\frac{1}{2} \\times (2x) \\times (\\frac{3}{2}) = \\frac{3}{2}x[\/latex]<\/center>Next, find the area of the rectangular door in square feet.\r\n\r\n<center>[latex]\\text{Area}_{\\text{door}} = lw = (x)(1) = x[\/latex]<\/center>The area of the front of the library can be found by adding the areas of the square and the triangle, and then subtracting the area of the rectangle.\r\n\r\n<center>[latex]\\begin{array}{rl}\r\n\\text{Area}_{\\text{front}} &amp; = \\text{Area}_{\\text{square}} + \\text{Area}_{\\text{triangle}} - \\text{Area}_{\\text{door}} \\\\[10pt]\r\n\\text{Area}_{\\text{front}} &amp; = 4x^2 + \\frac{3}{2}x - x \\ \\text{ft}^2 \\\\[10pt]\r\n\\text{Area}_{\\text{front}} &amp; = 4x^2 + \\frac{1}{2}x \\ \\text{ft}^2\r\n\\end{array}[\/latex]<\/center>[\/hidden-answer]\r\n\r\n<\/section><section class=\"textbox example\">Imagine that we are trying to find the area of a lawn so that we can determine how much grass seed to purchase. The lawn is the green portion in the figure below.\r\n\r\n[caption id=\"attachment_1366\" align=\"aligncenter\" width=\"300\"]<img class=\"wp-image-1366 size-medium\" src=\"https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/42\/2024\/05\/10175246\/Screenshot-2024-05-10-at-10.52.43%E2%80%AFAM-300x196.png\" alt=\"\" width=\"300\" height=\"196\" \/> Visual of lawn with labels[\/caption]\r\n\r\nCreate an expression that represents the area of the region that requires grass seed.\r\n\r\n[reveal-answer q=\"332598\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"332598\"]\r\n\r\nThe area of the entire region can be found using the formula for the area of a rectangle.\r\n\r\n<center>[latex]A = lw = (10x)(6x) = 60x^2[\/latex]<\/center>The areas of the portions that do not require grass seed need to be subtracted from the area of the entire region. The two square regions each have an area of\r\n\r\n<center>[latex]A = s^2 = 4^2 = 16 \\text{ units}^2[\/latex]<\/center>The other rectangular region has one side of length [latex]10x-8[\/latex] and one side of length [latex]4[\/latex], giving an area of\r\n\r\n<center>[latex]A = lw = 4(10x-8) = 40x-32 \\text{ units}^2[\/latex].<\/center>So the region that must be subtracted has an area of\r\n\r\n<center>[latex]2(16)+40x-32 = 40x \\text{ units}^2[\/latex].<\/center>Thus, the area of the region that requires grass seed is found by subtracting [latex]60x^2 - 40x \\text{ units}^2[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/section><section class=\"textbox tryIt\">[ohm2_question hide_question_numbers=1]18988[\/ohm2_question]<\/section><section class=\"textbox tryIt\">[ohm2_question hide_question_numbers=1]18989[\/ohm2_question]<\/section><section class=\"textbox example\">Find the dimensions of a shipping box given that the length is twice the width, the height is\u00a0 [latex]8[\/latex] inches, and the volume is [latex]1600 \\text{ in.}^3[\/latex].[reveal-answer q=\"137446\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"137446\"]Let [latex]w[\/latex] be the width of the box, [latex]l[\/latex] be the length of the box, and [latex]h[\/latex] be the height of the box.\r\nGiven relationships:\r\n<ul>\r\n \t<li>[latex]l = 2w[\/latex]<\/li>\r\n \t<li>[latex]h = 8 \\text{ in.}[\/latex]<\/li>\r\n \t<li>[latex]\\text{Volume: } V = lwh = 1600 \\text{ in.