{"id":42,"date":"2023-09-17T10:57:01","date_gmt":"2023-09-17T10:57:01","guid":{"rendered":"https:\/\/mathority.org\/pt\/problemas-de-otimizacao\/"},"modified":"2023-09-17T10:57:01","modified_gmt":"2023-09-17T10:57:01","slug":"problemas-de-otimizacao","status":"publish","type":"post","link":"https:\/\/mathority.org\/pt\/problemas-de-otimizacao\/","title":{"rendered":"Problemas de otimiza\u00e7\u00e3o"},"content":{"rendered":"<p>Aqui explicamos como os problemas de otimiza\u00e7\u00e3o de fun\u00e7\u00f5es s\u00e3o resolvidos em etapas. Al\u00e9m disso, voc\u00ea poder\u00e1 praticar com exerc\u00edcios resolvidos sobre problemas de otimiza\u00e7\u00e3o. <\/p>\n<h2 class=\"wp-block-heading\"><span class=\"ez-toc-section\" id=\"%c2%bfque-son-los-problemas-de-optimizacion\"><\/span> O que s\u00e3o problemas de otimiza\u00e7\u00e3o?<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p> <strong>Problemas de otimiza\u00e7\u00e3o<\/strong> s\u00e3o problemas nos quais \u00e9 necess\u00e1rio encontrar o m\u00e1ximo ou o m\u00ednimo de uma fun\u00e7\u00e3o. Por exemplo, um problema de otimiza\u00e7\u00e3o envolveria o c\u00e1lculo do m\u00e1ximo de uma fun\u00e7\u00e3o que define os lucros de uma empresa. <\/p>\n<h2 class=\"wp-block-heading\"><span class=\"ez-toc-section\" id=\"como-resolver-los-problemas-de-optimizacion\"><\/span> Como resolver problemas de otimiza\u00e7\u00e3o<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p> Etapas para resolver problemas de otimiza\u00e7\u00e3o de fun\u00e7\u00e3o:<\/p>\n<ol>\n<li> <strong>Defina a fun\u00e7\u00e3o<\/strong> que precisa ser otimizada.<\/li>\n<li> <strong>Derive a fun\u00e7\u00e3o a ser otimizada.<\/strong><\/li>\n<li> Encontre os <strong>pontos cr\u00edticos<\/strong> da fun\u00e7\u00e3o a ser otimizada. Para fazer isso, voc\u00ea precisa igualar a derivada da fun\u00e7\u00e3o a zero e resolver a equa\u00e7\u00e3o resultante.<\/li>\n<li> Estude a monotonicidade da fun\u00e7\u00e3o e <strong>determine o m\u00e1ximo ou m\u00ednimo da fun\u00e7\u00e3o.<\/strong> <\/li>\n<\/ol>\n<h2 class=\"wp-block-heading\"><span class=\"ez-toc-section\" id=\"ejemplo-de-problema-de-optimizacion\"><\/span> Exemplo de um problema de otimiza\u00e7\u00e3o<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p> Considerando a teoria dos problemas de otimiza\u00e7\u00e3o, resolveremos passo a passo um problema deste tipo para que voc\u00ea possa ver como eles s\u00e3o realizados.<\/p>\n<ul>\n<li> Entre todos os tri\u00e2ngulos ret\u00e2ngulos cujos catetos totalizam 10 cm, calcule as dimens\u00f5es daquele com maior \u00e1rea de superf\u00edcie.<\/li>\n<\/ul>\n<p> Para resolver o problema, chamaremos um ramo do tri\u00e2ngulo <em>de x<\/em> e o outro ramo <em>de y<\/em> : <\/p>\n<div class=\"wp-block-image\">\n<figure class=\"aligncenter size-large is-resized\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/uploads\/2023\/07\/triangle-optimisation-probleme.webp\" alt=\"problema de otimiza\u00e7\u00e3o de tri\u00e2ngulo\" class=\"wp-image-2463\" width=\"172\" height=\"156\" srcset=\"\" sizes=\"auto, \" data-src=\"\"><\/figure>\n<\/div>\n<p> <u style=\"text-decoration-color:#FF9B28;\"><strong><span style=\"color:#1976d2;\">Passo 1:<\/span><\/strong> Defina a fun\u00e7\u00e3o a ser otimizada.<\/u><\/p>\n<p> Queremos que a \u00e1rea do tri\u00e2ngulo seja m\u00e1xima, e a f\u00f3rmula para a \u00e1rea de um tri\u00e2ngulo \u00e9:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-d1fe72037bfa7f8a181f8f92fdeb5a93_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"A = \\cfrac{b \\cdot h}{2}\" title=\"Rendered by QuickLaTeX.com\" height=\"38\" width=\"70\" style=\"vertical-align: -12px;\"><\/p>\n<\/p>\n<p> No nosso caso, a base do tri\u00e2ngulo \u00e9 <em>x<\/em> e sua altura <em>\u00e9 y<\/em> . Ainda:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-fffb2f09f07fa450174fda22346dc38d_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"A = \\cfrac{x \\cdot y}{2}\" title=\"Rendered by QuickLaTeX.com\" height=\"34\" width=\"72\" style=\"vertical-align: -12px;\"><\/p>\n<\/p>\n<p> J\u00e1 temos a fun\u00e7\u00e3o de otimiza\u00e7\u00e3o, mas ela depende de duas vari\u00e1veis e s\u00f3 pode depender de uma. Por\u00e9m, o enunciado nos diz que as duas pernas devem totalizar 10 cm. Ainda:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-97009913e60b18a66b47683b142eaa14_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"x+ y = 10\" title=\"Rendered by QuickLaTeX.com\" height=\"16\" width=\"83\" style=\"vertical-align: -4px;\"><\/p>\n<\/p>\n<p> Resolvemos para <em>y<\/em> a partir desta equa\u00e7\u00e3o:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-9093174098ec3aeadce1fa9dd3724c27_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"y = 10 -x\" title=\"Rendered by QuickLaTeX.com\" height=\"16\" width=\"82\" style=\"vertical-align: -4px;\"><\/p>\n<\/p>\n<p> E substitu\u00edmos a express\u00e3o na fun\u00e7\u00e3o:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-d9ab3460a669f3a70a608b0a4c64f520_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"A = \\cfrac{x \\cdot y}{2} \\ \\xrightarrow{ y \\  = \\ 10 -x } \\ A = \\cfrac{x(10-x)}{2}\" title=\"Rendered by QuickLaTeX.com\" height=\"40\" width=\"278\" style=\"vertical-align: -12px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-28b512042d814bf88207c8fb0f9a5543_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"A(x) = \\cfrac{10x-x^2}{2}\" title=\"Rendered by QuickLaTeX.com\" height=\"41\" width=\"131\" style=\"vertical-align: -12px;\"><\/p>\n<\/p>\n<p> Agora temos a fun\u00e7\u00e3o de otimiza\u00e7\u00e3o planejada e ela depende apenas de uma vari\u00e1vel, ent\u00e3o podemos passar para a pr\u00f3xima etapa.<\/p>\n<p> <u style=\"text-decoration-color:#FF9B28;\"><strong><span style=\"color:#1976d2;\">Passo 2:<\/span><\/strong> Calcule a derivada da fun\u00e7\u00e3o a ser otimizada.<\/u><\/p>\n<p> \u00c9 uma fun\u00e7\u00e3o racional, ent\u00e3o aplicamos a f\u00f3rmula da derivada da divis\u00e3o para deriv\u00e1-la:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-0d555f298d7c31d09947787e4d294d59_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"A(x) = \\cfrac{10x-x^2}{2} \\ \\longrightarrow \\ A'(x) = \\cfrac{(10-2x) \\cdot 2 - (10x-x^2) \\cdot 0}{2^2}\" title=\"Rendered by QuickLaTeX.com\" height=\"41\" width=\"467\" style=\"vertical-align: -12px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-00979b72662361213ec4611a28e935a9_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"A'(x) = \\cfrac{20-4x}{4}\" title=\"Rendered by QuickLaTeX.com\" height=\"38\" width=\"126\" style=\"vertical-align: -12px;\"><\/p>\n<\/p>\n<p> <u style=\"text-decoration-color:#FF9B28;\"><strong><span style=\"color:#1976d2;\">Etapa 3:<\/span><\/strong> Encontre os pontos cr\u00edticos.<\/u><\/p>\n<p> Para encontrar os pontos cr\u00edticos da fun\u00e7\u00e3o, precisamos igualar a derivada a zero e resolver a equa\u00e7\u00e3o resultante:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-91e81f70cf7bd6388f6511629d203f7f_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"A'(x) = 0\" title=\"Rendered by QuickLaTeX.com\" height=\"19\" width=\"74\" style=\"vertical-align: -5px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-b5753c4fdb7aaae61182eac6fd15e52d_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\cfrac{20-4x}{4} =0\" title=\"Rendered by QuickLaTeX.com\" height=\"38\" width=\"91\" style=\"vertical-align: -12px;\"><\/p>\n<\/p>\n<p> O 4 divide todo o lado esquerdo, ent\u00e3o podemos multiplicar multiplicando todo o lado direito: <\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-23e63171262e022c8a63e717b31ed0b5_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"20-4x=0 \\cdot 4\" title=\"Rendered by QuickLaTeX.com\" height=\"12\" width=\"113\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-8e9b9b9a13bafb9de785e9cf5b3e86ca_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"20-4x=0\" title=\"Rendered by QuickLaTeX.com\" height=\"12\" width=\"91\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-e71409a99fca0883bc173a6df1b8c3af_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"-4x=-20\" title=\"Rendered by QuickLaTeX.com\" height=\"12\" width=\"87\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-66b399752305ae60043d99032ba112d9_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"x=\\cfrac{-20}{-4}\" title=\"Rendered by QuickLaTeX.com\" height=\"38\" width=\"76\" style=\"vertical-align: -12px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-8ddab230605c435eb8b7408a736d3e77_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"x=5\" title=\"Rendered by QuickLaTeX.com\" height=\"13\" width=\"42\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p> <u style=\"text-decoration-color:#FF9B28;\"><span style=\"color:#1976d2;\"><strong>Passo 4:<\/strong><\/span> Estude a monotonicidade da fun\u00e7\u00e3o e determine o m\u00e1ximo ou m\u00ednimo da fun\u00e7\u00e3o.<\/u><\/p>\n<p> Para estudar a monotonia da fun\u00e7\u00e3o, representamos o ponto cr\u00edtico encontrado \u00e0 direita: <\/p>\n<div class=\"wp-block-image\">\n<figure class=\"aligncenter size-large is-resized\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/uploads\/2023\/07\/nombre-ligne-5.webp\" alt=\"\" class=\"wp-image-2466\" width=\"227\" height=\"88\" srcset=\"\" sizes=\"auto, \" data-src=\"\"><\/figure>\n<\/div>\n<p> E agora avaliamos o sinal da derivada em cada intervalo para descobrir se a fun\u00e7\u00e3o \u00e9 crescente ou decrescente. Para fazer isso, pegamos um ponto em cada intervalo (nunca o ponto cr\u00edtico) e observamos qual sinal a derivada tem naquele ponto: <\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-00979b72662361213ec4611a28e935a9_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"A'(x) = \\cfrac{20-4x}{4}\" title=\"Rendered by QuickLaTeX.com\" height=\"38\" width=\"126\" style=\"vertical-align: -12px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-c9c9fbc0304871964748873af9e9918c_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"A'(0) = \\cfrac{20-4\\cdot0}{4} = \\cfrac{20}{4} = 5  \\  \\rightarrow \\ \\bm{+}\" title=\"Rendered by QuickLaTeX.com\" height=\"38\" width=\"264\" style=\"vertical-align: -12px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-05c98faeaeb88c5826a71a45b3e9c88a_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"A'(6) = \\cfrac{20-4\\cdot6}{4} = \\cfrac{20-24}{4} = \\cfrac{-4}{4} = -1   \\  \\rightarrow \\ \\bm{-}\" title=\"Rendered by QuickLaTeX.com\" height=\"38\" width=\"373\" style=\"vertical-align: -12px;\"><\/p>\n<\/p>\n<div class=\"wp-block-image\">\n<figure class=\"aligncenter size-large is-resized\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/uploads\/2023\/07\/ligne-numerique-5-positif-negatif.webp\" alt=\"\" class=\"wp-image-2467\" width=\"227\" height=\"160\" srcset=\"\" sizes=\"auto, \" data-src=\"\"><\/figure>\n<\/div>\n<p> Se a derivada for positiva, significa que a fun\u00e7\u00e3o est\u00e1 aumentando, e se a derivada for negativa, significa que a fun\u00e7\u00e3o est\u00e1 diminuindo. Portanto, os intervalos para aumentar e diminuir a fun\u00e7\u00e3o s\u00e3o:<\/p>\n<p class=\"has-text-align-center\"> <strong>Crescimento:<\/strong><\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-188a50f6279237f5c478c7116a509d49_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\bm{(-\\infty, 5)}\" title=\"Rendered by QuickLaTeX.com\" height=\"19\" width=\"60\" style=\"vertical-align: -5px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-center\"> <strong>Diminuir:<\/strong><\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-743f877080ce04389168296915ca6795_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\bm{(5,+\\infty)}\" title=\"Rendered by QuickLaTeX.com\" height=\"19\" width=\"61\" style=\"vertical-align: -5px;\"><\/p>\n<\/p>\n<p> Em x=5 a fun\u00e7\u00e3o vai de crescente para decrescente, ent\u00e3o <strong>x=5 \u00e9 um m\u00e1ximo relativo<\/strong> da fun\u00e7\u00e3o a ser otimizada <strong>.