<!doctype html public "-//w3c//dtd html 4.0 transitional//en"> <html> Dave Smith : 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. Dave Smith wrote: <blockquote TYPE=CITE>If I figure correctly, the diameter has to be 2.8 cm and the height of the rectangle is 1.4 cm.  Let me know if that's right or not, as I think I have blown enough time already, Ric. Dave Smith</blockquote> Quickly then... <img SRC="cid:part1.3FA23A59.74B4D3A5@grieg.uib.no" height=170 width=124> 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. So... 10 = pi r +  w + h + h. and you know that width is the same as 2 * r so 10 = pi r  + 2r + 2 h solving for h we get: h= 5 - 0.5pi r - r Now we write the formula for the area of this figure in terms of r Starting with 0.5pi r^2 + (width * height) and again remembering that width is 2r we get for Area A  = 0.5pi r^2 + 2r * ( 5 - 0.5pi r - r)     = 0.5pi r^2 + 10 r - pi r^2 - 2r^2     = 10r - 0.5pi r^2 - 2r^2 Now take the derivative of this to get A' = 10 - pi r - 4r A' = 0 gives the maximum point for the function of Area so 0 = 10 - pi r - 4r --> 10 = pi r + 4r --> 10 = r( pi + 4) -->   r = 10/(pi + 4), which is rounded off to 1.4 cm So knowing that the width of the rectangle is 2r we have that width = 20 / (pi + 4) .... or roughly  2.8 cm solving for the height of the rectangle is done by taking our first expression for height h = 5 - 0.5pi r - r and inserting 10 / (pi + 4) .... or our rounded 1.4 cm which yields h = r So for the maximum total area this figure can have is when width  = 20 / (pi + 4) cm height = r = 10 / (pi + 4) cm Cute eh ? 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. :) Cheers, and thanks for indulging me ! RicB   -- Richard Brekne RPT, N.P.T.F. UiB, Bergen, Norway <A HREF="mailto:rbrekne@broadpark.no">mailto:rbrekne@broadpark.no</A> <A HREF="http://home.broadpark.no/~rbrekne/ricmain.html">http://home.broadpark.no/~rbrekne/ricmain.html</A> <A HREF="http://www.hf.uib.no/grieg/personer/cv_RB.html">http://www.hf.uib.no/grieg/personer/cv_RB.html</A>  </html>