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[SPAM] RE: [obm-l] Perímetro Mínimo



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           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 pe=
r=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 40=
0cm=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 d=
o per=EDmetro.
=20
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?
=20
=20

_________________________________________________________________
Veja mapas e encontre as melhores rotas para fugir do tr=E2nsito com o Live=
 Search Maps!
http://www.livemaps.com.br/index.aspx?tr=3Dtrue=

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<html>
<head>
<style>
.hmmessage P
{
margin:0px;
padding:0px
}
body.hmmessage
{
FONT-SIZE: 10pt;
FONT-FAMILY:Tahoma
}
</style>
</head>
<body class=3D'hmmessage'>
<br><br><blockquote><hr>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nb=
sp;&nbsp; Ol=E1!<br>
&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; <br>
&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; Lembremo-nos, =
em primeiro momento, de que todo ret=E2ngulo =E9
um paralegramo. Conseguintemente, se denotarmos por a e b <br><br>&nbsp;&nb=
sp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; as medidas de dois de s=
eus lados consecutivos, a medida de se per=EDmetro ser=E1 P =3D 2(a + b) *<=
br>
<br>
&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; Para tal afirm=
a=E7=E3o recorremos =E0 seguinte propriedade dos
paralelogramos: qualquer par de lados opostos s=E3o congruentes.<br>
&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; <br>&nbsp;&nbsp;&nbs=
p;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; 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 <br>&nbsp;<br>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&=
nbsp;&nbsp; a =3D A/b ***.<br>
<br>
&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; Se substituirm=
os *** em *, obteremos a lei de correspond=EAncia da "fun=E7=E3o per=EDmetr=
o":<br>
<br>
&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; P(b) =3D 2(A +=
 b=B2)/b<br>
<br>
&nbsp; &nbsp; &nbsp; &nbsp; &nbsp;&nbsp; definida em R+ com imagens em R+.<=
br>
<br>
&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; &nbsp; A fun=E7=E3o deriva=
da de P =E9 definida pela seguinte lei<br>
<br>
&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; P'(b) =3D 2(b=
=B2 - A)/b<br>
<br>
&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; 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<br><br>&=
nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; vizinhan=E7a
V de b =3D 20 tal que P'(x) &lt; 0 se x &lt; b e P'(x) &gt; 0 se
x &gt; b. Por conseguinte P(b) =E9 o m=EDnimo absoluto de P, pois n=E3o<br>=
<br>&nbsp; &nbsp; &nbsp; &nbsp; &nbsp;&nbsp; existem outros extremantes.<br=
>
<br>
&nbsp; &nbsp; &nbsp; &nbsp; &nbsp;&nbsp; Assim, o per=EDmetro m=EDnimo do r=
et=E2ngulo cuja =E1rea vale 400cm=B2 =E9 P(20) =3D 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] P=
er=EDmetro M=EDnimo<br>From: oliveira-aline1983@xxxxxxxxxx<br>To: obm-l@mat=
.puc-rio.br<br><br><div style=3D"font-size: 12px; font-family: verdana,aria=
l;">

<div>
<p class=3D"EC_MsoNormal" style=3D"text-align: justify;"><span style=3D"fon=
t-size: 10pt; font-family: Arial;">Pessoal ,n=E3o consigo montar minha fun=
=E7=E3o quadr=E1tica em fun=E7=E3o do per=EDmetro.</span></p>
<p class=3D"EC_MsoNormal" style=3D"text-align: justify;"><span style=3D"fon=
t-size: 10pt; font-family: Arial;"></span>&nbsp;</p>
<p class=3D"EC_MsoNormal" style=3D"text-align: justify;"><span style=3D"fon=
t-size: 10pt; font-family: Arial;">Dentre os ret=E2ngulos com =E1rea 400cm<=
sup>2</sup>, existe um, cujo per=EDmetro =E9 o menor poss=EDvel. Qual o per=
=EDmetro deste ret=E2ngulo, em cm?</span></p>
<p class=3D"EC_MsoNormal" style=3D"text-align: justify;"><span style=3D"fon=
t-size: 10pt; font-family: Arial;"></span>&nbsp;</p>
<p class=3D"EC_MsoNormal" style=3D"text-align: justify;"><span style=3D"fon=
t-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 Search Maps! <a href=3D'=
http://www.livemaps.com.br/index.aspx?tr=3Dtrue' target=3D'_new'>Experiment=
e j=E1!</a></body>
</html>=

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