}^3[\/latex]<\/li>\r\n<\/ul>\r\nLet's substitute the known values and expressions into the volume formula and solve:\r\n\r\n<center>[latex]V = lwh = 2w \\cdot w \\cdot 8 = 1600[\/latex]<\/center><center>[latex]16w^2 = 1600[\/latex]<\/center><center>[latex]w^2 = \\frac{1600}{16} = 100[\/latex]<\/center><center>[latex]w = \\sqrt{100} = 10 \\text{ inches}[\/latex]<\/center>Find [latex]l[\/latex]:\r\n\r\n<center>[latex]l = 2w = 2 \\times 10 = 20 \\text{ inches}[\/latex]<\/center>Thus, the dimensions of the box are:\r\n\r\n<center>[latex]l = 20 \\text{ inches}[\/latex], [latex]w = 10 \\text{ inches}[\/latex], and [latex]h = 8 \\text{ inches}[\/latex]<\/center>[\/hidden-answer]\r\n\r\n<\/section><section class=\"textbox tryIt\">[ohm2_question hide_question_numbers=1]18990[\/ohm2_question]<\/section>","rendered":"<h2>Areas and Volumes<\/h2>\n<section class=\"textbox recall\">\n<ul>\n<li>Perimeter of a rectangle: [latex]P = 2L+2W[\/latex]<\/li>\n<li>Area of a square: [latex]A = s^2[\/latex]<\/li>\n<li>Area of a rectangle: [latex]A = lw[\/latex]<\/li>\n<li>Area of a triangle: [latex]A = \\dfrac{1}{2}bh[\/latex]<\/li>\n<li>Area of a cicle: [latex]A = \\pi r^2[\/latex]<\/li>\n<li>Volume of a box: [latex]V = LWH[\/latex]<\/li>\n<li>Volume of a sphere: [latex]V = \\dfrac{4}{3} \\pi r^3[\/latex]<\/li>\n<\/ul>\n<p>Maahi is building a little free library (a small house-shaped book repository), whose front is in the shape of a square topped with a triangle. There will be a rectangular door through which people can take and donate books. Maahi wants to find the area of the front of the library so that they can purchase the correct amount of paint. Note, Maahi is not painting the door.<\/p>\n<p>Using the measurements of the front of the house, shown below, <span style=\"font-family: 'Public Sans', -apple-system, BlinkMacSystemFont, 'Segoe UI', Roboto, Oxygen-Sans, Ubuntu, Cantarell, 'Helvetica Neue', sans-serif;\">create an expression that represents the area of the front of the library.<\/span><\/p>\n<figure id=\"attachment_6795\" aria-describedby=\"caption-attachment-6795\" style=\"width: 271px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-6795 size-full\" src=\"https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/42\/2024\/05\/17154248\/large-AreaDiagram1.png\" alt=\"A simple house-shaped diagram drawn in purple. The house has a rectangular base with a triangular roof on top. Inside the rectangular portion is a smaller rectangle that could represent a door. The diagram includes several measurements: the height of the door is labeled as &quot;x&quot;, the space above the door to the roof is labeled as &quot;1 foot&quot;, the height of the entire rectangular portion is marked as &quot;2x&quot;, and a horizontal measurement in the roof area is labeled as &quot;3\/2 feet&quot;.\" width=\"271\" height=\"234\" srcset=\"https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/42\/2024\/05\/17154248\/large-AreaDiagram1.png 271w, https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/42\/2024\/05\/17154248\/large-AreaDiagram1-65x56.png 65w, https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/42\/2024\/05\/17154248\/large-AreaDiagram1-225x194.png 225w\" sizes=\"(max-width: 271px) 100vw, 271px\" \/><figcaption id=\"caption-attachment-6795\" class=\"wp-caption-text\">House diagram with labels<\/figcaption><\/figure>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q7453\">Show Answer<\/button><\/p>\n<div id=\"q7453\" class=\"hidden-answer\" style=\"display: none\">\n<p>To calculate the area of the front of the little free library that Maahi needs to paint, we need to consider the areas of both the square and the triangular part of the front, minus the area of the door.<\/p>\n<p>First, find the area of the square in square feet.<\/p>\n<div style=\"text-align: center;\">[latex]\\text{Area}_{\\text{square}} = (2x)^2 = 4x^2[\/latex]<\/div>\n<p>Then, find the area of the triangle in square feet.