{"id":8280,"date":"2012-04-12T22:20:35","date_gmt":"2012-04-12T21:20:35","guid":{"rendered":"https:\/\/www.acasinhadamatematica.pt\/?p=8280"},"modified":"2022-01-13T23:36:34","modified_gmt":"2022-01-13T23:36:34","slug":"numa-empresa","status":"publish","type":"post","link":"https:\/\/www.acasinhadamatematica.pt\/?p=8280","title":{"rendered":"Numa empresa"},"content":{"rendered":"<p><ul id='GTTabs_ul_8280' class='GTTabs' style='display:none'>\n<li id='GTTabs_li_0_8280' class='GTTabs_curr'><a  id=\"8280_0\" onMouseOver=\"GTTabsShowLinks('Enunciado'); return true;\"  onMouseOut=\"GTTabsShowLinks();\"  class='GTTabsLinks'>Enunciado<\/a><\/li>\n<li id='GTTabs_li_1_8280' ><a  id=\"8280_1\" onMouseOver=\"GTTabsShowLinks('Resolu\u00e7\u00e3o'); return true;\"  onMouseOut=\"GTTabsShowLinks();\"  class='GTTabsLinks'>Resolu\u00e7\u00e3o<\/a><\/li>\n<\/ul>\n\n<div class='GTTabs_divs GTTabs_curr_div' id='GTTabs_0_8280'>\n<span class='GTTabs_titles'><b>Enunciado<\/b><\/span><\/p>\n<p>Numa empresa, o lucro originado pela produ\u00e7\u00e3o de $n$ pe\u00e7as \u00e9 dado, em milhares de contos, por $$L(n) = \\ln \\left( {100 + n} \\right) + k$$ com $k$ constante real.<\/p>\n<ol>\n<li>Sabendo que, n\u00e3o havendo produ\u00e7\u00e3o, n\u00e3o h\u00e1 lucro, determine $k$ e mostre que: $$L(n) = \\ln \\left( {1 + 0,01n} \\right)$$<\/li>\n<li>Qual \u00e9 o n\u00famero m\u00ednimo de pe\u00e7as que \u00e9 necess\u00e1rio produzir para que o lucro seja superior a 1 milhar de contos.<\/li>\n<li>Justifique que, apesar do lucro ir aumentando, \u00e0 medida que o n\u00famero de pe\u00e7as produzidas aumenta, essa varia\u00e7\u00e3o vai sendo feita de forma cada vez mais lenta.<\/li>\n<li>Calcule: $$\\mathop {\\lim }\\limits_{n \\to\u00a0 + \\infty } \\left[ {n \\times L\\left( {\\frac{1}{n}} \\right)} \\right]$$<\/li>\n<\/ol>\n<p><div class='GTTabsNavigation' style='display:none'><span class='GTTabs_nav_next'><a href='#GTTabs_ul_8280' onClick='GTTabs_show(1,8280)'>Resolu\u00e7\u00e3o &gt;&gt;<\/a><\/span><\/div><\/div>\n\n<div class='GTTabs_divs' id='GTTabs_1_8280'>\n<span class='GTTabs_titles'><b>Resolu\u00e7\u00e3o<\/b><\/span><!--more--><\/p>\n<blockquote>\n<p>Numa empresa, o lucro originado pela produ\u00e7\u00e3o de $n$ pe\u00e7as \u00e9 dado, em milhares de contos, por $$L(n) = \\ln \\left( {100 + n} \\right) + k$$ com $k$ constante real.<\/p>\n<\/blockquote>\n<p>\u00ad<\/p>\n<ol>\n<li>Sabendo que n\u00e3o havendo produ\u00e7\u00e3o n\u00e3o h\u00e1 lucro, ter-se-\u00e1 de verificar $L(0) = 0$.<br \/>\nOra, $$\\begin{array}{*{20}{l}}<br \/>\n{L(0) = 0}&amp; \\Leftrightarrow &amp;{\\ln \\left( {100 + 0} \\right) + k = 0} \\\\<br \/>\n{}&amp; \\Leftrightarrow &amp;{k =\u00a0 &#8211; \\ln 100}<br \/>\n\\end{array}$$<br \/>\nLogo, tem-se: $$\\begin{array}{*{20}{l}}<br \/>\n{L(n)}&amp; = &amp;{\\ln \\left( {100 + n} \\right) &#8211; \\ln 100} \\\\<br \/>\n{}&amp; = &amp;{\\ln \\frac{{100 + n}}{{100}}} \\\\<br \/>\n{}&amp; = &amp;{\\ln \\left( {1 + 0,01n} \\right)}<br \/>\n\\end{array}$$<br \/>\n\u00ad<\/li>\n<li>Ora, $$\\begin{array}{*{20}{l}}<br \/>\n{L(n) &gt; 1}&amp; \\Leftrightarrow &amp;{\\ln \\left( {1 + 0,01n} \\right) &gt; 1} \\\\<br \/>\n{}&amp; \\Leftrightarrow &amp;{\\ln \\left( {1 + 0,01n} \\right) &gt; \\ln e} \\\\<br \/>\n{}&amp; \\Leftrightarrow &amp;{\\begin{array}{*{20}{l}}<br \/>\n{1 + 0,01n &gt; e}&amp;{}&amp;{(*)}<br \/>\n\\end{array}} \\\\<br \/>\n{}&amp; \\Leftrightarrow &amp;{n &gt; 100\\left( {e &#8211; 1} \\right)}<br \/>\n\\end{array}$$<br \/>\n${(*)}$ pois a fun\u00e7\u00e3o $x \\to \\ln x$ \u00e9 estritamente crescente.<\/p>\n<p>Como $100\\left( {e &#8211; 1} \\right) \\approx 171,8$,\u00a0\u00e9 necess\u00e1rio produzir no m\u00ednimo 172 pe\u00e7as para que o lucro seja superior a 1 milhar de contos.<br \/>\n\u00ad<\/p>\n<\/li>\n<li>Consideremos a extens\u00e3o da fun\u00e7\u00e3o considerada a $\\mathbb{R}_0^ + $: $f(x) = \\ln \\left( {1 + 0,01x} \\right)$.