document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=881049>Question 881049</A>: <I><SPAN STYLE=\"word-break: break-all;\"> The area of a rectangle is 30cm squared <BR>\n" );
document.write( "its lenghth and width are whole numbers<BR>\n" );
document.write( "Find the smallest possible permimeter of this rectangle\r<BR>\n" );
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document.write( "This question is 4 marks </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #531920 by </B><A HREF=/tutors/aboutme.mpl?userid=Boyen><B>Boyen(12)</B></A><img src=\"/images/stars/stars-06.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=Boyen><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=531920\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> perimeter = (l+w)*2<BR>\n" );
document.write( "area = w*l\r<BR>\n" );
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document.write( "Out of this we learn that you want to look for 2 divisors that counted together are as low as possible<BR>\n" );
document.write( "all divisors of 30 :<BR>\n" );
document.write( "1*30 , 2*15 , 3*10, 5*6, 6*5, <BR>\n" );
document.write( "5*6 is the answer because 5+6 = only 11\r<BR>\n" );
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document.write( "so perimeter = 11*2 = 22 </SPAN>\n" );
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