<\/strong><\/p>\n<p> Portanto, x=5 \u00e9 o valor do ramo do tri\u00e2ngulo que possui a \u00e1rea m\u00e1xima. Basta calcular o valor da outra perna:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-d396031e1a0eb65570a9f95d6701f9c1_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"y = 10 -x \\ \\xrightarrow{x \\ = \\ 5} \\ y = 10-5= \\bm{5}\" title=\"Rendered by QuickLaTeX.com\" height=\"23\" width=\"265\" style=\"vertical-align: -4px;\"><\/p>\n<\/p>\n<p> Concluindo, os valores que maximizam a \u00e1rea m\u00e1xima do tri\u00e2ngulo s\u00e3o: <\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-85b25208de94c730a3de04d6d428dc59_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\bm{x=5} \\ \\mathbf{cm}\" title=\"Rendered by QuickLaTeX.com\" height=\"13\" width=\"75\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-e7e41736bd949777d18ec2a79b0beb93_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\bm{y=5} \\ \\mathbf{cm}\" title=\"Rendered by QuickLaTeX.com\" height=\"17\" width=\"74\" style=\"vertical-align: -4px;\"><\/p>\n<\/p>\n<h2 class=\"wp-block-heading\"><span class=\"ez-toc-section\" id=\"problemas-de-optimizacion-resueltos\"><\/span> Problemas de otimiza\u00e7\u00e3o resolvidos<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<h3 class=\"wp-block-heading\"> Problema 1<\/h3>\n<p> O medicamento \u00e9 dado a uma pessoa doente e<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-b4e3cbf5d4c5c6d9b702dd139f14c147_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"t\" title=\"Rendered by QuickLaTeX.com\" height=\"12\" width=\"6\" style=\"vertical-align: 0px;\"><\/p>\n<p> algumas horas depois, a concentra\u00e7\u00e3o sangu\u00ednea do princ\u00edpio ativo \u00e9 dada pela fun\u00e7\u00e3o<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-57f138f86e1400954205e4b91ed3103c_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"c(t) = te^{\u2212t\/2}\" title=\"Rendered by QuickLaTeX.com\" height=\"21\" width=\"84\" style=\"vertical-align: -5px;\"><\/p>\n<p> miligramas por mililitro. Determine o valor m\u00e1ximo de<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-c4be04707471a329c8bde249ab6cf526_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"c(t)\" title=\"Rendered by QuickLaTeX.com\" height=\"19\" width=\"27\" style=\"vertical-align: -5px;\"><\/p>\n<p> e indica quando esse valor \u00e9 alcan\u00e7ado. <\/p>\n<div class=\"wp-block-otfm-box-spoiler-start otfm-sp__wrapper otfm-sp__box js-otfm-sp-box__closed otfm-sp__E6F9EF\" role=\"button\" tabindex=\"0\" aria-expanded=\"false\" data-otfm-spc=\"#E6F9EF\" style=\"text-align:center\">\n<div class=\"otfm-sp__title\"> <strong>Veja a solu\u00e7\u00e3o<\/strong><\/div>\n<\/div>\n<p class=\"has-text-align-left\"> <strong>Passo 1: Defina a fun\u00e7\u00e3o a ser otimizada.<\/strong><\/p>\n<p class=\"has-text-align-left\"> Neste problema, eles j\u00e1 nos d\u00e3o a fun\u00e7\u00e3o proposta, que \u00e9<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-02fb8044e28a7e9ba53c1eb64bfec693_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle c(t) = t e^{-t\/2} .\" title=\"Rendered by QuickLaTeX.com\" height=\"22\" width=\"100\" style=\"vertical-align: -5px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> <strong>Passo 2: Calcule a derivada da fun\u00e7\u00e3o a ser otimizada.<\/strong><\/p>\n<p class=\"has-text-align-left\"> A fun\u00e7\u00e3o \u00e9 composta pelo produto de 2 fun\u00e7\u00f5es. Portanto, para calcular a derivada da fun\u00e7\u00e3o, devemos aplicar a regra para a derivada de um produto: <\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-1975d832fe40c4c12a2a4986244e3c24_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle c'(t)=1 \\cdot e^{-t\/2} + t \\cdot e^{-t\/2} \\cdot \\left( \\frac{-t}{2} \\right)'= e^{-t\/2} + t e^{-t\/2} \\cdot  \\frac{-1}{2}\" title=\"Rendered by QuickLaTeX.com\" height=\"46\" width=\"430\" style=\"vertical-align: -17px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-7d730042a2ff5df46d19bce35fe24392_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"c'(t)=e^{-t\/2} + \\cfrac{-1}{2}t  e^{-t\/2}\" title=\"Rendered by QuickLaTeX.com\" height=\"38\" width=\"193\" style=\"vertical-align: -12px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> <strong>Etapa 3: Encontre os pontos cr\u00edticos.<\/strong><\/p>\n<p class=\"has-text-align-left\"> Para encontrar os pontos cr\u00edticos da fun\u00e7\u00e3o, resolvemos <\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-626f673c4823e1bd527925bfe91d558f_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"c'(t)=0:\" title=\"Rendered by QuickLaTeX.com\" height=\"19\" width=\"74\" style=\"vertical-align: -5px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-64818981c87104d6e0b400434ce53533_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"c'(t)=0\" title=\"Rendered by QuickLaTeX.com\" height=\"19\" width=\"65\" style=\"vertical-align: -5px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-6d9ebad9fa6bc78dc99b358f4786999f_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle e^{-t\/2} + \\frac{-1}{2}t  e^{-t\/2}=0\" title=\"Rendered by QuickLaTeX.com\" height=\"36\" width=\"164\" style=\"vertical-align: -12px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Tomamos o fator comum para resolver a equa\u00e7\u00e3o:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-410004b26bf5de6dc03c12491bee91b9_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle e^{-t\/2} \\left(1 - \\frac{1}{2}t \\right) = 0\" title=\"Rendered by QuickLaTeX.com\" height=\"43\" width=\"150\" style=\"vertical-align: -17px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Para que a multiplica\u00e7\u00e3o seja igual a 0, um dos dois elementos da multiplica\u00e7\u00e3o deve ser zero. Portanto, definimos cada fator igual a 0:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-8390139724dbc4ad014db2a76e508290_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle e^{-t\/2}\\cdot \\left(1 - \\frac{1}{2}t \\right) = 0 \\longrightarrow \\begin{cases} e^{-t\/2}=0 \\ \\bm{\\times} \\\\[2ex]\\displaystyle 1 - \\frac{1}{2}t=0 \\ \\longrightarrow \\ 1= \\frac{1}{2}t \\ \\longrightarrow \\ \\bm{2=t} \\end{cases}\" title=\"Rendered by QuickLaTeX.com\" height=\"77\" width=\"486\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Um n\u00famero elevado a outro n\u00famero nunca pode dar 0, portanto,<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-10b6a1e4b24bf63d08aeff4da0374d25_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"e^{-t\/2}=0\" title=\"Rendered by QuickLaTeX.com\" height=\"16\" width=\"71\" style=\"vertical-align: 0px;\"><\/p>\n<p> N\u00e3o h\u00e1 solu\u00e7\u00e3o.<\/p>\n<p class=\"has-text-align-left\"> <strong>Passo 4: Estude a monotonicidade da fun\u00e7\u00e3o e determine o m\u00e1ximo ou m\u00ednimo da fun\u00e7\u00e3o.<\/strong><\/p>\n<p class=\"has-text-align-left\"> Para estudar a monotonia da fun\u00e7\u00e3o, representamos o ponto cr\u00edtico encontrado \u00e0 direita: <\/p>\n<div class=\"wp-block-image\">\n<figure class=\"aligncenter size-large is-resized\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/uploads\/2023\/07\/nombre-ligne-2.webp\" alt=\"\" class=\"wp-image-2469\" width=\"204\" height=\"79\" srcset=\"\" sizes=\"auto, \" data-src=\"\"><\/figure>\n<\/div>\n<p class=\"has-text-align-left\"> E agora avaliamos o sinal da derivada em cada intervalo, para descobrir se a fun\u00e7\u00e3o \u00e9 crescente ou decrescente. Portanto, pegamos um ponto em cada intervalo (nunca o ponto cr\u00edtico) e observamos qual sinal a derivada tem neste ponto:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-904d140578309f45e9c53d2a9a35e32d_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle c'(0)=e^{-0\/2} + \\frac{-1}{2}\\cdot 0 \\cdot e^{-0\/2} = e^0 +\\frac{-1}{2}\\cdot 0 \\cdot e^{0} = 1 + 0 = 1 \\ \\rightarrow \\ \\bm{+}\" title=\"Rendered by QuickLaTeX.com\" height=\"36\" width=\"508\" style=\"vertical-align: -12px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-ada4b79153afee560564a81df8d0c46f_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle c'(3)=e^{-3\/2} + \\frac{-1}{2}(3)e^{-3\/2} = 0,22-0,33 = -0,11 \\ \\rightarrow \\ \\bm{-}\" title=\"Rendered by QuickLaTeX.com\" height=\"36\" width=\"449\" style=\"vertical-align: -12px;\"><\/p>\n<\/p>\n<div class=\"wp-block-image\">\n<figure class=\"aligncenter size-large is-resized\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/uploads\/2023\/07\/droite-numerique-deux-positif-negatif.webp\" alt=\"\" class=\"wp-image-2472\" width=\"202\" height=\"143\" srcset=\"\" sizes=\"auto, \" data-src=\"\"><\/figure>\n<\/div>\n<p class=\"has-text-align-left\"> Se a derivada for positiva significa que a fun\u00e7\u00e3o aumenta, por outro lado se a derivada for negativa significa que a fun\u00e7\u00e3o diminui. Assim os intervalos de crescimento e diminui\u00e7\u00e3o da fun\u00e7\u00e3o a ser otimizada s\u00e3o:<\/p>\n<p class=\"has-text-align-center\"> <strong>Crescimento:<\/strong><\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-3868632c123ece40100b3a40c266cc25_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\bm{(-\\infty,2)}\" title=\"Rendered by QuickLaTeX.com\" height=\"19\" width=\"60\" style=\"vertical-align: -5px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-center\"> <strong>Diminuir:<\/strong><\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-4c9901d11cc4672cab7e1e2a6de08ce4_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\bm{(2,+\\infty)}\" title=\"Rendered by QuickLaTeX.com\" height=\"19\" width=\"61\" style=\"vertical-align: -5px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> A fun\u00e7\u00e3o vai de crescente a decrescente em t=2, ent\u00e3o <strong>t=2 \u00e9 o m\u00e1ximo<\/strong> da fun\u00e7\u00e3o. A concentra\u00e7\u00e3o m\u00e1xima ser\u00e1, portanto, alcan\u00e7ada em <strong>t=2 horas.