document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=245998>Question 245998</A>: <I><SPAN STYLE=\"word-break: break-all;\"> A carpenter is building a rectangular room with a fixed perimeter of 240 ft. What dimensions would yeild the maximum area? 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 #179689 by </B><A HREF=/tutors/aboutme.mpl?userid=oberobic><B>oberobic(2304)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=oberobic><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=179689\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> P = 240 = 2L + 2W<BR>\n" );
document.write( "A = L*W<BR>\n" );
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document.write( "By theorem (or from calculus), we know the maximum area is a square, so:  L=W.<BR>\n" );
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document.write( "P = 2L + 2W = 4L = 240<BR>\n" );
document.write( "L = 60<BR>\n" );
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document.write( "So a 60 by 60 room is 3600 sq ft in area.<BR>\n" );
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document.write( "We can test the area by testing another room with dimensions close to 60 by 60 and P = 240.<BR>\n" );
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document.write( "Let's try 59 by 61.  P = 2(59)+2(61) = 118+122 = 240.  OK.<BR>\n" );
document.write( "A = 59 * 61 = 3599, which is less than 3600.<BR>\n" );
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document.write( "Let's try 58 by 62.  P = 2(58) + 2(62) = 116 + 124 = 240. OK<BR>\n" );
document.write( "A = 58 * 63 =  3596, which is less than 3600 and less than 3599.<BR>\n" );
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document.write( "Let's try 57 by 63.  P = 240. OK<BR>\n" );
document.write( "A = 57 * 63 = 3591, which is less than 3600 and less than 3599 and less than 3596.  The area is decreasing as the values get further apart.<BR>\n" );
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document.write( "How about 1 by 119?  P = 2(1) + 2(119) = 2 + 238 = 240.<BR>\n" );
document.write( "A = 1*119 = 119, which is way less...<BR>\n" );
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document.write( "Another way to look at this problem is the define the perimeter = 2C, where C = L+W.<BR>\n" );
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document.write( "That way L = C-W, so the area will be (C-W) * W = CW - W^2.<BR>\n" );
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document.write( "The maximum value of CW - W^2 is C^2/4, which occurs at W = C/2.<BR>\n" );
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document.write( "So, if W = C/2, and C=L+W, then L=C/2, so we have a square. </SPAN>\n" );
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