document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=253415>Question 253415</A>: <I><SPAN STYLE=\"word-break: break-all;\"> 4. If the perimeter of a rectangle is 8&#8730;2 cm, what is the smallest possible value of the length of one of its diagonals in cm?\r<BR>\n" );
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document.write( "I've tried:<BR>\n" );
document.write( "let length be l, width be w, and diagonal be d<BR>\n" );
document.write( "let l be x, then w will be:<BR>\n" );
document.write( "(8&#8730;2-2x)/2 = 4&#8730;2-x\r<BR>\n" );
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document.write( "d^2=l^2+w^2<BR>\n" );
document.write( "      =x^2+(4&#8730;2-x)^2<BR>\n" );
document.write( "      =x^2+(32-8&#8730;2x-x^2)<BR>\n" );
document.write( "      =32-8&#8730;2x... </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #185684 by </B><A HREF=/tutors/aboutme.mpl?userid=stanbon><B>stanbon(75887)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=stanbon><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=185684\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> If the perimeter of a rectangle is 8&#8730;2 cm, what is the smallest possible value of the length of one of its diagonals in cm? <BR>\n" );
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document.write( "Perimeter = 2(L + W) = 8sqrt(2) cm<BR>\n" );
document.write( "L+W = 4sqrt(2) cm<BR>\n" );
document.write( "Note:<BR>\n" );
document.write( "W = 4sqrt(2)-L<BR>\n" );
document.write( "------------------------<BR>\n" );
document.write( "Diagonal = sqrt(L^2 + W^2)<BR>\n" );
document.write( "Substitute for W:<BR>\n" );
document.write( "D(L) = sqrt(L^2 + (4sqrt(2)-L)^2)<BR>\n" );
document.write( "D(L) = sqrt(L^2 + [L^2 - 8sqrt(2)L +32]<BR>\n" );
document.write( "D(L) = sqrt(2L^2 -8sqrt(2)L + 32)<BR>\n" );
document.write( "------<BR>\n" );
document.write( "You have a quadratic with a = 2 ; b = -8sqrt(2)<BR>\n" );
document.write( "Minimum occurs when L = -b/2a = 8sqrt(2)/(4) = 2sqrt(2)<BR>\n" );
document.write( "W = 4sqrt(2)-2sqrt(2) = 2sqrt(2)<BR>\n" );
document.write( "------------------------<BR>\n" );
document.write( "Minimum Diagonal:<BR>\n" );
document.write( "D^2 = L^2 + W^2<BR>\n" );
document.write( "D^2 = (2sqrt(2))^2 + (2sqrt(2))^2<BR>\n" );
document.write( "D^2 = 8 + 8<BR>\n" );
document.write( "D = 4 cm<BR>\n" );
document.write( "================<BR>\n" );
document.write( "Cheers,<BR>\n" );
document.write( "Stan H. </SPAN>\n" );
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