<\/strong><\/p>\n<p class=\"has-text-align-left\"> Finalmente, substitu\u00edmos o valor em que ocorre o m\u00e1ximo na fun\u00e7\u00e3o original, para encontrar o valor da concentra\u00e7\u00e3o m\u00e1xima: <\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-dd8123e6fd3fe714d1784775375011f2_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"c(2) = 2 \\cdot e^{-2\/2} = 2\\cdot e^{-1} = 2 \\cdot 0,37 = \\bm{0,74} \\ \\mathbf{mg\/ml}\" title=\"Rendered by QuickLaTeX.com\" height=\"21\" width=\"391\" style=\"vertical-align: -5px;\"><\/p>\n<\/p>\n<div class=\"wp-block-otfm-box-spoiler-end otfm-sp_end\"><\/div>\n<h3 class=\"wp-block-heading\">Problema 2<\/h3>\n<p> Uma loja espera vender 40 scooters el\u00e9tricas ao pre\u00e7o de 1.000 euros por scooter. Mas, de acordo com estudos de mercado, por cada redu\u00e7\u00e3o de 50\u20ac no pre\u00e7o das scooters, haver\u00e1 um aumento nas vendas das 10 scooters mais vendidas.<\/p>\n<p> Primeiro, escreva a fun\u00e7\u00e3o de receita da loja com base no n\u00famero de vezes que o pre\u00e7o original de US$ 1.000 da scooter \u00e9 reduzido em US$ 50. Em seguida, determine o pre\u00e7o da scooter para obter o lucro m\u00e1ximo e a receita obtida com esse pre\u00e7o. <\/p>\n<div class=\"wp-block-otfm-box-spoiler-start otfm-sp__wrapper otfm-sp__box js-otfm-sp-box__closed otfm-sp__E6F9EF\" role=\"button\" tabindex=\"0\" aria-expanded=\"false\" data-otfm-spc=\"#E6F9EF\" style=\"text-align:center\">\n<div class=\"otfm-sp__title\"> <strong>Veja a solu\u00e7\u00e3o<\/strong><\/div>\n<\/div>\n<p class=\"has-text-align-left\"> <strong>Passo 1: Defina a fun\u00e7\u00e3o a ser otimizada.<\/strong><\/p>\n<p class=\"has-text-align-left\"> A defini\u00e7\u00e3o do problema nos d\u00e1 uma pista, pois nos diz que a fun\u00e7\u00e3o deve depender do n\u00famero de vezes que o pre\u00e7o inicial \u00e9 reduzido em US$ 50. Chamaremos, portanto, de x o n\u00famero de vezes que o pre\u00e7o \u00e9 reduzido em 50\u20ac:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-7d0357e75c58ab161e387dae18d3a6f1_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"x= \\text{N\\'umero de veces que se rebaja el precio 50}\" title=\"Rendered by QuickLaTeX.com\" height=\"17\" width=\"368\" style=\"vertical-align: -4px;\"><\/p>\n<p> \u20ac<\/p>\n<p class=\"has-text-align-left\"> A fun\u00e7\u00e3o receita ser\u00e1 o n\u00famero de scooters vendidas multiplicado pelo pre\u00e7o de cada scooter:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-f04d6d19de5d7a7ad8dceadfbbdbdbe2_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"I(x)= \\text{N\\'umero patintetes vendidos} \\cdot \\text{Precio de cada patinete}\" title=\"Rendered by QuickLaTeX.com\" height=\"19\" width=\"470\" style=\"vertical-align: -5px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> O n\u00famero de scooters vendidas ser\u00e1 de 40 mais 10 scooters por cada redu\u00e7\u00e3o de pre\u00e7o de 50\u20ac. Ainda:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-88b7464372c9683d065acbdbe2598cbe_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\text{N\\'umero patintetes vendidos} = 40 + 10x\" title=\"Rendered by QuickLaTeX.com\" height=\"16\" width=\"308\" style=\"vertical-align: -4px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> O pre\u00e7o de cada scooter ser\u00e1 de 1000\u20ac no in\u00edcio e diminuir\u00e1 50\u20ac a cada descida de pre\u00e7o. Ainda:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-5c22a7a64d350ea748ad2b4b3decd217_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\text{Precio de cada patinete} = 1000 -50x\" title=\"Rendered by QuickLaTeX.com\" height=\"17\" width=\"291\" style=\"vertical-align: -4px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> A fun\u00e7\u00e3o para otimizar o problema \u00e9, portanto: <\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-f04d6d19de5d7a7ad8dceadfbbdbdbe2_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"I(x)= \\text{N\\'umero patintetes vendidos} \\cdot \\text{Precio de cada patinete}\" title=\"Rendered by QuickLaTeX.com\" height=\"19\" width=\"470\" style=\"vertical-align: -5px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-2cbffd131757d106f869b562d4e35c4c_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"I(x)= (40 + 10x) \\cdot (1000-50x)\" title=\"Rendered by QuickLaTeX.com\" height=\"19\" width=\"249\" style=\"vertical-align: -5px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-0ae987678d7542e5d127fc02f8bf4daf_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"I(x)= 40000-2000x+10000x-500x^2\" title=\"Rendered by QuickLaTeX.com\" height=\"20\" width=\"310\" style=\"vertical-align: -5px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-902cc0a5c209ed48e3d93432b5550b1d_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"I(x)= -500x^2+8000x+40000\" title=\"Rendered by QuickLaTeX.com\" height=\"20\" width=\"249\" style=\"vertical-align: -5px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> <strong>Passo 2: Calcule a derivada da fun\u00e7\u00e3o a ser otimizada.<\/strong><\/p>\n<p class=\"has-text-align-left\"> Por ser uma fun\u00e7\u00e3o polinomial, a derivada \u00e9 mais f\u00e1cil de calcular:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-acb8d705be4d3a30be83d2800f4d2403_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"I(x)= -500x^2+8000x+40000\\ \\longrightarrow \\ I'(x)= -1000x+8000\" title=\"Rendered by QuickLaTeX.com\" height=\"20\" width=\"477\" style=\"vertical-align: -5px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> <strong>Etapa 3: Encontre os pontos cr\u00edticos da fun\u00e7\u00e3o.<\/strong><\/p>\n<p class=\"has-text-align-left\"> Para encontrar os pontos cr\u00edticos da fun\u00e7\u00e3o, resolvemos <\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-b55b4fb6ca5af90f829cb082893a5519_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"I'(x)=0:\" title=\"Rendered by QuickLaTeX.com\" height=\"19\" width=\"79\" style=\"vertical-align: -5px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-8b48142b0c92fea1f9dd87af211ada6b_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"I'(x)=0\" title=\"Rendered by QuickLaTeX.com\" height=\"19\" width=\"70\" style=\"vertical-align: -5px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-5a0d58fd339d02e3f71b7fe031726cad_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"-1000x+8000=0\" title=\"Rendered by QuickLaTeX.com\" height=\"14\" width=\"148\" style=\"vertical-align: -2px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-a03bae4e019041d174c51fced30954ef_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"-1000x=-8000\" title=\"Rendered by QuickLaTeX.com\" height=\"12\" width=\"132\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-e67f8bf53c9c10c3384f729890f57b4d_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"x=\\cfrac{-8000}{-1000} = 8\" title=\"Rendered by QuickLaTeX.com\" height=\"38\" width=\"125\" style=\"vertical-align: -12px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> <strong>Passo 4: Estude a monotonicidade da fun\u00e7\u00e3o e determine o m\u00e1ximo ou m\u00ednimo da fun\u00e7\u00e3o.<\/strong><\/p>\n<p class=\"has-text-align-left\"> Para estudar a monotonicidade da fun\u00e7\u00e3o, representamos o ponto cr\u00edtico calculado na reta num\u00e9rica: <\/p>\n<div class=\"wp-block-image\">\n<figure class=\"aligncenter size-large is-resized\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/uploads\/2023\/07\/nombre-ligne-8.webp\" alt=\"\" class=\"wp-image-2481\" width=\"234\" height=\"91\" srcset=\"\" sizes=\"auto, \" data-src=\"\"><\/figure>\n<\/div>\n<p class=\"has-text-align-left\"> E agora avaliamos o sinal da derivada em cada intervalo, para descobrir se a fun\u00e7\u00e3o \u00e9 crescente ou decrescente. Portanto, pegamos um ponto em cada intervalo (nunca o ponto cr\u00edtico) e observamos qual sinal a derivada tem neste ponto:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-3cf47c71ae832d734c6e5f525efa05b0_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"I'(0)= -1000\\cdot 0+8000=8000 \\ \\rightarrow \\ \\bm{+}\" title=\"Rendered by QuickLaTeX.com\" height=\"19\" width=\"300\" style=\"vertical-align: -5px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-9df9f5fdff3d4ba6dcd7f17c68b16c2f_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"I'(10)= -1000\\cdot 10+8000=-2000 \\ \\rightarrow \\ \\bm{-}\" title=\"Rendered by QuickLaTeX.com\" height=\"19\" width=\"331\" style=\"vertical-align: -5px;\"><\/p>\n<\/p>\n<div class=\"wp-block-image\">\n<figure class=\"aligncenter size-large is-resized\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/uploads\/2023\/07\/ligne-numerique-8-positif-negatif.webp\" alt=\"\" class=\"wp-image-2482\" width=\"239\" height=\"169\" srcset=\"\" sizes=\"auto, \" data-src=\"\"><\/figure>\n<\/div>\n<p class=\"has-text-align-left\"> Se a derivada for positiva, significa que a fun\u00e7\u00e3o est\u00e1 aumentando, e se a derivada for negativa, significa que a fun\u00e7\u00e3o est\u00e1 diminuindo. Portanto, os intervalos de crescimento e decl\u00ednio s\u00e3o:<\/p>\n<p class=\"has-text-align-center\"> <strong>Crescimento:<\/strong><\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-d85d36c78a812c7c5a8f127bb5145c42_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\bm{(-\\infty,8)}\" title=\"Rendered by QuickLaTeX.com\" height=\"19\" width=\"60\" style=\"vertical-align: -5px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-center\"> <strong>Diminuir:<\/strong><\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-ba6980946255650ee8fe4dbc4c87c681_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\bm{(8,+\\infty)}\" title=\"Rendered by QuickLaTeX.com\" height=\"19\" width=\"61\" style=\"vertical-align: -5px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> A fun\u00e7\u00e3o vai de crescente a decrescente em x=8, ent\u00e3o <strong>x=8 \u00e9 o m\u00e1ximo<\/strong> da fun\u00e7\u00e3o. Portanto, o rendimento m\u00e1ximo ser\u00e1 obtido fazendo <strong>8 vezes a redu\u00e7\u00e3o de 50\u20ac.