document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=416512>Question 416512</A>: <I><SPAN STYLE=\"word-break: break-all;\"> A farmer wants to build a pen with one side attached to his barn. He has 800metres of fencing. What is the max area that the pen can be?\r<BR>\n" );
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document.write( "p = 2L + W<BR>\n" );
document.write( "800 = 2L + W<BR>\n" );
document.write( "W = 800 - 2L\r<BR>\n" );
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document.write( " </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #291872 by </B><A HREF=/tutors/aboutme.mpl?userid=htmentor><B>htmentor(1343)</B></A><img src=\"/images/stars/stars-12.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=htmentor><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=htmentor><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=291872\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> This is an optimization problem requiring derivatives.  You are on the right track.<BR>\n" );
document.write( "The next step is to maximize the area<BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=A+=+L%2AW+=+%28800-2L%29%2AL+=+800L+-+2L%5E2 ALIGN=MIDDLE  ALT=\"A+=+L%2AW+=+%28800-2L%29%2AL+=+800L+-+2L%5E2\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "Now take the derivative and set = 0<BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=dA%2FdL+=+0+=+800+-+4L ALIGN=MIDDLE  ALT=\"dA%2FdL+=+0+=+800+-+4L\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "Solving for L gives L = 200 m<BR>\n" );
document.write( "And W = 800 - 2L -> W = 400 m<BR>\n" );
document.write( "So the max. area = A = L*W = 200*400 = 80000 m^2 </SPAN>\n" );
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