document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=476600>Question 476600</A>: <I><SPAN STYLE=\"word-break: break-all;\"> If the area is 64 square feet what is the smallest perimeter </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #326900 by </B><A HREF=/tutors/aboutme.mpl?userid=solver91311><B>solver91311(24713)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=solver91311><IMG SRC=http://www.algebra.com/images/aboutme-small.gif BORDER=0 alt=\"About Me\" VALIGN=MIDDLE></A>&nbsp;<IMG SRC=http://www.algebra.com/cgi-bin/show-face.mpl?user=solver91311><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=326900\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> <font face=\"Times New Roman\" size=\"+2\">\r<BR>\n" );
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document.write( "The area of a rectangle is given by:\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE \ \ \ \ \ \ \ \ \ \ A\ =\ lw\">\r<BR>\n" );
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document.write( "so we can say:\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE \ \ \ \ \ \ \ \ \ \ l\ =\ \frac{A}{w}\">\r<BR>\n" );
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document.write( "The perimeter is given by:\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE \ \ \ \ \ \ \ \ \ \ P\ =\ 2l\ +\ 2w\">\r<BR>\n" );
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document.write( "Substituting we can write a function for the perimeter in terms of <IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\Large w\">:\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE \ \ \ \ \ \ \ \ \ \ P(w)\ =\ \frac{2A}{w}\ +\ 2w\">\r<BR>\n" );
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document.write( "Extremum where the first derivative is equal to zero:\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE \ \ \ \ \ \ \ \ \ \ \frac{dP}{dx}\ =\ 2\ -\ \frac{2A}{w^2}\">\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE \ \ \ \ \ \ \ \ \ \ 2\ =\ \frac{2A}{w^2}\">\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE \ \ \ \ \ \ \ \ \ \ w\ =\ \pm\sqrt{A}\">\r<BR>\n" );
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document.write( "But discard the negative root, hence:\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE \ \ \ \ \ \ \ \ \ \ w\ =\ \sqrt{A}\">\r<BR>\n" );
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document.write( "The extremum is a minimum if the second derivative is positive at the extreme point.\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE \ \ \ \ \ \ \ \ \ \ \frac{d^2P}{dx^2}\ =\ \frac{4A}{w^3}\">\r<BR>\n" );
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document.write( "Since <IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\Large w\"> and <IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\Large A\"> are both positive quantities, the second derivative is positive for all values of <IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\Large w\"> and therefore the extremum is a minimum.\r<BR>\n" );
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document.write( "Take the positive square root of your given area and you have your answer.\r<BR>\n" );
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document.write( "John<BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE e^{i\pi} + 1 = 0\"><BR>\n" );
document.write( "My calculator said it, I believe it, that settles it<BR>\n" );
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