<\/strong><\/p>\n<p class=\"has-text-align-left\"> Substitu\u00edmos agora o valor em que o rendimento m\u00e1ximo aparece na fun\u00e7\u00e3o original, para encontrar o valor do rendimento m\u00e1ximo:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-902cc0a5c209ed48e3d93432b5550b1d_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"I(x)= -500x^2+8000x+40000\" title=\"Rendered by QuickLaTeX.com\" height=\"20\" width=\"249\" style=\"vertical-align: -5px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-fce0d14898ab44824c50e758de24099d_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"I(8)= -500\\cdot 8^2+8000\\cdot 8+40000 = \\bm{72000}\" title=\"Rendered by QuickLaTeX.com\" height=\"20\" width=\"339\" style=\"vertical-align: -5px;\"><\/p>\n<p> <strong>\u20ac<\/strong><\/p>\n<p class=\"has-text-align-left\"> E o pre\u00e7o de cada scooter depois de ter feito o desconto de 50\u20ac 8 vezes ser\u00e1:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-5c22a7a64d350ea748ad2b4b3decd217_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\text{Precio de cada patinete} = 1000 -50x\" title=\"Rendered by QuickLaTeX.com\" height=\"17\" width=\"291\" style=\"vertical-align: -4px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-a4646c3d785d36e70ba549b9672b697c_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\text{Precio de cada patinete} = 1000 -50\\cdot 8=\\bm{600}\" title=\"Rendered by QuickLaTeX.com\" height=\"17\" width=\"353\" style=\"vertical-align: -4px;\"><\/p>\n<p> <strong>\u20ac<\/strong><\/p>\n<div class=\"wp-block-otfm-box-spoiler-end otfm-sp_end\"><\/div>\n<h3 class=\"wp-block-heading\"> Problema 3<\/h3>\n<p> A fun\u00e7\u00e3o custo (em milhares de euros) de uma empresa pode ser determinada atrav\u00e9s da seguinte express\u00e3o:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-5c84c0900acd506ad979886e5b5f10ff_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"f(x)=40-6x+x^2, \\quad x \\ge  0\" title=\"Rendered by QuickLaTeX.com\" height=\"20\" width=\"224\" style=\"vertical-align: -5px;\"><\/p>\n<\/p>\n<p> Ouro<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-ede05c264bba0eda080918aaa09c4658_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"x\" title=\"Rendered by QuickLaTeX.com\" height=\"8\" width=\"10\" style=\"vertical-align: 0px;\"><\/p>\n<p> representa os milhares de unidades produzidas de um determinado item.<\/p>\n<p> Determine quanto deve ser produzido para que o custo seja m\u00ednimo, qual \u00e9 esse custo e qual seria o custo se nenhum desses itens fosse produzido. <\/p>\n<div class=\"wp-block-otfm-box-spoiler-start otfm-sp__wrapper otfm-sp__box js-otfm-sp-box__closed otfm-sp__E6F9EF\" role=\"button\" tabindex=\"0\" aria-expanded=\"false\" data-otfm-spc=\"#E6F9EF\" style=\"text-align:center\">\n<div class=\"otfm-sp__title\"> <strong>Veja a solu\u00e7\u00e3o<\/strong><\/div>\n<\/div>\n<p class=\"has-text-align-left\"> <strong>Passo 1: Defina a fun\u00e7\u00e3o a ser otimizada.<\/strong><\/p>\n<p class=\"has-text-align-left\"> A declara\u00e7\u00e3o do problema j\u00e1 nos fornece a fun\u00e7\u00e3o a ser otimizada, que \u00e9<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-6dc6eb92c2776f8026c348e7e5824d09_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle f(x)=40-6x+x^2 .\" title=\"Rendered by QuickLaTeX.com\" height=\"21\" width=\"160\" style=\"vertical-align: -5px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> <strong>Passo 2: Calcule a derivada da fun\u00e7\u00e3o a ser otimizada.<\/strong><\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-5056826599d14c0ff8aa3ec134a68b0f_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"f(x)=40-6x+x^2 \\ \\longrightarrow \\ f'(x)=-6+2x\" title=\"Rendered by QuickLaTeX.com\" height=\"20\" width=\"332\" style=\"vertical-align: -5px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> <strong>Etapa 3: Encontre os pontos cr\u00edticos.<\/strong><\/p>\n<p class=\"has-text-align-left\"> Para encontrar os pontos cr\u00edticos da fun\u00e7\u00e3o, resolvemos <\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-58dcd049349f740f082d583dfd9e364c_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"f'(x)=0:\" title=\"Rendered by QuickLaTeX.com\" height=\"19\" width=\"80\" style=\"vertical-align: -5px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-36700780d306ccf4975387990b1949fb_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"f'(x)=0\" title=\"Rendered by QuickLaTeX.com\" height=\"19\" width=\"72\" style=\"vertical-align: -5px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-ecec21f50b0875af5ebb5c55ef5e2502_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"-6+2x=0\" title=\"Rendered by QuickLaTeX.com\" height=\"14\" width=\"95\" style=\"vertical-align: -2px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-d47687d0a8cb82ab26be7eadf0d7f3c8_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"2x=6\" title=\"Rendered by QuickLaTeX.com\" height=\"12\" width=\"52\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-6f1ffe509b34b0d560397fbc1859cb64_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"x=\\cfrac{6}{2} = 3\" title=\"Rendered by QuickLaTeX.com\" height=\"38\" width=\"77\" style=\"vertical-align: -12px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> <strong>Passo 4: Estude a monotonicidade da fun\u00e7\u00e3o e determine o m\u00e1ximo ou m\u00ednimo da fun\u00e7\u00e3o.<\/strong><\/p>\n<p class=\"has-text-align-left\"> Representamos o ponto cr\u00edtico encontrado \u00e0 direita: <\/p>\n<div class=\"wp-block-image\">\n<figure class=\"aligncenter size-large is-resized\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/uploads\/2023\/07\/ligne-numerique-3.webp\" alt=\"\" class=\"wp-image-2485\" width=\"231\" height=\"90\" srcset=\"\" sizes=\"auto, \" data-src=\"\"><\/figure>\n<\/div>\n<p class=\"has-text-align-left\"> E agora avaliamos o sinal da derivada em cada intervalo, para descobrir se a fun\u00e7\u00e3o \u00e9 crescente ou decrescente. Portanto, pegamos um ponto em cada intervalo (nunca o ponto cr\u00edtico) e observamos qual sinal a derivada tem neste ponto:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-2647164feab3d1bdc3e7cdb14c123294_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"f'(0)=-6+2\\cdot 0=-6\\ \\rightarrow \\ \\bm{-}\" title=\"Rendered by QuickLaTeX.com\" height=\"19\" width=\"235\" style=\"vertical-align: -5px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-7fe3f1add3273a0ba4d0418a3b9788a1_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"f'(4)=-6+2\\cdot 4=-6+8=2\\ \\rightarrow \\ \\bm{+}\" title=\"Rendered by QuickLaTeX.com\" height=\"19\" width=\"298\" style=\"vertical-align: -5px;\"><\/p>\n<\/p>\n<div class=\"wp-block-image\">\n<figure class=\"aligncenter size-large is-resized\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/uploads\/2023\/07\/ligne-numerique-3-negatif-positif.webp\" alt=\"\" class=\"wp-image-2486\" width=\"231\" height=\"163\" srcset=\"\" sizes=\"auto, \" data-src=\"\"><\/figure>\n<\/div>\n<p class=\"has-text-align-left\"> Se a derivada for maior que zero, a fun\u00e7\u00e3o aumenta nesse intervalo. Por outro lado, se a derivada for menor que zero, a fun\u00e7\u00e3o diminui neste intervalo. Assim, os intervalos de aumento e diminui\u00e7\u00e3o da fun\u00e7\u00e3o s\u00e3o:<\/p>\n<p class=\"has-text-align-center\"> <strong>Crescimento:<\/strong><\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-fbc0e31e2b07080a4f83c2a34053c0f1_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\bm{(3,+\\infty)}\" title=\"Rendered by QuickLaTeX.com\" height=\"19\" width=\"61\" style=\"vertical-align: -5px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-center\"> <strong>Diminuir:<\/strong><\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-a890e45091b2c4c9115051361c0a1a2c_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\bm{(-\\infty,3)}\" title=\"Rendered by QuickLaTeX.com\" height=\"19\" width=\"60\" style=\"vertical-align: -5px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> A fun\u00e7\u00e3o vai de decrescente para crescente em x=3, ent\u00e3o <strong>x=3 \u00e9 o m\u00ednimo<\/strong> da fun\u00e7\u00e3o. Portanto, o custo m\u00ednimo ser\u00e1 alcan\u00e7ado com a produ\u00e7\u00e3o <strong>de 3.000 unidades.<\/strong><\/p>\n<p class=\"has-text-align-left\"> Agora substitu\u00edmos o valor pelo qual o custo m\u00ednimo \u00e9 alcan\u00e7ado na fun\u00e7\u00e3o original para encontrar o valor do custo m\u00ednimo:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-4a1c84d3c81fd4c7ebbecc922cc2288e_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"f(3)=40-6\\cdot 3+3^2=\\bm{31}\" title=\"Rendered by QuickLaTeX.com\" height=\"20\" width=\"206\" style=\"vertical-align: -5px;\"><\/p>\n<p> milh\u00f5es de euros.<\/p>\n<p class=\"has-text-align-left\"> Por outro lado, perguntam-nos qual seria o custo se nada fosse produzido, ou seja, quando<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-aa5140cb12e100167e56a99c53750148_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"x= 0 .\" title=\"Rendered by QuickLaTeX.com\" height=\"12\" width=\"47\" style=\"vertical-align: 0px;\"><\/p>\n<p> \u00c9 necess\u00e1rio, portanto, calcular<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-8f92d7beea0ed3a053927c2d429d3450_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"f(0):\" title=\"Rendered by QuickLaTeX.com\" height=\"19\" width=\"42\" style=\"vertical-align: -5px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-a7051466b3981f72509f3e5e2aa1d2f7_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"f(0)=40-6\\cdot 0+0^2=   \\bm{40}\" title=\"Rendered by QuickLaTeX.com\" height=\"20\" width=\"207\" style=\"vertical-align: -5px;\"><\/p>\n<p> milh\u00f5es de euros.<\/p>\n<div class=\"wp-block-otfm-box-spoiler-end otfm-sp_end\"><\/div>\n<h3 class=\"wp-block-heading\"> Problema 4<\/h3>\n<p> Queremos construir uma moldura retangular de madeira que delimite uma \u00e1rea de 2 m <sup>2<\/sup> . Sabemos que o pre\u00e7o da madeira \u00e9 de 7,5\u20ac\/m para as laterais horizontais e de 12,5\u20ac\/m para as laterais verticais. Determine as dimens\u00f5es que o ret\u00e2ngulo deve ter para que o custo total da moldura seja o m\u00ednimo poss\u00edvel e esse custo seja m\u00ednimo. <\/p>\n<div class=\"wp-block-otfm-box-spoiler-start otfm-sp__wrapper otfm-sp__box js-otfm-sp-box__closed otfm-sp__E6F9EF\" role=\"button\" tabindex=\"0\" aria-expanded=\"false\" data-otfm-spc=\"#E6F9EF\" style=\"text-align:center\">\n<div class=\"otfm-sp__title\"> <strong>Veja a solu\u00e7\u00e3o<\/strong><\/div>\n<\/div>\n<p class=\"has-text-align-left\"> <strong>Passo 1: Defina a fun\u00e7\u00e3o a ser otimizada.