I think the answer may have vaguely biblical connotations?


A square has the largest area for any given perimeter, so assume D=H.

Then volume = L*D^2, where L=2000-4D

volume = (2000-4D)*(D^2) = 2000D^2 - 4*D^3

Take the 1st differential and solve for zero:

  4000D - 12D^2 = 0

Simplify
  4000 - 12D = 0


12D = 4000 so D = 333.3

L = 2000 - (4*333.3) = 666.6


That took a few minutes but only because I haven't performed any calculus for a long while.  Then again, it might be wrong!

Regards

Mike





>-----Original Message-----
>From: piclist-bounces@MIT.EDU [mailto:piclist-bounces@MIT.EDU] 
>On Behalf Of Russell McMahon
>Sent: 11 June 2007 12:09
>To: PIC List
>Subject: [OT]:: Today's problem - send a parcel to Kristin
>
>
>A diversion for those who can be bothered:
>
>Kristin is studying in Australia.
>The weather is getting cold lately.
>Russell & Valerie want to send her a box full of clothes, 
>eiderdown, text books and more. The largest box that can be 
>sent by parcel post is
>
>    Largest side <= 1000 mm
>    Sum of L +     2 x (D + H)  <= 2000 mm = Volume.weight = VW
>
>        L     length
>        D    depth
>        H    Height
>
>True volume is of course L x D x H
>
>Assignment for students:
>
>
>                What box dimensions maximise the true volume 
>that can be sent?
>                Assume a rectangular box.
>
>                Report time to solve.
>
>
>The answer seems somehow obvious in retrospect but was 
>annoyingly unintuitive (to me).
>
>Clue - of sorts:        John Crook will tell me that the answer is
>obvious on inspection using information theory (he'll be right).
>
>

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