document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1083089>Question 1083089</A>: <I><SPAN STYLE=\"word-break: break-all;\"> The perimeter of a rectangle is 64m, what are three possible measurements for the length and width? what is the area of these rectangles?<BR>\n" );
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document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #697092 by </B><A HREF=/tutors/aboutme.mpl?userid=Theo><B>Theo(13342)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=Theo><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=Theo><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=697092\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> P = 2L + 2W\r<BR>\n" );
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document.write( "P is the perimeter<BR>\n" );
document.write( "L is the length<BR>\n" );
document.write( "W is the width\r<BR>\n" );
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document.write( "you can solve for L in terms of W and P as follows:\r<BR>\n" );
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document.write( "start with P = 2L + 2W<BR>\n" );
document.write( "subtract 2W from both sides to get P - 2W = 2L<BR>\n" );
document.write( "divide both sides fo 2 to get (P-2W)/2 = L\r<BR>\n" );
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document.write( "since you know P = 64, this becomes (64 - 2W)/2 = L\r<BR>\n" );
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document.write( "simplify to get 32 - W = L\r<BR>\n" );
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document.write( "the factors of 32 are 1, 2, 4, 8, 16, 32\r<BR>\n" );
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document.write( "W can't be 32, because then L would be 0 and that's not allowed.\r<BR>\n" );
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document.write( "if W = 16, then L = 16 and P = 2*16 + 2*16 = 32 + 32 = 64<BR>\n" );
document.write( "if W = 8, then L = 24 and P = 2*8 + 2*24 = 16 + 48 = 64<BR>\n" );
document.write( "if W = 4, then L = 28 and P = 2*4 + 2*28 = 8 + 56 = 64<BR>\n" );
document.write( "if W = 2, then L = 30 and P = 2*2 + 2*30 = 4 + 60 = 64<BR>\n" );
document.write( "if W = 1 then L = 31 and P = 2*1 + 2*31 = 2 + 62 = 64\r<BR>\n" );
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document.write( "all of these possible integer values of L and W yield P = 64.\r<BR>\n" );
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document.write( "the area, however, will be different for each one.\r<BR>\n" );
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document.write( "let A represent area.\r<BR>\n" );
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document.write( "the area of a rectangle is based on the formula of A = L * W\r<BR>\n" );
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document.write( "if W = 16, then L = 16 and A = 16 * 16 = 256<BR>\n" );
document.write( "if W = 8, then L = 24 and A = 8 * 24 = 192<BR>\n" );
document.write( "if W = 4, then L = 28 and A = 4 * 28 = 112<BR>\n" );
document.write( "if W = 2, then L = 30 and A = 2 * 30 = 60<BR>\n" );
document.write( "if W = 1 then L = 31 and A = 1 * 31 = 31\r<BR>\n" );
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document.write( "you will find that the greatest area of the rectangle is when the length and width are closest to each other in value.\r<BR>\n" );
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document.write( "this occurs when L = 16 and W = 16.\r<BR>\n" );
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document.write( "this can be seen graphically when you allow y to be the area and x to be the width.\r<BR>\n" );
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document.write( "your formula for the area becomes y = x * (32 - x)\r<BR>\n" );
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document.write( "here's what the graph looks like.\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" src = \"http://theo.x10hosting.com/2017/052903.jpg\" alt=\"$$$\" </>\r<BR>\n" );
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document.write( "the graph of the perimeter becomes y = 2x + 2*(23-x)\r<BR>\n" );
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document.write( "here's what the graph looks like:\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" src = \"http://theo.x10hosting.com/2017/052904.jpg\" alt=\"$$$\" </>\r<BR>\n" );
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document.write( "as you can see, the graph of the area forms a parabola which has a maximum value when x = (32 - x), while the graph of the perimeter is always equal to 64.\r<BR>\n" );
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