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[SPAM] [obm-l] Re: [obm-l] Perímetro Mínimo



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A fun=E7=E3o correta do per=EDmetro em fun=E7=E3o de um dos lados =E9

P(b) =3D 2(400/b +b), ou

P(b) =3D 800/b + 2b, que N=C3O =E9 quadr=E1tica. Sua derivada =E9 P'(b) =
=3D -800/b=B2 + 2b, que zera quando b vale raiz c=FAbica de 400. Nesse =
ponto deve estar o m=EDnimo (claro, admitindo tacitamente que, j=E1 que =
o problema afirmou que tem m=EDnimo, esse ponto ser=E1 mesmo de =
m=EDnimo)

Abra=E7o,

Jo=E3o Lu=EDs
  ----- Original Message -----=20
  From: Tales Prates Correia=20
  To: obm-l@xxxxxxxxxxxxxx=20
  Sent: Friday, November 16, 2007 2:16 PM
  Subject: RE: [obm-l] Per=EDmetro M=EDnimo






-------------------------------------------------------------------------=
---
               Ol=E1!
              =20
               Lembremo-nos, em primeiro momento, de que todo =
ret=E2ngulo =E9 um paralegramo. Conseguintemente, se denotarmos por a e =
b=20

               as medidas de dois de seus lados consecutivos, a medida =
de se per=EDmetro ser=E1 P =3D 2(a + b) *

               Para tal afirma=E7=E3o recorremos =E0 seguinte =
propriedade dos paralelogramos: qualquer par de lados opostos s=E3o =
congruentes.
             =20
               Sabemos ainda que a =E1rea do ret=E2ngulo d=E1-se pelo =
produto das medidas de seus lados: A =3D ab **. Logo, =E9 v=E1lido =
afirmar que=20
    =20
               a =3D A/b ***.

               Se substituirmos *** em *, obteremos a lei de =
correspond=EAncia da "fun=E7=E3o per=EDmetro":

               P(b) =3D 2(A + b=B2)/b

               definida em R+ com imagens em R+.

               A fun=E7=E3o derivada de P =E9 definida pela seguinte lei

               P'(b) =3D 2(b=B2 - A)/b

               A =FAnica raiz dessa fun=E7=E3o =E9 b =3D 20, uma vez que =
D(P') =3D R+. Com um pouco de paci=EAncia, =E9 poss=EDvel mostrar que =
existe uma

               vizinhan=E7a V de b =3D 20 tal que P'(x) < 0 se x < b e =
P'(x) > 0 se x > b. Por conseguinte P(b) =E9 o m=EDnimo absoluto de P, =
pois n=E3o

               existem outros extremantes.

               Assim, o per=EDmetro m=EDnimo do ret=E2ngulo cuja =E1rea =
vale 400cm=B2 =E9 P(20) =3D 80cm.

               Acho que =E9 isso.


-------------------------------------------------------------------------=
-
      Date: Fri, 16 Nov 2007 12:33:39 -0200
      Subject: [obm-l] Per=EDmetro M=EDnimo
      From: oliveira-aline1983@xxxxxxxxxx
      To: obm-l@xxxxxxxxxxxxxx


      Pessoal ,n=E3o consigo montar minha fun=E7=E3o quadr=E1tica em =
fun=E7=E3o do per=EDmetro.



      Dentre os ret=E2ngulos com =E1rea 400cm2, existe um, cujo =
per=EDmetro =E9 o menor poss=EDvel. Qual o per=EDmetro deste =
ret=E2ngulo, em cm?







