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posted by  Aerowen on 11/7/2009 12:06:47 PM  |  status: Closed  |  Earned Karma: 35

Optimization

Course Textbook Chapter Problem Needs by
Calculus N/A N/A N/A 11/8/2009 at 4:00:00 PM
Question Details:
I actually enjoy these, but I'm a little stuck on a few, here is the first:
Let me set it up, not sure if I can get the picture to show. We have a rectangle piece of cardboard 24 X 36" , we want to make it so that we can make an open top box. To do this we cut squares out of each corner of the piece so that we can fold up the edges. How long should the sides of these squares be to get the largest possible volume out of our box?
"You cannot depend on your eyes when your imagination is out of focus." - Mark Twain
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Oracle
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posted by anonymousxyz on 11/7/2009 12:59:33 PM  |  status: Live
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Response Details:


x  =  the length of a side of each square in the diagram


0  <  x  <  12


The dimensions of the box are:


L  =  36  -  2x

w  =  24  -  2x

h  =  x


V

      =  Lwh

     =  ( 36  -  2x )( 24  -  2x )( x )

     =  ( 36  -  2x )( 24x  -  2x2 )

     =  864x  -  72x2  -  48x2  +  4x3

     =  4x3  -  120x2  +  864x


Maximize V:


V'  =  12x2  -  240x  +  864


V'  =  0  =>

12x2  -  240x  +  864  =  0  =>

x2  -  20x  +  72  =  0  =>

x  =  [    20  ±  sqrt(  400  -  ( 4 )( 1 )( 72 )  )    ]  /  [  2  ]  =>

x  =  [    20  ±  sqrt(  112  )    ]  /  [  2  ]  =>

x  =  [    20  ±  4√7    ]  /  [  2  ]  =>

x  =  10  ±  2√7  =>

x  =  4.70849738    or    x  =  15.2915026


0  <  x  <  12    =>    x  ˜  4.708 in


The maximum volume of the box is:


V( 4.70849738 )  =  1825.29659
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