document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=391158>Question 391158</A>: <I><SPAN STYLE=\"word-break: break-all;\"> what is the maximum area of a rectangle with a perimeter of 100 feet </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #277475 by </B><A HREF=/tutors/aboutme.mpl?userid=richard1234><B>richard1234(7193)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=richard1234><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=richard1234><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=277475\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Many different ways to solve this.\r<BR>\n" );
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document.write( "Solution 1<BR>\n" );
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document.write( "If the sides of the rectangle are x and 50-x (so that the perimeter is 100), the area is <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=x%2850-x%29+=+-x%5E2+%2B+50x ALIGN=MIDDLE  ALT=\"x%2850-x%29+=+-x%5E2+%2B+50x\"   DISABLED_style= \"max-width:90%; height:auto;\" >, which is a parabola in terms of x. The vertex occurs at <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=-b%2F2a ALIGN=MIDDLE  ALT=\"-b%2F2a\"   DISABLED_style= \"max-width:90%; height:auto;\" >, or x = 25. It points downward, so x = 25 maximizes the area, so the area is 625.\r<BR>\n" );
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document.write( "Solution 2<BR>\n" );
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document.write( "Another way is finding the extrema of <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=y+=+-x%5E2+%2B+50x ALIGN=MIDDLE  ALT=\"y+=+-x%5E2+%2B+50x\"   DISABLED_style= \"max-width:90%; height:auto;\" >. We have <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=dy%2Fdx+=+-2x+%2B+50 ALIGN=MIDDLE  ALT=\"dy%2Fdx+=+-2x+%2B+50\"   DISABLED_style= \"max-width:90%; height:auto;\" > which is equal to zero when x = 25 (this is actually where the -b/2a rule comes from).\r<BR>\n" );
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document.write( "Solution 3<BR>\n" );
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document.write( "Yet another way comes from an unusual theorem: AM-GM inequality. Suppose we have two terms <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=a%5B1%5D+=+x ALIGN=MIDDLE  ALT=\"a%5B1%5D+=+x\"   DISABLED_style= \"max-width:90%; height:auto;\" > and <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=a%5B2%5D+=+50-x ALIGN=MIDDLE  ALT=\"a%5B2%5D+=+50-x\"   DISABLED_style= \"max-width:90%; height:auto;\" >. Then by AM-GM,\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=%28a%5B1%5D+%2B+a%5B2%5D%29%2F2+%3E=+sqrt%28a%5B1%5Da%5B2%5D%29 ALIGN=MIDDLE  ALT=\"%28a%5B1%5D+%2B+a%5B2%5D%29%2F2+%3E=+sqrt%28a%5B1%5Da%5B2%5D%29\"   DISABLED_style= \"max-width:90%; height:auto;\" > Since <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=a%5B1%5D+%2B+a%5B2%5D+=+50 ALIGN=MIDDLE  ALT=\"a%5B1%5D+%2B+a%5B2%5D+=+50\"   DISABLED_style= \"max-width:90%; height:auto;\" >, <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=50%2F2+=+25 ALIGN=MIDDLE  ALT=\"50%2F2+=+25\"   DISABLED_style= \"max-width:90%; height:auto;\" >,\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=25+%3E=+sqrt%28a%5B1%5Da%5B2%5D%29 ALIGN=MIDDLE  ALT=\"25+%3E=+sqrt%28a%5B1%5Da%5B2%5D%29\"   DISABLED_style= \"max-width:90%; height:auto;\" >\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=625+%3E=+a%5B1%5Da%5B2%5D ALIGN=MIDDLE  ALT=\"625+%3E=+a%5B1%5Da%5B2%5D\"   DISABLED_style= \"max-width:90%; height:auto;\" >\r<BR>\n" );
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document.write( "The AM-GM inequality says the equality occurs if and only if all the <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=a%5Bi%5D ALIGN=MIDDLE  ALT=\"a%5Bi%5D\"   DISABLED_style= \"max-width:90%; height:auto;\" >'s are equal, that is, <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=a%5B1%5D+=+a%5B2%5D+=+25 ALIGN=MIDDLE  ALT=\"a%5B1%5D+=+a%5B2%5D+=+25\"   DISABLED_style= \"max-width:90%; height:auto;\" >, and the optimal area is 625.\r<BR>\n" );
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document.write( "It's pretty rare you'll see a solution like the last one. However AM-GM can be used to prove many similar theorems, i.e. proving that the rectangular solid of fixed surface area that has maximum volume is a cube. </SPAN>\n" );
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