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-----
  Veja mapas e encontre as melhores rotas para fugir do tr=E2nsito com o =
Live Search Maps! Experimente j=E1! 
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<HTML><HEAD>
<META http-equiv=3DContent-Type content=3D"text/html; =
charset=3Diso-8859-1">
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<BODY class=3Dhmmessage bgColor=3D#ffffff>
<DIV>
<DIV><FONT face=3DVerdana>A fun=E7=E3o correta do per=EDmetro em =
fun=E7=E3o de um dos lados=20
=E9</FONT></DIV>
<DIV><FONT face=3DVerdana></FONT>&nbsp;</DIV>
<DIV><FONT face=3DVerdana>P(b) =3D 2(400/b +b), ou</FONT></DIV>
<DIV><FONT face=3DVerdana></FONT>&nbsp;</DIV>
<DIV><FONT face=3DVerdana>P(b) =3D 800/b + 2b, que N=C3O =E9 =
quadr=E1tica. Sua derivada =E9=20
P'(b) =3D -800/b=B2 + 2b, que zera quando b vale raiz c=FAbica de 400. =
Nesse ponto=20
deve estar o m=EDnimo (claro, admitindo tacitamente que, j=E1 que o =
problema afirmou=20
que tem m=EDnimo, esse ponto ser=E1 mesmo de m=EDnimo)</FONT></DIV>
<DIV><FONT face=3DVerdana></FONT>&nbsp;</DIV>
<DIV><FONT face=3DVerdana>Abra=E7o,</FONT></DIV>
<DIV><FONT face=3DVerdana></FONT>&nbsp;</DIV>
<DIV><FONT face=3DVerdana>Jo=E3o Lu=EDs</FONT></DIV></DIV>
<BLOCKQUOTE=20
style=3D"PADDING-RIGHT: 0px; PADDING-LEFT: 5px; MARGIN-LEFT: 5px; =
BORDER-LEFT: #000000 2px solid; MARGIN-RIGHT: 0px">
  <DIV style=3D"FONT: 10pt arial">----- Original Message ----- </DIV>
  <DIV=20
  style=3D"BACKGROUND: #e4e4e4; FONT: 10pt arial; font-color: =
black"><B>From:</B>=20
  <A title=3Dtales1337@xxxxxxxxxxx =
href=3D"mailto:tales1337@xxxxxxxxxxx";>Tales=20
  Prates Correia</A> </DIV>
  <DIV style=3D"FONT: 10pt arial"><B>To:</B> <A =
title=3Dobm-l@xxxxxxxxxxxxxx=20
  href=3D"mailto:obm-l@xxxxxxxxxxxxxx";>obm-l@xxxxxxxxxxxxxx</A> </DIV>
  <DIV style=3D"FONT: 10pt arial"><B>Sent:</B> Friday, November 16, 2007 =
2:16=20
  PM</DIV>
  <DIV style=3D"FONT: 10pt arial"><B>Subject:</B> RE: [obm-l] =
Per=EDmetro=20
  M=EDnimo</DIV>
  <DIV><BR></DIV><BR><BR>
  <BLOCKQUOTE>
    <HR>
    &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;=20
    =
Ol=E1!<BR>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;=20
    <BR>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;=20
    Lembremo-nos, em primeiro momento, de que todo ret=E2ngulo =E9 um =
paralegramo.=20
    Conseguintemente, se denotarmos por a e b=20
    <BR><BR>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; =
as=20
    medidas de dois de seus lados consecutivos, a medida de se =
per=EDmetro ser=E1 P=20
    =3D 2(a + b)=20
    =
*<BR><BR>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; =
Para=20
    tal afirma=E7=E3o recorremos =E0 seguinte propriedade dos =
paralelogramos: qualquer=20
    par de lados opostos s=E3o=20
    =
congruentes.<BR>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;=20
    <BR>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; =
Sabemos=20
    ainda que a =E1rea do ret=E2ngulo d=E1-se pelo produto das medidas =
de seus lados:=20
    A =3D ab **. Logo, =E9 v=E1lido afirmar que=20
    =
<BR>&nbsp;<BR>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp=
; a=20
    =3D A/b=20
    =
***.<BR><BR>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; =
Se=20
    substituirmos *** em *, obteremos a lei de correspond=EAncia da =
"fun=E7=E3o=20
    =
per=EDmetro":<BR><BR>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbs=
p;&nbsp;=20
    P(b) =3D 2(A + b=B2)/b<BR><BR>&nbsp; &nbsp; &nbsp; &nbsp; =
&nbsp;&nbsp; definida=20
    em R+ com imagens em=20
    R+.<BR><BR>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; &nbsp; A =
fun=E7=E3o=20
    derivada de P =E9 definida pela seguinte=20
    =
lei<BR><BR>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;=20
    P'(b) =3D 2(b=B2 -=20
    =
A)/b<BR><BR>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; =
A=20
    =FAnica raiz dessa fun=E7=E3o =E9 b =3D 20, uma vez que D(P') =3D =
R+. Com um pouco de=20
    paci=EAncia, =E9 poss=EDvel mostrar que existe=20
    =
uma<BR><BR>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;=20
    vizinhan=E7a V de b =3D 20 tal que P'(x) &lt; 0 se x &lt; b e P'(x) =
&gt; 0 se x=20
    &gt; b. Por conseguinte P(b) =E9 o m=EDnimo absoluto de P, pois=20
    n=E3o<BR><BR>&nbsp; &nbsp; &nbsp; &nbsp; &nbsp;&nbsp; existem outros =

