document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1491>Question 1491</A>: <I><SPAN STYLE=\"word-break: break-all;\"> find the length and width of a rectangle having a perimeter of 200 meteres that will maximize its area. </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #523 by </B><A HREF=/tutors/aboutme.mpl?userid=khwang><B>khwang(438)</B></A><img src=\"/images/stars/stars-10.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=khwang><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/images/nopicture.jpg><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=523\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\">  Let L& W be the length &  width of the rectangle.<BR>\n" );
document.write( " The perimeter = 2(L+W) = 200, so L+W =100 or W = 100 -L <BR>\n" );
document.write( " Its area = LW = L(100 -L) = 100 L - L^2<BR>\n" );
document.write( "          = - (L^2 - 100 L + (100/2)^2)  + (100/2)^2 [Complete square]<BR>\n" );
document.write( "          = 2500 - (L -50)^2\r<BR>\n" );
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document.write( " We see that when L= 50,the area has the maximum value 2500.<BR>\n" );
document.write( " Also, when L = 50,W = 100 -50 = 50.\r<BR>\n" );
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document.write( " Answer: when L= 50,and W = 50, we attain the  maximum area 2500. </SPAN>\n" );
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