<br \/>\nOra, $$\\begin{array}{*{20}{l}}<br \/>\n{f'(x)}&amp; = &amp;{\\left( {\\ln \\left( {1 + 0,01x} \\right)} \\right)&#8217;} \\\\<br \/>\n{}&amp; = &amp;{\\frac{{0,01}}{{1 + 0,01x}}}<br \/>\n\\end{array}$$<br \/>\ne<br \/>\n\\[\\begin{array}{*{20}{l}}<br \/>\n{f&#8221;(x)}&amp; = &amp;{{{\\left( {\\frac{{0,01}}{{1 + 0,01x}}} \\right)}^\\prime }} \\\\<br \/>\n{}&amp; = &amp;{\\frac{{ &#8211; {{0,01}^2}}}{{{{\\left( {1 + 0,01x} \\right)}^2}}}}<br \/>\n\\end{array}\\]<\/p>\n<p>Como $f'(x) &gt; 0,\\forall x \\in \\mathbb{R}_0^ + $, ent\u00e3o a fun\u00e7\u00e3o $f$ \u00e9 estritamente crescente.<br \/>\nComo $f&#8221;(x) &lt; 0,\\forall x \\in \\mathbb{R}_0^ + $, ent\u00e3o a fun\u00e7\u00e3o $f&#8217;$ \u00e9 estritamente decrescente.<\/p>\n<p>Id\u00eantica conclus\u00e3o se retirar\u00e1 relativamente\u00a0\u00e0 fun\u00e7\u00e3o $L$. Consequentemente, apesar do lucro ir aumentando, \u00e0 medida que o n\u00famero de pe\u00e7as produzidas aumenta, essa varia\u00e7\u00e3o vai sendo feita de forma cada vez mais lenta, pois a restri\u00e7\u00e3o de $f&#8217;$ a ${\\mathbb{N}_0}$\u00a0\u00e9 decrescente.<br \/>\n\u00ad<\/p>\n<\/li>\n<li>Ora,<br \/>\n$$\\begin{array}{*{20}{l}}<br \/>\n{\\mathop {\\lim }\\limits_{n \\to\u00a0 + \\infty } \\left[ {n \\times L\\left( {\\frac{1}{n}} \\right)} \\right]}&amp; = &amp;{\\mathop {\\lim }\\limits_{n \\to\u00a0 + \\infty } \\left[ {n \\times \\ln \\left( {1 + 0,01 \\times \\frac{1}{n}} \\right)} \\right]} \\\\<br \/>\n{}&amp; = &amp;{\\mathop {\\lim }\\limits_{n \\to\u00a0 + \\infty } \\left[ {n \\times \\ln \\left( {1 + \\frac{1}{{100n}}} \\right)} \\right]} \\\\<br \/>\n{}&amp; = &amp;{\\mathop {\\lim }\\limits_{n \\to\u00a0 + \\infty } \\left[ {\\ln {{\\left( {1 + \\frac{1}{{100n}}} \\right)}^n}} \\right]} \\\\<br \/>\n{}&amp; = &amp;{\\ln {{\\left[ {\\underbrace {\\mathop {\\lim }\\limits_{n \\to\u00a0 + \\infty } {{\\left( {1 + \\frac{1}{{100n}}} \\right)}^{100n}}}_e} \\right]}^{\\frac{1}{{100}}}}} \\\\<br \/>\n{}&amp; = &amp;{\\ln {e^{\\frac{1}{{100}}}}} \\\\<br \/>\n{}&amp; = &amp;{\\frac{1}{{100}}}<br \/>\n\\end{array}$$<\/li>\n<\/ol>\n<div class='GTTabsNavigation' style='display:none'><span class='GTTabs_nav_prev'><a href='#GTTabs_ul_8280' onClick='GTTabs_show(0,8280)'>&lt;&lt; Enunciado<\/a><\/span><\/div><\/div>\n\n","protected":false},"excerpt":{"rendered":"<p>Enunciado Resolu\u00e7\u00e3o Enunciado Numa empresa, o lucro originado pela produ\u00e7\u00e3o de $n$ pe\u00e7as \u00e9 dado, em milhares de contos, por $$L(n) = \\ln \\left( {100 + n} \\right) + k$$ com $k$ constante real.&#46;&#46;&#46;<\/p>\n","protected":false},"author":1,"featured_media":19178,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_jetpack_memberships_contains_paid_content":false,"footnotes":""},"categories":[226,97,292],"tags":[427,136,286],"series":[],"class_list":["post-8280","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-12--ano","category-aplicando","category-calculo-diferencial","tag-12-o-ano","tag-derivada","tag-limites"],"views":2238,"jetpack_featured_media_url":"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2021\/12\/Mat69.png","jetpack_sharing_enabled":true,"jetpack_likes_enabled":true,"_links":{"self":[{"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/posts\/8280","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fcomments&post=8280"}],"version-history":[{"count":0,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/posts\/8280\/revisions"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/media\/19178"}],"wp:attachment":[{"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fmedia&parent=8280"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fcategories&post=8280"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Ftags&post=8280"},{"taxonomy":"series","embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fseries&post=8280"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}