{"id":2900,"date":"2024-06-10T19:17:38","date_gmt":"2024-06-10T19:17:38","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus1\/?post_type=chapter&#038;p=2900"},"modified":"2024-08-05T01:52:53","modified_gmt":"2024-08-05T01:52:53","slug":"derivatives-as-rates-of-change-learn-it-4","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus1\/chapter\/derivatives-as-rates-of-change-learn-it-4\/","title":{"raw":"Derivatives as Rates of Change: Learn It 4","rendered":"Derivatives as Rates of Change: Learn It 4"},"content":{"raw":"<h2>Rate of Change Applications Cont.<\/h2>\r\n<h3>Changes in Cost and Revenue<\/h3>\r\n<p>In addition to analyzing motion along a line and population growth, derivatives are useful in analyzing changes in cost, revenue, and profit. The concept of a marginal function is common in the fields of business and economics and implies the use of derivatives.<\/p>\r\n<ul>\r\n\t<li>The<strong> marginal cost<\/strong> is the derivative of the cost function.<\/li>\r\n\t<li>The <strong>marginal revenue <\/strong>is the derivative of the revenue function.<\/li>\r\n\t<li>The <strong>marginal profit <\/strong>is the derivative of the profit function, which is based on the cost function and the revenue function.<\/li>\r\n<\/ul>\r\n<section class=\"textbox keyTakeaway\">\r\n<h3>Marginal Cost, Marginal Revenue and Marginal Profit<\/h3>\r\n<ul>\r\n\t<li id=\"fs-id1169736618496\">If [latex]C(x)[\/latex] is the cost of producing [latex]x[\/latex] items, then the <strong>marginal cost<\/strong> [latex]MC(x)[\/latex] is <br \/>\r\n<center>[latex]MC(x)=C^{\\prime}(x)[\/latex].<\/center><\/li>\r\n\t<li>If [latex]R(x)[\/latex] is the revenue obtained from selling [latex]x[\/latex] items, then the <strong>marginal revenue<\/strong> [latex]MR(x)[\/latex] is <br \/>\r\n<center>[latex]MR(x)=R^{\\prime}(x)[\/latex].<\/center><\/li>\r\n\t<li>If [latex]P(x)=R(x)-C(x)[\/latex] is the profit obtained from selling [latex]x[\/latex] items, then the <strong>marginal profit<\/strong> [latex]MP(x)[\/latex] is defined to be <br \/>\r\n<center>[latex]MP(x)=P^{\\prime}(x)=MR(x)-MC(x)=R^{\\prime}(x)-C^{\\prime}(x)[\/latex].<\/center><\/li>\r\n<\/ul>\r\n<\/section>\r\n<p id=\"fs-id1169739299256\">We can roughly approximate<\/p>\r\n<div id=\"fs-id1165040641522\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]MC(x)=C^{\\prime}(x)=\\underset{h\\to 0}{\\lim}\\dfrac{C(x+h)-C(x)}{h}[\/latex]<\/div>\r\n<p id=\"fs-id1169739007166\">by choosing an appropriate value for [latex]h[\/latex].<\/p>\r\n<p>Since [latex]x[\/latex] represents objects, a reasonable and small value for [latex]h[\/latex] is [latex]1[\/latex]. Thus, by substituting [latex]h=1[\/latex], we get the approximation [latex]MC(x)=C^{\\prime}(x)\\approx C(x+1)-C(x)[\/latex].<\/p>\r\n<p>Consequently, [latex]C^{\\prime}(x)[\/latex] for a given value of [latex]x[\/latex] can be thought of as the change in cost associated with producing one additional item. In a similar way, [latex]MR(x)=R^{\\prime}(x)[\/latex] approximates the revenue obtained by selling one additional item, and [latex]MP(x)=P^{\\prime}(x)[\/latex] approximates the profit obtained by producing and selling one additional item.<\/p>\r\n<section class=\"textbox example\">\r\n<p id=\"fs-id1169739274279\">Assume that the number of barbeque dinners that can be sold, [latex]x[\/latex], can be related to the price charged, [latex]p[\/latex], by the equation<\/p>\r\n<p style=\"text-align: center;\">[latex]p(x)=9-0.03x, \\, 0\\le x\\le 300[\/latex].<\/p>\r\n<p id=\"fs-id1169739305211\">In this case, the revenue in dollars obtained by selling [latex]x[\/latex] barbeque dinners is given by<\/p>\r\n<div id=\"fs-id1169739212801\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]R(x)=xp(x)=x(9-0.03x)=-0.03x^2+9x[\/latex] for [latex]0\\le x\\le 300[\/latex].<\/div>\r\n<p id=\"fs-id1169739194211\">Use the marginal revenue function to estimate the revenue obtained from selling the [latex]101[\/latex]<sup>st<\/sup> barbeque dinner.<\/p>\r\n<p>Compare this to the actual revenue obtained from the sale of this dinner.