    extremantes.<BR><BR>&nbsp; &nbsp; &nbsp; &nbsp; &nbsp;&nbsp; Assim, =
o=20
    per=EDmetro m=EDnimo do ret=E2ngulo cuja =E1rea vale 400cm=B2 =E9 =
P(20) =3D=20
    80cm.<BR><BR>&nbsp; &nbsp; &nbsp; &nbsp; &nbsp;&nbsp; Acho que =E9 =
isso.<BR>
    <BLOCKQUOTE>
      <HR>
      Date: Fri, 16 Nov 2007 12:33:39 -0200<BR>Subject: [obm-l] =
Per=EDmetro=20
      M=EDnimo<BR>From: oliveira-aline1983@xxxxxxxxxx<BR>To:=20
      obm-l@xxxxxxxxxxxxxx<BR><BR>
      <DIV style=3D"FONT-SIZE: 12px; FONT-FAMILY: verdana,arial">
      <DIV>
      <P class=3DEC_MsoNormal style=3D"TEXT-ALIGN: justify"><SPAN=20
      style=3D"FONT-SIZE: 10pt; FONT-FAMILY: Arial">Pessoal ,n=E3o =
consigo montar=20
      minha fun=E7=E3o quadr=E1tica em fun=E7=E3o do =
per=EDmetro.</SPAN></P>
      <P class=3DEC_MsoNormal style=3D"TEXT-ALIGN: justify"><SPAN=20
      style=3D"FONT-SIZE: 10pt; FONT-FAMILY: Arial"></SPAN>&nbsp;</P>
      <P class=3DEC_MsoNormal style=3D"TEXT-ALIGN: justify"><SPAN=20
      style=3D"FONT-SIZE: 10pt; FONT-FAMILY: Arial">Dentre os =
ret=E2ngulos com =E1rea=20
      400cm<SUP>2</SUP>, existe um, cujo per=EDmetro =E9 o menor =
poss=EDvel. Qual o=20
      per=EDmetro deste ret=E2ngulo, em cm?</SPAN></P>
      <P class=3DEC_MsoNormal style=3D"TEXT-ALIGN: justify"><SPAN=20
      style=3D"FONT-SIZE: 10pt; FONT-FAMILY: Arial"></SPAN>&nbsp;</P>
      <P class=3DEC_MsoNormal style=3D"TEXT-ALIGN: justify"><SPAN=20
      style=3D"FONT-SIZE: 10pt; FONT-FAMILY: =
Arial"></SPAN>&nbsp;</P></DIV></DIV></BLOCKQUOTE></BLOCKQUOTE><BR>
  <HR>
  Veja mapas e encontre as melhores rotas para fugir do tr=E2nsito com o =
Live=20
  Search Maps! <A =
href=3D"http://www.livemaps.com.br/index.aspx?tr=3Dtrue"=20
  target=3D_new>Experimente j=E1!</A> </BLOCKQUOTE></BODY></HTML>

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