Question 1087412
Let {{{ w }}} = the width in cm
{{{ w + 2 }}} = the length in cm
{{{ 2*( w + 2 ) + 5 }}} = the height in cm
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{{{ V = w*( w + 2 )*( 2*( w + 2 ) + 5 ) }}}
{{{ V = ( w^2 + 2w )*( 2w + 4 + 5 ) }}}
{{{ V = ( w^2 + 2w )*( 2w + 9 ) }}}
{{{ 33 = ( w^2 + 2w )*( 2w + 9 ) }}}
{{{ 2w^3 + 4w^2 + 9w^2 + 18w = 33 }}}
{{{ 2w^3 + 13w^2 + 18w - 33 = 0 }}}
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Let {{{ w = 1 }}}
{{{ 2*1^3 + 13*1^2 + 18*1 - 33 = 0 }}}
{{{ 2 + 13 + 18 - 33 = 0 }}}
{{{ 33 - 33 = 0 }}}
{{{ w + 2 = 3 }}}
{{{ 2*( w + 2 ) + 5 = 2*3 + 5 }}}
{{{ 2*( w + 2 ) + 5 = 11 }}}
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The dimensions are:
1 x 3 x 11
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I just guessed at {{{ w=1 }}} 
There may be another way to do this, but
I dont know what that would be
You can get another opinion