<\/strong><\/p>\n<p class=\"has-text-align-left\"> Para resolver o problema, chamaremos o lado horizontal <em>de x<\/em> e o lado vertical <em>de y<\/em> : <\/p>\n<div class=\"wp-block-image\">\n<figure class=\"aligncenter size-large is-resized\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/uploads\/2023\/07\/rectangle-fonction-optimisation-problemes.webp\" alt=\"problemas de otimiza\u00e7\u00e3o de fun\u00e7\u00e3o ret\u00e2ngulo\" class=\"wp-image-2490\" width=\"166\" height=\"188\" srcset=\"\" sizes=\"auto, \" data-src=\"\"><\/figure>\n<\/div>\n<p class=\"has-text-align-left\"> Comprar um lado horizontal custa 7,5\u20ac e comprar um lado vertical custa 12,5\u20ac. Al\u00e9m disso, para cada quadro precisamos de dois lados horizontais e dois lados verticais. Portanto, o custo do quadro pode ser determinado com a seguinte fun\u00e7\u00e3o:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-8485280ab046816dfbd587b192901fb9_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"C(x,y)= 7,5\\cdot 2x+12,5 \\cdot 2y = 15x +25y\" title=\"Rendered by QuickLaTeX.com\" height=\"19\" width=\"323\" style=\"vertical-align: -5px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> J\u00e1 temos a fun\u00e7\u00e3o de otimizar. Mas depende de duas vari\u00e1veis quando s\u00f3 pode depender de uma. Por\u00e9m, o comunicado diz-nos que a superf\u00edcie da moldura deve ser de 2 m <sup>2<\/sup> . Ainda:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-13ea4da7e9b4674fdd1a733e0e3cd26e_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"x \\cdot y = 2\" title=\"Rendered by QuickLaTeX.com\" height=\"16\" width=\"64\" style=\"vertical-align: -4px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Exclu\u00edmos a vari\u00e1vel <em>y<\/em> :<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-e00a5c6da8de49964d27f2941dac8a4b_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"y =\\cfrac{2}{x}\" title=\"Rendered by QuickLaTeX.com\" height=\"38\" width=\"46\" style=\"vertical-align: -12px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> E substitu\u00edmos a express\u00e3o encontrada na fun\u00e7\u00e3o a ser otimizada: <\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-2cb9add65869ac746532ac4bf1537eb1_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"C(x,y)= 15x +25y\\ \\xrightarrow{y \\ = \\ \\frac{2}{x} } \\ C(x)= 15x+25\\left(\\cfrac{2}{x} \\right)\" title=\"Rendered by QuickLaTeX.com\" height=\"54\" width=\"396\" style=\"vertical-align: -23px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-e232dbf6025d3b8d6593e4157cd09bff_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"C(x)= 15x+\\cfrac{50}{x} =15x +50x^{-1}\" title=\"Rendered by QuickLaTeX.com\" height=\"39\" width=\"249\" style=\"vertical-align: -12px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> <strong>Passo 2: Calcule a derivada da fun\u00e7\u00e3o a ser otimizada.<\/strong><\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-4ef3647de078c1f760dc37a5d3d6b68e_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"C(x)=15x +50x^{-1} \\ \\longrightarrow \\ C'(x)=15 -50x^{-2}\" title=\"Rendered by QuickLaTeX.com\" height=\"20\" width=\"359\" style=\"vertical-align: -5px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> <strong>Etapa 3: Encontre os pontos cr\u00edticos.<\/strong><\/p>\n<p class=\"has-text-align-left\"> Para encontrar os pontos cr\u00edticos da fun\u00e7\u00e3o, resolvemos <\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-97f234a562c87b78e257fc7953dee7ea_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"C'(x)=0:\" title=\"Rendered by QuickLaTeX.com\" height=\"19\" width=\"84\" style=\"vertical-align: -5px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-e8732228827d41a41af6935b1f0bdc0d_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"C'(x)=0\" title=\"Rendered by QuickLaTeX.com\" height=\"19\" width=\"75\" style=\"vertical-align: -5px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-cdaf71028a32e2818a7ec3376338df75_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"15 -50x^{-2}=0\" title=\"Rendered by QuickLaTeX.com\" height=\"15\" width=\"117\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-8bf1bee39e8e6e8f54b838f7dc8810d8_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"15 =50x^{-2}\" title=\"Rendered by QuickLaTeX.com\" height=\"15\" width=\"86\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-d4f9337813bfb8c7bf0d09f9889514c3_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"15=\\cfrac{50}{x^2}\" title=\"Rendered by QuickLaTeX.com\" height=\"39\" width=\"60\" style=\"vertical-align: -12px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-0aea588515cc1c895b2ca5030cd55ffc_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\cfrac{15}{1}=\\cfrac{50}{x^2}\" title=\"Rendered by QuickLaTeX.com\" height=\"39\" width=\"61\" style=\"vertical-align: -12px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Multiplicamos transversalmente para resolver a equa\u00e7\u00e3o com fra\u00e7\u00f5es: <\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-99a602636bc09eb630e637ec6454c8ce_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"15 \\cdot x^2 = 50 \\cdot 1\" title=\"Rendered by QuickLaTeX.com\" height=\"15\" width=\"110\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-dcdd7918a1af31e63925cb694da7d6be_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"x^2 = \\cfrac{50}{15}\" title=\"Rendered by QuickLaTeX.com\" height=\"39\" width=\"61\" style=\"vertical-align: -12px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-e9090b3956470b16769ec2ae441f2d30_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"x^2 = 3,33\" title=\"Rendered by QuickLaTeX.com\" height=\"19\" width=\"76\" style=\"vertical-align: -4px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-92abf359db44a399822919b1bb62e7cf_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\sqrt{x^2} = \\sqrt{3,33}\" title=\"Rendered by QuickLaTeX.com\" height=\"21\" width=\"105\" style=\"vertical-align: -4px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-b87ffa5999c0bbea78cc07d6a412460a_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"x = 1,83\" title=\"Rendered by QuickLaTeX.com\" height=\"16\" width=\"69\" style=\"vertical-align: -4px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> <strong>Passo 4: Estude a monotonicidade da fun\u00e7\u00e3o e determine o m\u00e1ximo ou m\u00ednimo da fun\u00e7\u00e3o.<\/strong><\/p>\n<p class=\"has-text-align-left\"> Representamos o ponto cr\u00edtico encontrado para analisar a monotonia da fun\u00e7\u00e3o na reta: <\/p>\n<div class=\"wp-block-image\">\n<figure class=\"aligncenter size-large is-resized\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/uploads\/2023\/07\/nombre-ligne-183.webp\" alt=\"\" class=\"wp-image-2491\" width=\"206\" height=\"80\" srcset=\"\" sizes=\"auto, \" data-src=\"\"><\/figure>\n<\/div>\n<p class=\"has-text-align-left\"> E agora avaliamos o sinal da derivada em cada intervalo, para descobrir se a fun\u00e7\u00e3o \u00e9 crescente ou decrescente. Portanto, pegamos um ponto em cada intervalo (nunca o ponto cr\u00edtico) e observamos qual sinal a derivada tem neste ponto:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-64f2221e9103ecdcb21609567e990135_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"f'(1)=15 -50\\cdot 1^{-2} = 15-50 = -35 \\ \\rightarrow \\ \\bm{-}\" title=\"Rendered by QuickLaTeX.com\" height=\"20\" width=\"347\" style=\"vertical-align: -5px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-ad12501e06c6997d1555df684480d38b_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"f'(2)=15 -50\\cdot 2^{-2} = 15-12,5 = 2,5 \\ \\rightarrow \\ \\bm{+}\" title=\"Rendered by QuickLaTeX.com\" height=\"20\" width=\"358\" style=\"vertical-align: -5px;\"><\/p>\n<\/p>\n<div class=\"wp-block-image\">\n<figure class=\"aligncenter size-large is-resized\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/uploads\/2023\/07\/droite-numerique-183-negatif-positif.webp\" alt=\"\" class=\"wp-image-2493\" width=\"206\" height=\"146\" srcset=\"\" sizes=\"auto, \" data-src=\"\"><\/figure>\n<\/div>\n<p class=\"has-text-align-left\"> Se a derivada for positiva, significa que a fun\u00e7\u00e3o est\u00e1 aumentando, e se a derivada for negativa, significa que a fun\u00e7\u00e3o est\u00e1 diminuindo. Portanto, os intervalos de crescimento e decl\u00ednio s\u00e3o:<\/p>\n<p class=\"has-text-align-center\"> <strong>Crescimento:<\/strong><\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-8d8e4f82a46ac576b10134beda86f0bb_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\bm{(1,83,+\\infty)}\" title=\"Rendered by QuickLaTeX.com\" height=\"19\" width=\"86\" style=\"vertical-align: -5px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-center\"> <strong>Diminuir:<\/strong><\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-57f0911e594b424ad7908294a15f1b84_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\bm{(-\\infty,1,83)}\" title=\"Rendered by QuickLaTeX.com\" height=\"19\" width=\"86\" style=\"vertical-align: -5px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> A fun\u00e7\u00e3o muda de decrescente para crescente em x=1,83, ent\u00e3o <strong>x=1,83 \u00e9 o m\u00ednimo<\/strong> da fun\u00e7\u00e3o.<\/p>\n<p class=\"has-text-align-left\"> Portanto, x=1,83 \u00e9 o valor do lado horizontal que representa o custo m\u00ednimo. Agora vamos calcular o valor do lado vertical:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-325134d449963e4d814cdd19a87bdfd4_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"y =\\cfrac{2}{x} \\ \\longrightarrow \\ y =\\cfrac{2}{1,83} = \\bm{1,09}\" title=\"Rendered by QuickLaTeX.com\" height=\"42\" width=\"223\" style=\"vertical-align: -16px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Assim, os valores que comp\u00f5em o custo m\u00ednimo do enquadramento s\u00e3o:<\/p>\n<p class=\"has-text-align-center\"> lado horizontal<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-c8b9148690c3a5b5e441df6c02e23e85_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"= x = \\bm{1,83} \\ \\mathbf{m}\" title=\"Rendered by QuickLaTeX.com\" height=\"16\" width=\"110\" style=\"vertical-align: -4px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-center\"> lado vertical<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-0ad15cb0e6023282190abbb21c5f8a47_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"= y = \\bm{1,09} \\ \\mathbf{m}\" title=\"Rendered by QuickLaTeX.com\" height=\"16\" width=\"109\" style=\"vertical-align: -4px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> E o custo m\u00ednimo alcan\u00e7ado com estes valores \u00e9:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-bef42a21eacaeb4cd25b5e43c15e268b_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"C= 15\\cdot 1,83+25\\cdot 1,09=\\bm{54,70}\" title=\"Rendered by QuickLaTeX.com\" height=\"17\" width=\"256\" style=\"vertical-align: -4px;\"><\/p>\n<p> <strong>\u20ac<\/strong><\/p>\n<div class=\"wp-block-otfm-box-spoiler-end otfm-sp_end\"><\/div>\n<h3 class=\"wp-block-heading\"> Problema 5<\/h3>\n<p> A porta de uma catedral \u00e9 formada por um arco semicircunferencial sustentado por duas colunas, conforme mostra a figura a seguir: <\/p>\n<div class=\"wp-block-image\">\n<figure class=\"aligncenter size-large is-resized\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/uploads\/2023\/07\/problemes-d-optimisation-de-la-geometrie.webp\" alt=\"problemas de otimiza\u00e7\u00e3o de geometria\" class=\"wp-image-2500\" width=\"182\" height=\"268\" srcset=\"\" sizes=\"auto, \" data-src=\"\"><\/figure>\n<\/div>\n<p> Se o per\u00edmetro da porta for 20 m, determine as medidas<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-ede05c264bba0eda080918aaa09c4658_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"x\" title=\"Rendered by QuickLaTeX.com\" height=\"8\" width=\"10\" style=\"vertical-align: 0px;\"><\/p>\n<p> E<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-0af556714940c351c933bba8cf840796_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"y\" title=\"Rendered by QuickLaTeX.com\" height=\"12\" width=\"9\" style=\"vertical-align: -4px;\"><\/p>\n<p> o que maximiza a \u00e1rea de superf\u00edcie de toda a porta. <\/p>\n<div class=\"wp-block-otfm-box-spoiler-start otfm-sp__wrapper otfm-sp__box js-otfm-sp-box__closed otfm-sp__E6F9EF\" role=\"button\" tabindex=\"0\" aria-expanded=\"false\" data-otfm-spc=\"#E6F9EF\" style=\"text-align:center\">\n<div class=\"otfm-sp__title\"> <strong>Veja a solu\u00e7\u00e3o<\/strong><\/div>\n<\/div>\n<p class=\"has-text-align-left\"> <strong>Passo 1: Defina a fun\u00e7\u00e3o a ser otimizada.