OT Trig Puzzle

[Dave Smith]

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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. =20

Dave Smith
SW FL
  ----- Original Message -----=20
  From: Richard Brekne=20
  To: Pianotech=20
  Sent: Friday, October 31, 2003 5:32 AM
  Subject: Re: OT Trig Puzzle


  Dave Smith :=20
  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.=20

  Dave Smith wrote:=20

    If I figure correctly, the diameter has to be 2.8 cm and the height =
of the=20
    rectangle is 1.4 cm.  Let me know if that's right or not, as I think =
I have=20
    blown enough time already, Ric.=20
    Dave Smith

  Quickly then...=20
  =20

  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.=20

  So... 10 =3D pi r +  w + h + h.=20

  and you know that width is the same as 2 * r=20

  so=20

  10 =3D pi r  + 2r + 2 h=20

  solving for h we get: h=3D 5 - 0.5pi r - r=20

  Now we write the formula for the area of this figure in terms of r=20

  Starting with 0.5pi r^2 + (width * height) and again remembering that =
width is 2r we get for Area=20

  A  =3D 0.5pi r^2 + 2r * ( 5 - 0.5pi r - r)=20
      =3D 0.5pi r^2 + 10 r - pi r^2 - 2r^2=20

      =3D 10r - 0.5pi r^2 - 2r^2=20

  Now take the derivative of this to get=20

  A' =3D 10 - pi r - 4r=20

  A' =3D 0 gives the maximum point for the function of Area so=20

  0 =3D 10 - pi r - 4r=20
  --> 10 =3D pi r + 4r=20
  --> 10 =3D r( pi + 4)=20
  -->   r =3D 10/(pi + 4), which is rounded off to 1.4 cm=20

  So knowing that the width of the rectangle is 2r we have that=20

  width =3D 20 / (pi + 4) .... or roughly  2.8 cm=20

  solving for the height of the rectangle is done by taking our first =
expression for height=20

  h =3D 5 - 0.5pi r - r and inserting 10 / (pi + 4) .... or our rounded =
1.4 cm=20

  which yields h =3D r=20

  So for the maximum total area this figure can have is when=20

  width  =3D 20 / (pi + 4) cm=20
  height =3D r =3D 10 / (pi + 4) cm=20

  Cute eh ?=20

  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. :)=20

  Cheers, and thanks for indulging me !=20

  RicB=20
   =20

  --=20
  Richard Brekne=20
  RPT, N.P.T.F.=20
  UiB, Bergen, Norway=20
  mailto:rbrekne@broadpark.no=20
  http://home.broadpark.no/~rbrekne/ricmain.html=20
  http://www.hf.uib.no/grieg/personer/cv_RB.html=20
   =20


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