document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=629991>Question 629991</A>: <I><SPAN STYLE=\"word-break: break-all;\"> David has 440 yards of fencing and wishes to enclose a rectangular area. 1. Express the area A of the rectangle as a function of the width W of the rectangle.<BR>\n" );
document.write( "2. For what value of W is the area largest?<BR>\n" );
document.write( "3. What is the maximum area? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #396652 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=396652\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> The perimeter, P = 2(l+w)<BR>\n" );
document.write( "Expressing the length in terms of the width, we have<BR>\n" );
document.write( "l = P/2 - w<BR>\n" );
document.write( "The area, A = w*l = w(P/2 - w) = (P/2)w - w^2<BR>\n" );
document.write( "The area is maximized when dA/dw = 0:<BR>\n" );
document.write( "dA/dw = 0 = P/2 - 2w -> w = P/4, which means the area is maximized when l=w, ie. a square<BR>\n" );
document.write( "Inserting the value for P gives w = 440/4 = 110 yd<BR>\n" );
document.write( "A = 110*110 = 12,100 sq. yd.<BR>\n" );
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