<\/strong><\/p>\n<p class=\"has-text-align-left\"> A \u00e1rea de um c\u00edrculo \u00e9 calculada pela f\u00f3rmula<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-1ab0b6b8d454bae5631e7a82caed58d8_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\pi r^2.\" title=\"Rendered by QuickLaTeX.com\" height=\"15\" width=\"31\" style=\"vertical-align: 0px;\"><\/p>\n<p> Portanto, a \u00e1rea de toda a porta ser\u00e1 a \u00e1rea do ret\u00e2ngulo mais metade da \u00e1rea da circunfer\u00eancia: <\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-95c1f9b6152c4cbc623aeb8a0a5757b7_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"A(x,y)= x\\cdot y + \\cfrac{1}{2} \\left[ \\pi r ^2 \\right]\" title=\"Rendered by QuickLaTeX.com\" height=\"38\" width=\"184\" style=\"vertical-align: -12px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-baa2e14b061cf14a657782db8fe91b92_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"A(x,y)= x y + \\cfrac{1}{2} \\left[ \\pi \\left(\\cfrac{x}{2}\\right)^2 \\right]\" title=\"Rendered by QuickLaTeX.com\" height=\"64\" width=\"212\" style=\"vertical-align: -27px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-9196d8284edebe6450d49aa5a0b6a3e1_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"A(x,y)= x y + \\cfrac{1}{2} \\left[ \\pi \\cdot \\cfrac{x^2}{4} \\right]\" title=\"Rendered by QuickLaTeX.com\" height=\"54\" width=\"193\" style=\"vertical-align: -23px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-b25e1d2c661585be998d1596d6650c01_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"A(x,y)= x y +\\cfrac{1}{2} \\left[  \\cfrac{\\pi \\cdot x^2}{4} \\right]\" title=\"Rendered by QuickLaTeX.com\" height=\"54\" width=\"193\" style=\"vertical-align: -23px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-64a0c7fa480320a1b564fec5f9ff8265_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"A(x,y)= xy +\\cfrac{\\pi x^2}{8}\" title=\"Rendered by QuickLaTeX.com\" height=\"41\" width=\"150\" style=\"vertical-align: -12px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> J\u00e1 temos a fun\u00e7\u00e3o de otimizar. Mas depende de duas vari\u00e1veis quando s\u00f3 pode depender de uma.<\/p>\n<p class=\"has-text-align-left\"> Por\u00e9m, o comunicado informa que o per\u00edmetro de todo o port\u00e3o \u00e9 de 20m. O per\u00edmetro de um c\u00edrculo \u00e9 calculado pela f\u00f3rmula<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-e815d233e188a7121eef89639e48fe75_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"2 \\pi r.\" title=\"Rendered by QuickLaTeX.com\" height=\"12\" width=\"31\" style=\"vertical-align: 0px;\"><\/p>\n<p> Portanto, o per\u00edmetro de toda a porta ser\u00e1:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-7709d0c72bf84a17ac83bc46f5cce002_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"P= x +2y +\\cfrac{1}{2} \\left[ 2 \\pi \\left( \\cfrac{x}{2}\\right) \\right] = x+2y + \\cfrac{2 \\pi x }{2 \\cdot 2} = x+2y + \\cfrac{ \\pi x }{2 }\" title=\"Rendered by QuickLaTeX.com\" height=\"54\" width=\"452\" style=\"vertical-align: -23px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> O per\u00edmetro deve ser de 20 m. Portanto, definimos a express\u00e3o anterior igual a 20 para encontrar a rela\u00e7\u00e3o entre<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-ede05c264bba0eda080918aaa09c4658_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"x\" title=\"Rendered by QuickLaTeX.com\" height=\"8\" width=\"10\" style=\"vertical-align: 0px;\"><\/p>\n<p> E<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-a70e6a4387a816f153e8597195143f54_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"y :\" title=\"Rendered by QuickLaTeX.com\" height=\"12\" width=\"18\" style=\"vertical-align: -4px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-282aa8aebdd04ba4989ba7466fdd694d_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"x+2y + \\cfrac{ \\pi x }{2 } = 20\" title=\"Rendered by QuickLaTeX.com\" height=\"34\" width=\"136\" style=\"vertical-align: -12px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Multiplicamos todos os termos por 2 para eliminar fra\u00e7\u00f5es: <\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-a185cfda7fb65f370636fa469b0de2c4_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"2\\cdot x+2\\cdot 2y + 2 \\cdot \\cfrac{ \\pi x }{2 } = 2 \\cdot 20\" title=\"Rendered by QuickLaTeX.com\" height=\"34\" width=\"223\" style=\"vertical-align: -12px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-d67cd2fbdde0f884603e75bf270cb52c_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"2x+4y +  \\pi x = 40\" title=\"Rendered by QuickLaTeX.com\" height=\"16\" width=\"143\" style=\"vertical-align: -4px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> N\u00f3s limpamos <\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-a70e6a4387a816f153e8597195143f54_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"y :\" title=\"Rendered by QuickLaTeX.com\" height=\"12\" width=\"18\" style=\"vertical-align: -4px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-740ff13a6824a3ee18cb1de0cbf52660_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"4y  = 40-2x- \\pi x\" title=\"Rendered by QuickLaTeX.com\" height=\"16\" width=\"143\" style=\"vertical-align: -4px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-db16e16a94223af02d63ea70eae3db7d_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"y   = \\cfrac{40-2x- \\pi x}{4}\" title=\"Rendered by QuickLaTeX.com\" height=\"38\" width=\"136\" style=\"vertical-align: -12px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> E substitu\u00edmos a express\u00e3o encontrada na fun\u00e7\u00e3o a ser otimizada: <\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-16c3b1d1a3ac443d4640ae2b72c164e6_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"A(x,y)= x y +\\cfrac{\\pi x^2}{8}\\ \\xrightarrow{y \\ = \\ \\frac{40-2x- \\pi x}{4} }\" title=\"Rendered by QuickLaTeX.com\" height=\"41\" width=\"262\" style=\"vertical-align: -12px;\"><\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-f463f9095e8028624678c06a08696a96_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"A(x)= x \\cdot \\cfrac{40-2x- \\pi x}{4}+\\cfrac{\\pi x^2}{8}\" title=\"Rendered by QuickLaTeX.com\" height=\"41\" width=\"239\" style=\"vertical-align: -12px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-095586f92b592c3c9c50750f8a6eb7d4_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"A(x)= \\cfrac{40x-2x^2-\\pi x^2}{4}+\\cfrac{\\pi x^2}{8}\" title=\"Rendered by QuickLaTeX.com\" height=\"41\" width=\"242\" style=\"vertical-align: -12px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> <strong>Passo 2: Calcule a derivada da fun\u00e7\u00e3o a ser otimizada.<\/strong> <\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-42ce3330af662099ad431ee1187d26f1_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"A'(x)=\\cfrac{(40-4x-2\\pi x)\\cdot 4 +(40x-2x^2- \\pi x^2)\\cdot 0 }{4^2} +\\cfrac{2\\pi x \\cdot 8 + \\pi x^2 \\cdot 0}{8^2}\" title=\"Rendered by QuickLaTeX.com\" height=\"41\" width=\"544\" style=\"vertical-align: -12px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-1a75cddf1efd4f17b6a3211093f2e0b5_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"A'(x)=\\cfrac{160-16x-8\\pi x }{16} +\\cfrac{16\\pi x}{64}\" title=\"Rendered by QuickLaTeX.com\" height=\"38\" width=\"257\" style=\"vertical-align: -12px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> <strong>Etapa 3: Encontre os pontos cr\u00edticos.<\/strong><\/p>\n<p class=\"has-text-align-left\"> Para encontrar os pontos cr\u00edticos da fun\u00e7\u00e3o, resolvemos <\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-a8f5a921f7e1978b553d11a76e0962c7_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"A'(x)=0:\" title=\"Rendered by QuickLaTeX.com\" height=\"19\" width=\"83\" style=\"vertical-align: -5px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-424f3d5d91120e872bebd055792d71a1_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"A'(x)=0\" title=\"Rendered by QuickLaTeX.com\" height=\"19\" width=\"74\" style=\"vertical-align: -5px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-89a04d2a2a532f36f15042d2d2c79d41_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\cfrac{160-16x-8\\pi x }{16} +\\cfrac{16\\pi x}{64} = 0\" title=\"Rendered by QuickLaTeX.com\" height=\"38\" width=\"222\" style=\"vertical-align: -12px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Esta \u00e9 uma equa\u00e7\u00e3o com fra\u00e7\u00f5es, ent\u00e3o multiplicamos cada termo pelo lcm dos denominadores para eliminar as fra\u00e7\u00f5es: <\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-87d60f06d42276c79b3a4f33b00fdb7b_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"64 \\cdot \\cfrac{160-16x-8\\pi x }{16} +64 \\cdot \\cfrac{16\\pi x}{64} = 0\" title=\"Rendered by QuickLaTeX.com\" height=\"38\" width=\"285\" style=\"vertical-align: -12px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-b2a7b3bbdad1ea1bf9f12861b1dbdf0a_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"4\\cdot ( 160-16x-8\\pi x) +1\\cdot 16\\pi x= 0\" title=\"Rendered by QuickLaTeX.com\" height=\"19\" width=\"278\" style=\"vertical-align: -5px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-61a4622d4b65310580e066941ae13fac_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"640-64x -32 \\pi x +16\\pi x= 0\" title=\"Rendered by QuickLaTeX.com\" height=\"14\" width=\"229\" style=\"vertical-align: -2px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-94c6efd66ea5f2f1c68ab09d59fb5432_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"-64x -32 \\pi x +16\\pi x= -640\" title=\"Rendered by QuickLaTeX.com\" height=\"14\" width=\"226\" style=\"vertical-align: -2px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-bdaf73c6d74c3d4f605b5aea3cdeb1ad_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"-64x -16 \\pi x = -640\" title=\"Rendered by QuickLaTeX.com\" height=\"12\" width=\"166\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-4bfe710d37882f664daed4de23f2eb7a_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"(-64 -16 \\pi) x = -640\" title=\"Rendered by QuickLaTeX.com\" height=\"19\" width=\"169\" style=\"vertical-align: -5px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-b551560fd86665e639dbc88aeb88a2bb_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"x=\\cfrac{-640}{-64 -16 \\pi}  = 5,6\" title=\"Rendered by QuickLaTeX.com\" height=\"38\" width=\"167\" style=\"vertical-align: -12px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> <strong>Passo 4: Estude a monotonicidade da fun\u00e7\u00e3o e determine o m\u00e1ximo ou m\u00ednimo da fun\u00e7\u00e3o.