Question

Um arame de 15cm de comprimento deve ser cortado em dois pedaços, um dos quais será torcido de
modo a formar um quadrado e o outro, a formar uma circunferência. De que modo deverá ser cortado
para que a soma das áreas das regiões limitadas pelas figuras obtidas sejam mínima?




Quem responder com gracinha vou denunciar!​

Answer 1

$\Large\boxed{\begin{array}{l}\sf Seja~\ell~o~lado~do~quadrado~e~r~o~raio\\\sf do~c\acute irculo.\\\sf a~soma~dos~comprimentos\\\sf~dever\acute a~ser~igual~a~15~cm. DA\acute I\\\sf 4\ell+2\pi r=15\\\sf A~\acute area~a~ser~minimizada~\acute e\\\sf A=\ell^2+\pi r^2\\\sf isolando~\ell~na~1^a~equac_{\!\!,}\tilde ao~temos:\\\sf \ell=\dfrac{15-2\pi r}{4} \\\sf substitua~na~2^a~equac_{\!\!,}\tilde ao: \\\sf A=\bigg(\dfrac{15-2\pi r}{4}\bigg)^2+\pi r^2\end{array}}$

$\Large\boxed{\begin{array}{l}\sf Para~otimizar, vamos~encontrar~\\\sf a~derivada~e~em~seguida~igualar~a~zero.\\\sf A=\dfrac{225-60\pi r+4\pi^2 r^2}{16}+\pi r^2\\\\\sf A=\dfrac{225-60\pi r+4\pi^2 r^2+16\pi r^2}{16}\\\\\sf A =\dfrac{1}{16}\bigg[225-60\pi r+4\pi^2 r^2+16\pi r^2\bigg]\\\\\sf\dfrac{dA}{dr}=\dfrac{1}{16}\bigg[0-60\pi+8\pi^2r+32\pi r\bigg]\\\\\sf \dfrac{dA}{dr}=0\iff \dfrac{1}{16}\bigg[ -60\pi+8\pi^2r+32\pi r\bigg]=0\end{array}}$

$\Large\boxed{\begin{array}{l}\sf-60\pi+8\pi^2 r^2+32\pi r=0\div4\pi\\\sf -15+2\pi r^2+32=0\\\sf admita~\pi=3,14\\\sf -15+2\cdot 3,14 r^2+32r=0\\\sf 6,28r^2+32 r-15=0\\\sf \Delta=1024+376,8=1400,8\\\sf \sqrt{\Delta}=37,42\\\sf r=\dfrac{-32+37,42}{12,56}\approxeq0,43\end{array}}$

$\Large\boxed{\begin{array}{l}\sf \ell=\dfrac{15-2\cdot 3,14\cdot 0,43}{4}\\\\\sf \ell\approxeq3,07~cm\\\sf P_{quadrado}=4\cdot3,07=12,28\\\sf P_{circunfer\hat encia}=2\cdot3,14\cdot0,43\\\sf P_{circunfer\hat encia}=2,70~cm\end{array}}$

$\Large\boxed{\begin{array}{l}\rm Dever\acute a~ser~cortado~de~modo~que\\\rm o~per\acute imetro~do~quadrado~seja\\\rm 12,28~cm\\\rm e~o~comprimento~da~circunfer\hat encia\\\rm seja~2,70~cm\end{array}}$