<\/p>\r\n<p>[reveal-answer q=\"fs-id1169739351550\"]Show Solution[\/reveal-answer]<br \/>\r\n[hidden-answer a=\"fs-id1169739351550\"]<\/p>\r\n<p id=\"fs-id1169739351550\">First, find the marginal revenue function: [latex]MR(x)=R^{\\prime}(x)=-0.06x+9[\/latex].<\/p>\r\n<p id=\"fs-id1169739305060\">Next, use [latex]R^{\\prime}(100)[\/latex] to approximate [latex]R(101)-R(100)[\/latex], the revenue obtained from the sale of the [latex]101[\/latex]<sup>st<\/sup>\u00a0 dinner. Since [latex]R^{\\prime}(100)=3[\/latex], the revenue obtained from the sale of the [latex]101[\/latex]<sup>st<\/sup>\u00a0 dinner is approximately [latex]$3[\/latex].<\/p>\r\n<p id=\"fs-id1169736611332\">The actual revenue obtained from the sale of the [latex]101[\/latex]<sup>st<\/sup> dinner is<\/p>\r\n<div id=\"fs-id1169737285423\" class=\"equation unnumbered\">[latex]R(101)-R(100)=602.97-600=2.97[\/latex], or [latex]\\$2.97[\/latex].<\/div>\r\n<p id=\"fs-id1169739298517\">The marginal revenue is a fairly good estimate in this case and has the advantage of being easy to compute.<\/p>\r\n<p>Watch the following video to see the worked solution to this example.<\/p>\r\n<center><iframe title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/XOgb95xhWF0?controls=0&amp;start=1168&amp;end=1320&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\" data-mce-fragment=\"1\"><\/iframe><\/center>\r\n<p>For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.<\/p>\r\n<p>You can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus\/Calculus1+Videos\/3.4DerivativesAsRatesOfChange1168to1320_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of \"3.4 Derivatives as Rates of Change\" here (opens in new window)<\/a>.<\/p>\r\n<p>[\/hidden-answer]<\/p>\r\n<\/section>","rendered":"<h2>Rate of Change Applications Cont.<\/h2>\n<h3>Changes in Cost and Revenue<\/h3>\n<p>In addition to analyzing motion along a line and population growth, derivatives are useful in analyzing changes in cost, revenue, and profit. The concept of a marginal function is common in the fields of business and economics and implies the use of derivatives.<\/p>\n<ul>\n<li>The<strong> marginal cost<\/strong> is the derivative of the cost function.<\/li>\n<li>The <strong>marginal revenue <\/strong>is the derivative of the revenue function.<\/li>\n<li>The <strong>marginal profit <\/strong>is the derivative of the profit function, which is based on the cost function and the revenue function.<\/li>\n<\/ul>\n<section class=\"textbox keyTakeaway\">\n<h3>Marginal Cost, Marginal Revenue and Marginal Profit<\/h3>\n<ul>\n<li id=\"fs-id1169736618496\">If [latex]C(x)[\/latex] is the cost of producing [latex]x[\/latex] items, then the <strong>marginal cost<\/strong> [latex]MC(x)[\/latex] is \n<div style=\"text-align: center;\">[latex]MC(x)=C^{\\prime}(x)[\/latex].<\/div>\n<\/li>\n<li>If [latex]R(x)[\/latex] is the revenue obtained from selling [latex]x[\/latex] items, then the <strong>marginal revenue<\/strong> [latex]MR(x)[\/latex] is \n<div style=\"text-align: center;\">[latex]MR(x)=R^{\\prime}(x)[\/latex].<\/div>\n<\/li>\n<li>If [latex]P(x)=R(x)-C(x)[\/latex] is the profit obtained from selling [latex]x[\/latex] items, then the <strong>marginal profit<\/strong> [latex]MP(x)[\/latex] is defined to be \n<div style=\"text-align: center;\">[latex]MP(x)=P^{\\prime}(x)=MR(x)-MC(x)=R^{\\prime}(x)-C^{\\prime}(x)[\/latex].<\/div>\n<\/li>\n<\/ul>\n<\/section>\n<p id=\"fs-id1169739299256\">We can roughly approximate<\/p>\n<div id=\"fs-id1165040641522\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]MC(x)=C^{\\prime}(x)=\\underset{h\\to 0}{\\lim}\\dfrac{C(x+h)-C(x)}{h}[\/latex]<\/div>\n<p id=\"fs-id1169739007166\">by choosing an appropriate value for [latex]h[\/latex].<\/p>\n<p>Since [latex]x[\/latex] represents objects, a reasonable and small value for [latex]h[\/latex] is [latex]1[\/latex]. Thus, by substituting [latex]h=1[\/latex], we get the approximation [latex]MC(x)=C^{\\prime}(x)\\approx C(x+1)-C(x)[\/latex].