<\/strong><\/p>\n<p class=\"has-text-align-left\"> Para estudar a monotonia da fun\u00e7\u00e3o, representamos o ponto cr\u00edtico encontrado \u00e0 direita: <\/p>\n<div class=\"wp-block-image\">\n<figure class=\"aligncenter size-large is-resized\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/uploads\/2023\/07\/nombre-ligne-56.webp\" alt=\"\" class=\"wp-image-2501\" width=\"214\" height=\"83\" srcset=\"\" sizes=\"auto, \" data-src=\"\"><\/figure>\n<\/div>\n<p class=\"has-text-align-left\"> E agora avaliamos o sinal da derivada em cada intervalo, para descobrir se a fun\u00e7\u00e3o \u00e9 crescente ou decrescente. Portanto, pegamos um ponto em cada intervalo (nunca o ponto cr\u00edtico) e observamos qual sinal a derivada tem neste ponto: <\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-e8b0c6b03e591be56505c8853595dd0a_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"A'(0)=\\cfrac{160-16\\cdot 0-8\\pi \\cdot 0 }{16} +\\cfrac{16\\pi \\cdot 0}{64} = 10 +0 = 10 \\ \\rightarrow \\ \\bm{+}\" title=\"Rendered by QuickLaTeX.com\" height=\"38\" width=\"456\" style=\"vertical-align: -12px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-91c7e9012e4937dad79ec792910df916_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"A'(6)=\\cfrac{160-16\\cdot 6-8\\pi \\cdot 6 }{16} +\\cfrac{16\\pi \\cdot 6}{64} = -5,42 +4,71 = -0,71 \\ \\rightarrow \\ \\bm{-}\" title=\"Rendered by QuickLaTeX.com\" height=\"38\" width=\"543\" style=\"vertical-align: -12px;\"><\/p>\n<\/p>\n<div class=\"wp-block-image\">\n<figure class=\"aligncenter size-large is-resized\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/uploads\/2023\/07\/droite-numerique-56-positif-negatif.webp\" alt=\"\" class=\"wp-image-2503\" width=\"218\" height=\"155\" srcset=\"\" sizes=\"auto, \" data-src=\"\"><\/figure>\n<\/div>\n<p class=\"has-text-align-left\"> Se a derivada for positiva, significa que a fun\u00e7\u00e3o est\u00e1 aumentando, e se a derivada for negativa, significa que a fun\u00e7\u00e3o est\u00e1 diminuindo. Portanto, os intervalos de crescimento e decl\u00ednio s\u00e3o:<\/p>\n<p class=\"has-text-align-center\"> <strong>Crescimento:<\/strong><\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-375f15fd5182721ee68f606765b66e89_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\bm{(-\\infty , 5,6)}\" title=\"Rendered by QuickLaTeX.com\" height=\"19\" width=\"77\" style=\"vertical-align: -5px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-center\"> <strong>Diminuir:<\/strong><\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-ce3629c3ea3fa84fd3413a15913976f1_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\bm{(5,6,+\\infty)}\" title=\"Rendered by QuickLaTeX.com\" height=\"19\" width=\"77\" style=\"vertical-align: -5px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> A fun\u00e7\u00e3o vai de crescente a decrescente em x=5,6, ent\u00e3o <strong>x=5,6 \u00e9 o m\u00e1ximo<\/strong> da fun\u00e7\u00e3o.<\/p>\n<p class=\"has-text-align-left\"> Ainda,<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-c0d40793b19725cb50b88536bdaf3239_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"x=5,6\" title=\"Rendered by QuickLaTeX.com\" height=\"17\" width=\"60\" style=\"vertical-align: -4px;\"><\/p>\n<p> \u00e9 o valor que constitui a superf\u00edcie m\u00e1xima. Agora calculamos o valor de<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-a70e6a4387a816f153e8597195143f54_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"y :\" title=\"Rendered by QuickLaTeX.com\" height=\"12\" width=\"18\" style=\"vertical-align: -4px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-af3ff0bf06cb82b6661c9ead44cabaa6_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"y = \\cfrac{40-2\\cdot 5,6- \\pi \\cdot 5,6}{4} = 2,80\" title=\"Rendered by QuickLaTeX.com\" height=\"39\" width=\"251\" style=\"vertical-align: -12px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Assim, os valores que comp\u00f5em a superf\u00edcie m\u00e1xima s\u00e3o: <\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-25cd4efdc51c2a6fb453056df61536fa_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\bm{x = 5,60} \\ \\mathbf{m}\" title=\"Rendered by QuickLaTeX.com\" height=\"17\" width=\"91\" style=\"vertical-align: -4px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-9b69c85499698e995f01f60b2b78ef20_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\bm{y = 2,80} \\ \\mathbf{m}\" title=\"Rendered by QuickLaTeX.com\" height=\"16\" width=\"90\" style=\"vertical-align: -4px;\"><\/p>\n<\/p>\n<div class=\"wp-block-otfm-box-spoiler-end otfm-sp_end\"><\/div>\n<h3 class=\"wp-block-heading\">Problema 6<\/h3>\n<p> Queremos construir um tanque em forma de cilindro com \u00e1rea de 54 cm <sup>2<\/sup> . Determine o raio da base e a altura do cilindro para que o volume seja m\u00e1ximo. <\/p>\n<div class=\"wp-block-otfm-box-spoiler-start otfm-sp__wrapper otfm-sp__box js-otfm-sp-box__closed otfm-sp__E6F9EF\" role=\"button\" tabindex=\"0\" aria-expanded=\"false\" data-otfm-spc=\"#E6F9EF\" style=\"text-align:center\">\n<div class=\"otfm-sp__title\"> <strong>Veja a solu\u00e7\u00e3o<\/strong><\/div>\n<\/div>\n<p class=\"has-text-align-left\"> <strong>Passo 1: Defina a fun\u00e7\u00e3o a ser otimizada.<\/strong><\/p>\n<p class=\"has-text-align-left\"> O volume de um cilindro \u00e9 calculado pela seguinte f\u00f3rmula:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-fbe33001d54c8245539894746beb9eac_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"V= A_{base}\\cdot h\" title=\"Rendered by QuickLaTeX.com\" height=\"16\" width=\"101\" style=\"vertical-align: -3px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> A \u00e1rea da base \u00e9 um c\u00edrculo, ent\u00e3o sua f\u00f3rmula \u00e9<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-fc935e01b9c94505406438dea94cf121_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"A_{\\text{base}}=\\pi r^2\" title=\"Rendered by QuickLaTeX.com\" height=\"18\" width=\"90\" style=\"vertical-align: -3px;\"><\/p>\n<p> . A f\u00f3rmula para o volume do cilindro \u00e9, portanto:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-df6ffff5ee54af4f7b10424871d1af92_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"V= \\pi r^2 \\cdot h\" title=\"Rendered by QuickLaTeX.com\" height=\"15\" width=\"88\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> J\u00e1 temos a fun\u00e7\u00e3o de otimizar. Mas depende de duas vari\u00e1veis (<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-c409433a9e2dfcdb83360a974d243f18_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"r\" title=\"Rendered by QuickLaTeX.com\" height=\"8\" width=\"8\" style=\"vertical-align: 0px;\"><\/p>\n<p> E<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-14b463d0ecd5b350ced6cf1d6a12eef3_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"h\" title=\"Rendered by QuickLaTeX.com\" height=\"12\" width=\"10\" style=\"vertical-align: 0px;\"><\/p>\n<p> ) embora s\u00f3 possa depender de um. Por\u00e9m, o enunciado nos diz que a \u00e1rea do cilindro deve ser 54 cm <sup>2<\/sup> , ent\u00e3o aproveitaremos esta condi\u00e7\u00e3o para encontrar a rela\u00e7\u00e3o entre<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-c409433a9e2dfcdb83360a974d243f18_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"r\" title=\"Rendered by QuickLaTeX.com\" height=\"8\" width=\"8\" style=\"vertical-align: 0px;\"><\/p>\n<p> E<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-6f0bf134a8aafe0dfaa4d711f78b8b1f_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"h .\" title=\"Rendered by QuickLaTeX.com\" height=\"12\" width=\"14\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Para calcular a \u00e1rea de um cilindro voc\u00ea deve somar sua \u00e1rea lateral com as \u00e1reas das duas bases: <\/p>\n<div class=\"wp-block-image\">\n<figure class=\"aligncenter size-large is-resized\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/uploads\/2023\/07\/problemes-optimisation-des-fonctions-verins.webp\" alt=\"problemas de otimiza\u00e7\u00e3o de fun\u00e7\u00f5es de cilindro.png\" class=\"wp-image-2507\" width=\"557\" height=\"271\" srcset=\"\" sizes=\"auto, \" data-src=\"\"><\/figure>\n<\/div>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-edd9578a95d32143e6eaf0830db3854b_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"A_{cilindro} = A_{lateral}+2A_{base} = 2\\pi r h + 2\\pi r^2\" title=\"Rendered by QuickLaTeX.com\" height=\"18\" width=\"330\" style=\"vertical-align: -3px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> A \u00e1rea do cilindro deve ser 54 cm <sup>2<\/sup> , ent\u00e3o igualamos a express\u00e3o anterior a 54 para obter a rela\u00e7\u00e3o entre<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-c409433a9e2dfcdb83360a974d243f18_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"r\" title=\"Rendered by QuickLaTeX.com\" height=\"8\" width=\"8\" style=\"vertical-align: 0px;\"><\/p>\n<p> E<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-f6868abdfd034048aa59644d2ac62353_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"h :\" title=\"Rendered by QuickLaTeX.com\" height=\"12\" width=\"19\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-0ef4f7f8554876495f8907d4f71e1152_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"A_{cilindro} =2\\pi r h + 2\\pi r^2 = 54\" title=\"Rendered by QuickLaTeX.com\" height=\"18\" width=\"223\" style=\"vertical-align: -3px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> N\u00f3s limpamos <\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-f6868abdfd034048aa59644d2ac62353_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"h :\" title=\"Rendered by QuickLaTeX.com\" height=\"12\" width=\"19\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-dd73c08d458b0ab2061031782d125ab9_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"2\\pi