A box has integer dimensions, and when each one is increased by 2, its volume doubles. What is the largest possible dimension?

I call this one somewhat analytical because I use a graphing program to derive one part and have yet to show it analyticaly.

start with
(x+2)(y+2)(z+2)=2xyz
(x+2)(y+2)z+2(x+2)(y+2)=2xyz
2(x+2)(y+2)=(2xy-(x+2)(y+2))z
z=2(x+2)(y+2)/[2xy-(x+2)(y+2)]
no we can arbitrarily set 0<=x<=y<=z
so we now have
y<=2(x+2)(y+2)/[2xy-(x+2)(y+2)]
y>=x
x>=0

graphing these 3 innequalities it is easy to see that for integer values of x we are stuck with 3,4,5,6

now if x=3 then we have
y<=10(y+2)/(y-10)
y(y-10)<=10(y+2)
y^2-10y<=10y+20
y^2-20y-20<=0
now this restricts y to the interval [3,20]
and by simply trying each of these y values we get the
largest z value when y=11 and z=130

if x=4 we have
y<=12(y+2)/[2y-12]
y<=6(y+2)/(y-6)
y(y-6)<=6(y+2)
y^2-6y<=6y+12
y^2-12y-12<=0
and this restricts y to the interval [4,12]
and the largest z out of these is z=54 when y=7

and for x=5 by similar method we restrict y to the interval
[5,10]
and this gives the largest z at z=98 when y=5

and finally for x=6 we restrict y to [6,8]
and this gives largest z at z=16 and y=6

and thus the largest z over all is 130 when y=11 and x=3 which matches the solution found in previous posts.

I am working on a analytical proof that x is restricted to [3,6]

Posted by Daniel on 2009-01-02 21:41:48