<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.0 Transitional//EN"> <HTML><HEAD> <META http-equiv=Content-Type content="text/html; = charset=iso-8859-1"> <META content="MSHTML 5.50.4728.2300" name=GENERATOR> <STYLE></STYLE> </HEAD> <BODY bgColor=#ffffff> <DIV><FONT face=Arial size=2>Thanks, Ric.  Everything I know, I = owe to my piano technology coach and mentor, Phil Bondi, RPT,  except trig, = Calculus and assorted other marginally useful things.  </FONT></DIV> <DIV><FONT face=Arial size=2></FONT> </DIV> <DIV><FONT face=Arial size=2>Dave Smith</FONT></DIV> <DIV><FONT face=Arial size=2>SW FL</FONT></DIV> <BLOCKQUOTE dir=ltr style="PADDING-RIGHT: 0px; PADDING-LEFT: 5px; MARGIN-LEFT: 5px; = BORDER-LEFT: #000000 2px solid; MARGIN-RIGHT: 0px"> <DIV style="FONT: 10pt arial">----- Original Message ----- </DIV> <DIV style="BACKGROUND: #e4e4e4; FONT: 10pt arial; font-color: = black"><B>From:</B> <A title=Richard.Brekne@grieg.uib.no href="mailto:Richard.Brekne@grieg.uib.no">Richard Brekne</A> </DIV> <DIV style="FONT: 10pt arial"><B>To:</B> <A = title=pianotech@ptg.org href="mailto:pianotech@ptg.org">Pianotech</A> </DIV> <DIV style="FONT: 10pt arial"><B>Sent:</B> Friday, October 31, 2003 = 5:32 AM</DIV> <DIV style="FONT: 10pt arial"><B>Subject:</B> Re: OT Trig = Puzzle</DIV> <DIV><BR></DIV>Dave Smith : <P>Correcto !! Mark Davidson also sent me the correct answer off line. = Bill Ballard was correct in noting that a bit of Calculus had to be used as = well. <P>Dave Smith wrote: <BLOCKQUOTE TYPE="CITE">If I figure correctly, the diameter has to = be 2.8 cm and the height of the <BR>rectangle is 1.4 cm.  Let me know if = that's right or not, as I think I have <BR>blown enough time already, Ric. <P>Dave Smith</P></BLOCKQUOTE>Quickly then... <P><IMG height=170 src="cid:006101c39fba$88b1e840$6501a8c0@pineis01.fl.comcast.net" = width=124> <P>The border is 10 cm. That plus the figure is all you have to go on. = But you know that the border is given by the circumfrence of the half circle, = added to the three remaining sides of the rectangle... i.e. width + height + = height. <P>So... 10 = pi r +  w + h + h. <P>and you know that width is the same as 2 * r <P>so <P>10 = pi r  + 2r + 2 h <P>solving for h we get: h= 5 - 0.5pi r - r <P>Now we write the formula for the area of this figure in terms of r <P>Starting with 0.5pi r^2 + (width * height) and again remembering = that width is 2r we get for Area <P>A  = 0.5pi r^2 + 2r * ( 5 - 0.5pi r - r) = <BR>    = 0.5pi r^2 + 10 r - pi r^2 - 2r^2 <P>    = 10r - 0.5pi r^2 - 2r^2 <P>Now take the derivative of this to get <P>A' = 10 - pi r - 4r <P>A' = 0 gives the maximum point for the function of Area so <P>0 = 10 - pi r - 4r <BR>--> 10 = pi r + 4r <BR>--> 10 = = r( pi + 4) <BR>-->   r = 10/(pi + 4), which is rounded off to 1.4 = cm <P>So knowing that the width of the rectangle is 2r we have that <P>width = 20 / (pi + 4) .... or roughly  2.8 cm <P>solving for the height of the rectangle is done by taking our first = expression for height <P>h = 5 - 0.5pi r - r and inserting 10 / (pi + 4) .... or our = rounded 1.4 cm <P>which yields h = r <P>So for the maximum total area this figure can have is when <P>width  = 20 / (pi + 4) cm <BR>height = r = 10 / (pi + 4) = cm <P>Cute eh ? <P>Nice efforts all the way around, and Congrats to Dave and Mark for = keeping their high school maths together !! As for those of you who got it = wrong... you may take solice in the fact that I stumbled around for 3 hours = last nite until I saw the solution... then it took a few minutes to execute... = but just so. :) <P>Cheers, and thanks for indulging me ! <P>RicB <BR>  <P>-- <BR>Richard Brekne <BR>RPT, N.P.T.F. <BR>UiB, Bergen, Norway = <BR><A href="mailto:rbrekne@broadpark.no">mailto:rbrekne@broadpark.no</A> = <BR><A = href="http://home.broadpark.no/~rbrekne/ricmain.html">http://home.broad= park.no/~rbrekne/ricmain.html</A> <BR><A = href="http://www.hf.uib.no/grieg/personer/cv_RB.html">http://www.hf.uib= .no/grieg/personer/cv_RB.html</A> <BR>  </P></BLOCKQUOTE></BODY></HTML>