r h = 54 - 2\\pi r^2\" title=\"Rendered by QuickLaTeX.com\" height=\"15\" width=\"136\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-abf53b37d2a3ae4784079bfd67d3007f_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"h = \\cfrac{54 - 2\\pi r^2}{2\\pi r}\" title=\"Rendered by QuickLaTeX.com\" height=\"41\" width=\"112\" style=\"vertical-align: -12px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> E substitu\u00edmos a express\u00e3o encontrada na fun\u00e7\u00e3o a ser otimizada: <\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-e2c47f99ce14b444ed73661b2b568e8b_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"V=  \\pi r^2 \\cdot h \\xrightarrow{h \\ = \\ \\frac{54 - 2\\pi r^2}{2\\pi r} } V = \\pi r^2 \\cdot \\cfrac{54 - 2\\pi r^2}{2\\pi r}\" title=\"Rendered by QuickLaTeX.com\" height=\"42\" width=\"346\" style=\"vertical-align: -12px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-81d564fcd7cdf880066ed3eb6eb9a18a_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"V = \\cancel{\\pi}  r^{\\cancel{2}} \\cdot \\cfrac{54 - 2\\pi r^2}{2 \\cancel{\\pi} \\cancel{r}} =r \\cdot \\cfrac{54 - 2\\pi r^2}{2}\" title=\"Rendered by QuickLaTeX.com\" height=\"41\" width=\"278\" style=\"vertical-align: -12px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-f95e7e1fc6088ba84a7b3885fe929d58_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"V(r) = r \\cdot (27 - \\pi r^2)= 27r - \\pi r^3\" title=\"Rendered by QuickLaTeX.com\" height=\"20\" width=\"259\" style=\"vertical-align: -5px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> <strong>Passo 2: Calcule a derivada da fun\u00e7\u00e3o a ser otimizada.<\/strong><\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-58e3403f2812564067058ece1bcd83ec_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"V(r)=27r - \\pi r^3\\ \\longrightarrow \\ V'(r)= 27-3 \\pi r^2\" title=\"Rendered by QuickLaTeX.com\" height=\"20\" width=\"324\" style=\"vertical-align: -5px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> <strong>Etapa 3: Encontre os pontos cr\u00edticos.<\/strong><\/p>\n<p class=\"has-text-align-left\"> Para encontrar os pontos cr\u00edticos da fun\u00e7\u00e3o, resolvemos <\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-e868c9ed52907fbe13197b00c29b7d21_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"V'(r)=0:\" title=\"Rendered by QuickLaTeX.com\" height=\"19\" width=\"83\" style=\"vertical-align: -5px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-d3de17aa96e46530402559cf14e8db4d_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"V'(r)=0\" title=\"Rendered by QuickLaTeX.com\" height=\"19\" width=\"74\" style=\"vertical-align: -5px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-10638a98b4a67b085c0854f95c86a65b_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"27-3 \\pi r^2=0\" title=\"Rendered by QuickLaTeX.com\" height=\"15\" width=\"108\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-8e31d335383816f692cd02cb2ca8f55b_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"-3 \\pi r^2=-27\" title=\"Rendered by QuickLaTeX.com\" height=\"15\" width=\"104\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-ff57ef2a3aa07b3c9f9f76a9cb00f484_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"r^2=\\cfrac{-27}{-3\\pi }\" title=\"Rendered by QuickLaTeX.com\" height=\"39\" width=\"82\" style=\"vertical-align: -12px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-7d9e55eb22dc67a402b53bbb9ca17298_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"r^2=2,86\" title=\"Rendered by QuickLaTeX.com\" height=\"19\" width=\"75\" style=\"vertical-align: -4px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-8c7c4c4acf5b028e2dbba11dfc83d20b_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\sqrt{r^2}=\\sqrt{2,86}\" title=\"Rendered by QuickLaTeX.com\" height=\"21\" width=\"103\" style=\"vertical-align: -4px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-01af3fa2bc910bd2da923aa2c83e1d7e_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"r=1,69\" title=\"Rendered by QuickLaTeX.com\" height=\"16\" width=\"67\" style=\"vertical-align: -4px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> <strong>Passo 4: Estude a monotonicidade da fun\u00e7\u00e3o e determine o m\u00e1ximo ou m\u00ednimo da fun\u00e7\u00e3o.<\/strong><\/p>\n<p class=\"has-text-align-left\"> Para estudar a monotonicidade da fun\u00e7\u00e3o, representamos o ponto cr\u00edtico encontrado na reta num\u00e9rica: <\/p>\n<div class=\"wp-block-image\">\n<figure class=\"aligncenter size-large is-resized\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/uploads\/2023\/07\/ligne-numerique-169.webp\" alt=\"\" class=\"wp-image-2508\" width=\"232\" height=\"90\" srcset=\"\" sizes=\"auto, \" data-src=\"\"><\/figure>\n<\/div>\n<p class=\"has-text-align-left\"> E agora avaliamos o sinal da derivada em cada intervalo, para descobrir se a fun\u00e7\u00e3o \u00e9 crescente ou decrescente. Portanto, pegamos um ponto em cada intervalo (nunca o ponto cr\u00edtico) e observamos qual sinal a derivada tem neste ponto: <\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-8d2aa45b6cfc986ec1e33cdd5e9266e3_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"V'(0)= 27-3 \\pi\\cdot 0^2 = 27-0 = +27 \\ \\rightarrow \\ \\bm{+}\" title=\"Rendered by QuickLaTeX.com\" height=\"20\" width=\"333\" style=\"vertical-align: -5px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-2c599118454ccdcc0e318f01ac61b84d_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"V'(2)= 27-3 \\pi \\cdot 2^2 = 27-37,70 = -10,70 \\ \\rightarrow \\ \\bm{-}\" title=\"Rendered by QuickLaTeX.com\" height=\"20\" width=\"393\" style=\"vertical-align: -5px;\"><\/p>\n<\/p>\n<div class=\"wp-block-image\">\n<figure class=\"aligncenter size-large is-resized\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/uploads\/2023\/07\/ligne-numerique-169-positif-negatif.webp\" alt=\"\" class=\"wp-image-2509\" width=\"249\" height=\"176\" srcset=\"\" sizes=\"auto, \" data-src=\"\"><\/figure>\n<\/div>\n<p class=\"has-text-align-left\"> Se a derivada for positiva, significa que a fun\u00e7\u00e3o est\u00e1 aumentando, e se a derivada for negativa, significa que a fun\u00e7\u00e3o est\u00e1 diminuindo. Portanto, os intervalos de crescimento e decl\u00ednio s\u00e3o:<\/p>\n<p class=\"has-text-align-center\"> <strong>Crescimento:<\/strong><\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-b5e2776182fc7c0a5138adfee1aafaaf_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\bm{(-\\infty,1,69)}\" title=\"Rendered by QuickLaTeX.com\" height=\"19\" width=\"86\" style=\"vertical-align: -5px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-center\"> <strong>Diminuir:<\/strong><\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-f4ac44f6269498e03069a77e1989c8be_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\bm{(1,69,+\\infty)}\" title=\"Rendered by QuickLaTeX.com\" height=\"19\" width=\"86\" style=\"vertical-align: -5px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> A fun\u00e7\u00e3o vai de crescente para decrescente em r=1,69, ent\u00e3o <strong>r=1,69 cm \u00e9 o m\u00e1ximo<\/strong> da fun\u00e7\u00e3o.<\/p>\n<p class=\"has-text-align-left\"> Portanto, r=1,69 \u00e9 o valor do raio que perfaz o volume m\u00e1ximo. Agora calculamos a altura:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-381f3c5eb279572f486e28e68f0c5af2_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"h = \\cfrac{54 - 2\\pi \\cdot 1,69^2}{2\\pi \\cdot1,69} = \\cfrac{54 - 17,94}{10,62} = 3,39\" title=\"Rendered by QuickLaTeX.com\" height=\"45\" width=\"316\" style=\"vertical-align: -16px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Ent\u00e3o os valores que fazem o volume m\u00e1ximo s\u00e3o:<\/p>\n<p class=\"has-text-align-center\"> <strong>R\u00e1dio<\/strong><\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-5c99f9d93c8a300be7afef76dfb71379_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\bm{= r = 1,69} \\ \\mathbf{cm}\" title=\"Rendered by QuickLaTeX.com\" height=\"16\" width=\"117\" style=\"vertical-align: -4px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-center\"> <strong>Altura<\/strong><\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-5ee2564c5e4b2a4e4409c392e5d01627_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\bm{= h = 3,39} \\ \\mathbf{cm}\" title=\"Rendered by QuickLaTeX.com\" height=\"16\" width=\"119\" style=\"vertical-align: -4px;\"><\/p>\n<\/p>\n<div class=\"wp-block-otfm-box-spoiler-end otfm-sp_end\"><\/div>\n","protected":false},"excerpt":{"rendered":"<p>Aqui explicamos como os problemas de otimiza\u00e7\u00e3o de fun\u00e7\u00f5es s\u00e3o resolvidos em etapas. Al\u00e9m disso, voc\u00ea poder\u00e1 praticar com exerc\u00edcios resolvidos sobre problemas de otimiza\u00e7\u00e3o. O que s\u00e3o problemas de otimiza\u00e7\u00e3o? Problemas de otimiza\u00e7\u00e3o s\u00e3o problemas nos quais \u00e9 necess\u00e1rio encontrar o m\u00e1ximo ou o m\u00ednimo de uma fun\u00e7\u00e3o. Por exemplo, um problema de otimiza\u00e7\u00e3o &hellip;<\/p>\n<p class=\"read-more\"> <a class=\"\" href=\"https:\/\/mathority.org\/pt\/problemas-de-otimizacao\/\"> <span class=\"screen-reader-text\">Problemas de otimiza\u00e7\u00e3o<\/span> Leia mais &raquo;<\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"","ast-site-content-layout":"","site-content-style":"default","site-sidebar-style":"default","ast-global-header-display":"","ast-banner-title-visibility":"","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","theme-transparent-header-meta":"","adv-header-id-meta":"","stick-header-meta":"","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","astra-migrate-meta-layouts":"","footnotes":""},"categories":[22],"tags":[],"class_list":["post-42","post","type-post","status-publish","format-standard","hentry","category-representacao-de-funcao"],"yoast_head":"<!-- This site is optimized with the Yoast SEO plugin v21.2 - https:\/\/yoast.com\/wordpress\/plugins\/seo\/ -->\n<title>Problemas de otimiza\u00e7\u00e3o -<\/title>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/mathority.org\/pt\/problemas-de-otimizacao\/\" \/>\n<meta property=\"og:locale\" content=\"pt_BR\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"Problemas de otimiza\u00e7\u00e3o -\" \/>\n<meta property=\"og:description\" content=\"Aqui explicamos como os problemas de otimiza\u00e7\u00e3o de fun\u00e7\u00f5es s\u00e3o resolvidos em etapas. Al\u00e9m disso, voc\u00ea poder\u00e1 praticar com exerc\u00edcios resolvidos sobre problemas de otimiza\u00e7\u00e3o. O que s\u00e3o problemas de otimiza\u00e7\u00e3o? Problemas de otimiza\u00e7\u00e3o s\u00e3o problemas nos quais \u00e9 necess\u00e1rio encontrar o m\u00e1ximo ou o m\u00ednimo de uma fun\u00e7\u00e3o. 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