<\/p>\n<p>Consequently, [latex]C^{\\prime}(x)[\/latex] for a given value of [latex]x[\/latex] can be thought of as the change in cost associated with producing one additional item. In a similar way, [latex]MR(x)=R^{\\prime}(x)[\/latex] approximates the revenue obtained by selling one additional item, and [latex]MP(x)=P^{\\prime}(x)[\/latex] approximates the profit obtained by producing and selling one additional item.<\/p>\n<section class=\"textbox example\">\n<p id=\"fs-id1169739274279\">Assume that the number of barbeque dinners that can be sold, [latex]x[\/latex], can be related to the price charged, [latex]p[\/latex], by the equation<\/p>\n<p style=\"text-align: center;\">[latex]p(x)=9-0.03x, \\, 0\\le x\\le 300[\/latex].<\/p>\n<p id=\"fs-id1169739305211\">In this case, the revenue in dollars obtained by selling [latex]x[\/latex] barbeque dinners is given by<\/p>\n<div id=\"fs-id1169739212801\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]R(x)=xp(x)=x(9-0.03x)=-0.03x^2+9x[\/latex] for [latex]0\\le x\\le 300[\/latex].<\/div>\n<p id=\"fs-id1169739194211\">Use the marginal revenue function to estimate the revenue obtained from selling the [latex]101[\/latex]<sup>st<\/sup> barbeque dinner.<\/p>\n<p>Compare this to the actual revenue obtained from the sale of this dinner.<\/p>\n<p><div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"qfs-id1169739351550\">Show Solution<\/button><\/p>\n<div id=\"qfs-id1169739351550\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1169739351550\">First, find the marginal revenue function: [latex]MR(x)=R^{\\prime}(x)=-0.06x+9[\/latex].<\/p>\n<p id=\"fs-id1169739305060\">Next, use [latex]R^{\\prime}(100)[\/latex] to approximate [latex]R(101)-R(100)[\/latex], the revenue obtained from the sale of the [latex]101[\/latex]<sup>st<\/sup>\u00a0 dinner. Since [latex]R^{\\prime}(100)=3[\/latex], the revenue obtained from the sale of the [latex]101[\/latex]<sup>st<\/sup>\u00a0 dinner is approximately [latex]$3[\/latex].<\/p>\n<p id=\"fs-id1169736611332\">The actual revenue obtained from the sale of the [latex]101[\/latex]<sup>st<\/sup> dinner is<\/p>\n<div id=\"fs-id1169737285423\" class=\"equation unnumbered\">[latex]R(101)-R(100)=602.97-600=2.97[\/latex], or [latex]\\$2.97[\/latex].<\/div>\n<p id=\"fs-id1169739298517\">The marginal revenue is a fairly good estimate in this case and has the advantage of being easy to compute.<\/p>\n<p>Watch the following video to see the worked solution to this example.<\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/XOgb95xhWF0?controls=0&amp;start=1168&amp;end=1320&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\" data-mce-fragment=\"1\"><\/iframe><\/div>\n<p>For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.<\/p>\n<p>You can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus\/Calculus1+Videos\/3.4DerivativesAsRatesOfChange1168to1320_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of &#8220;3.4 Derivatives as Rates of Change&#8221; here (opens in new window)<\/a>.<\/p>\n<\/div>\n<\/div>\n<\/section>\n","protected":false},"author":15,"menu_order":26,"template":"","meta":{"_candela_citation":"[]","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"part":503,"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\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/2900"}],"collection":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/users\/15"}],"version-history":[{"count":6,"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/2900\/revisions"}],"predecessor-version":[{"id":2922,"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/2900\/revisions\/2922"}],"part":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/parts\/503"}],"metadata":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/2900\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/media?parent=2900"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapter-type?post=2900"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/contributor?post=2900"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/license?post=2900"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}