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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 u=
m paralegramo. Conseguintemente, se denotarmos por a e b as medidas

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

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

           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 e=
xtremantes.

           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
=20
Aline Marques

_________________________________________________________________
Receba GR=C1TIS as mensagens do Messenger no seu celular quando voc=EA esti=
ver offline. Conhe=E7a  o MSN Mobile!
http://mobile.live.com/signup/signup2.aspx?lc=3Dpt-br=

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<html>
<head>
<style>
.hmmessage P
{
margin:0px;
padding:0px
}
body.hmmessage
{
FONT-SIZE: 10pt;
FONT-FAMILY:Tahoma
}
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<body class=3D'hmmessage'>
&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; Ol=E1!<br>&nbs=
p;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; <br>&nbsp;&nbsp;&n=
bsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; Lembremo-nos, em primeiro mo=
mento, de que todo ret=E2ngulo =E9 um paralegramo. Conseguintemente, se den=
otarmos por a e b as medidas<br><br>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nb=
sp;&nbsp;&nbsp;&nbsp; de dois de seus lados consecutivos, a medida de se pe=
r=EDmetro ser=E1 P =3D 2(a + b) *<br><br>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbs=
p;&nbsp;&nbsp;&nbsp;&nbsp; Para tal afirma=E7=E3o recorremos =E0 seguinte p=
ropriedade dos paralelogramos: qualquer par de lados opostos s=E3o congruen=
tes.<br>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; <br>&nbsp;&n=
bsp;&nbsp;&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 a =3D A/b ***.<br><br>&nbsp;&nbsp;&n=
bsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; Se substituirmos *** em *, o=
bteremos a lei de correspond=EAncia da "fun=E7=E3o per=EDmetro":<br><br>&nb=
sp;&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 i=
magens em R+.<br><br>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; &nbsp=
; A fun=E7=E3o derivada de P =E9 definida pela seguinte lei<br><br>&nbsp;&n=
bsp;&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 =FA=
nica raiz dessa fun=E7=E3o =E9 b =3D 20, uma vez que D(P') =3D R+. Com um p=
ouco de paci=EAncia, =E9 poss=EDvel mostrar que existe uma vizinhan=E7a V <=
br><br>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; de b =
=3D 20 tal que P'(x) &lt; 0 se x &lt; b e P'(x) &gt; 0 se x &gt; b. Por con=
seguinte P(b) =E9 o m=EDnimo absoluto de P, pois n=E3o existem outros extre=
mantes.<br><br>&nbsp; &nbsp; &nbsp; &nbsp; &nbsp;&nbsp; Assim, o per=EDmetr=
o m=EDnimo do ret=E2ngulo cuja =E1rea vale 400cm=B2 =E9 P(20) =3D 80cm.<br>=
<br>&nbsp; &nbsp; &nbsp; &nbsp; &nbsp;&nbsp; Acho que =E9 isso.<br><blockqu=
ote><hr>Date: Fri, 16 Nov 2007 12:33:39 -0200<br>Subject: [obm-l] Per=EDmet=
ro M=EDnimo<br>From: oliveira-aline1983@xxxxxxxxxx<br>To: obm-l@xxxxxxxxxxx=
.br<br><br><div style=3D"font-size: 12px; font-family: verdana,arial;">
<div></div>
<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>
<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;">Aline Marques</span></p></div></div>
</blockquote><br /><hr />Receba GR=C1TIS as mensagens do Messenger no seu c=
elular quando voc=EA estiver offline. Conhe=E7a  o MSN Mobile! <a href=3D'h=
ttp://mobile.live.com/signup/signup2.aspx?lc=3Dpt-br' target=3D'_new'>Cadas=
tre-se j=E1!</a></body>
</html>=

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