<\/p>\n<div style=\"text-align: center;\">[latex]\\text{Area}_{\\text{triangle}} = \\frac{1}{2} \\times \\text{base} \\times \\text{height} = \\frac{1}{2} \\times (2x) \\times (\\frac{3}{2}) = \\frac{3}{2}x[\/latex]<\/div>\n<p>Next, find the area of the rectangular door in square feet.<\/p>\n<div style=\"text-align: center;\">[latex]\\text{Area}_{\\text{door}} = lw = (x)(1) = x[\/latex]<\/div>\n<p>The area of the front of the library can be found by adding the areas of the square and the triangle, and then subtracting the area of the rectangle.<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{rl}  \\text{Area}_{\\text{front}} & = \\text{Area}_{\\text{square}} + \\text{Area}_{\\text{triangle}} - \\text{Area}_{\\text{door}} \\\\[10pt]  \\text{Area}_{\\text{front}} & = 4x^2 + \\frac{3}{2}x - x \\ \\text{ft}^2 \\\\[10pt]  \\text{Area}_{\\text{front}} & = 4x^2 + \\frac{1}{2}x \\ \\text{ft}^2  \\end{array}[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox example\">Imagine that we are trying to find the area of a lawn so that we can determine how much grass seed to purchase. The lawn is the green portion in the figure below.<\/p>\n<figure id=\"attachment_1366\" aria-describedby=\"caption-attachment-1366\" style=\"width: 300px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-1366 size-medium\" src=\"https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/42\/2024\/05\/10175246\/Screenshot-2024-05-10-at-10.52.43%E2%80%AFAM-300x196.png\" alt=\"\" width=\"300\" height=\"196\" srcset=\"https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/42\/2024\/05\/10175246\/Screenshot-2024-05-10-at-10.52.43%E2%80%AFAM-300x196.png 300w, https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/42\/2024\/05\/10175246\/Screenshot-2024-05-10-at-10.52.43%E2%80%AFAM-65x43.png 65w, https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/42\/2024\/05\/10175246\/Screenshot-2024-05-10-at-10.52.43%E2%80%AFAM-225x147.png 225w, https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/42\/2024\/05\/10175246\/Screenshot-2024-05-10-at-10.52.43%E2%80%AFAM-350x229.png 350w, https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/42\/2024\/05\/10175246\/Screenshot-2024-05-10-at-10.52.43%E2%80%AFAM.png 498w\" sizes=\"(max-width: 300px) 100vw, 300px\" \/><figcaption id=\"caption-attachment-1366\" class=\"wp-caption-text\">Visual of lawn with labels<\/figcaption><\/figure>\n<p>Create an expression that represents the area of the region that requires grass seed.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q332598\">Show Answer<\/button><\/p>\n<div id=\"q332598\" class=\"hidden-answer\" style=\"display: none\">\n<p>The area of the entire region can be found using the formula for the area of a rectangle.<\/p>\n<div style=\"text-align: center;\">[latex]A = lw = (10x)(6x) = 60x^2[\/latex]<\/div>\n<p>The areas of the portions that do not require grass seed need to be subtracted from the area of the entire region. The two square regions each have an area of<\/p>\n<div style=\"text-align: center;\">[latex]A = s^2 = 4^2 = 16 \\text{ units}^2[\/latex]<\/div>\n<p>The other rectangular region has one side of length [latex]10x-8[\/latex] and one side of length [latex]4[\/latex], giving an area of<\/p>\n<div style=\"text-align: center;\">[latex]A = lw = 4(10x-8) = 40x-32 \\text{ units}^2[\/latex].<\/div>\n<p>So the region that must be subtracted has an area of<\/p>\n<div style=\"text-align: center;\">[latex]2(16)+40x-32 = 40x \\text{ units}^2[\/latex].<\/div>\n<p>Thus, the area of the region that requires grass seed is found by subtracting [latex]60x^2 - 40x \\text{ units}^2[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox tryIt\"><iframe loading=\"lazy\" id=\"ohm18988\" class=\"resizable\" src=\"https:\/\/ohm.one.lumenlearning.com\/multiembedq.php?id=18988&theme=lumen&iframe_resize_id=ohm18988&source=tnh\" width=\"100%\" height=\"150\"><\/iframe><\/section>\n<section class=\"textbox tryIt\"><iframe loading=\"lazy\" id=\"ohm18989\" class=\"resizable\" src=\"https:\/\/ohm.one.lumenlearning.com\/multiembedq.php?id=18989&theme=lumen&iframe_resize_id=ohm18989&source=tnh\" width=\"100%\" height=\"150\"><\/iframe><\/section>\n<section class=\"textbox example\">Find the dimensions of a shipping box given that the length is twice the width, the height is\u00a0 [latex]8[\/latex] inches, and the volume is [latex]1600 \\text{ in.}^3[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q137446\">Show Answer<\/button><\/p>\n<div id=\"q137446\" class=\"hidden-answer\" style=\"display: none\">Let [latex]w[\/latex] be the width of the box, [latex]l[\/latex] be the length of the box, and [latex]h[\/latex] be the height of the box.<br \/>\nGiven relationships:<\/p>\n<ul>\n<li>[latex]l = 2w[\/latex]<\/li>\n<li>[latex]h = 8 \\text{ in.}[\/latex]<\/li>\n<li>[latex]\\text{Volume: } V = lwh = 1600 \\text{ in.}^3[\/latex]<\/li>\n<\/ul>\n<p>Let&#8217;s substitute the known values and expressions into the volume formula and solve:<\/p>\n<div style=\"text-align: center;\">[latex]V = lwh = 2w \\cdot w \\cdot 8 = 1600[\/latex]<\/div>\n<div style=\"text-align: center;\">[latex]16w^2 = 1600[\/latex]<\/div>\n<div style=\"text-align: center;\">[latex]w^2 = \\frac{1600}{16} = 100[\/latex]<\/div>\n<div style=\"text-align: center;\">[latex]w = \\sqrt{100} = 10 \\text{ inches}[\/latex]<\/div>\n<p>Find [latex]l[\/latex]:<\/p>\n<div style=\"text-align: center;\">[latex]l = 2w = 2 \\times 10 = 20 \\text{ inches}[\/latex]<\/div>\n<p>Thus, the dimensions of the box are:<\/p>\n<div style=\"text-align: center;\">[latex]l = 20 \\text{ inches}[\/latex], [latex]w = 10 \\text{ inches}[\/latex], and [latex]h = 8 \\text{ inches}[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox tryIt\"><iframe loading=\"lazy\" id=\"ohm18990\" class=\"resizable\" src=\"https:\/\/ohm.one.lumenlearning.com\/multiembedq.php?id=18990&theme=lumen&iframe_resize_id=ohm18990&source=tnh\" width=\"100%\" height=\"150\"><\/iframe><\/section>\n","protected":false},"author":12,"menu_order":20,"template":"","meta":{"_candela_citation":"[]","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"part":92,"module-header":"learn_it","content_attributions":[],"internal_book_links":[],"video_content":null,"cc_video_embed_content":{"cc_scripts":"","media_targets":[]},"try_it_collection":null,"_links":{"self":[{"href":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/1362"}],"collection":[{"href":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/wp-json\/wp\/v2\/users\/12"}],"version-history":[{"count":15,"href":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/1362\/revisions"}],"predecessor-version":[{"id":7617,"href":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/1362\/revisions\/7617"}],"part":[{"href":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/wp-json\/pressbooks\/v2\/parts\/92"}],"metadata":[{"href":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/1362\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/wp-json\/wp\/v2\/media?parent=1362"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/wp-json\/pressbooks\/v2\/chapter-type?post=1362"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/wp-json\/wp\/v2\/contributor?post=1362"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/wp-json\/wp\/v2\/license?post=1362"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}