If the question is to maximize surface area, then, as is pointed out in solutions above, we find that SA = 18*Pi*r, so the surface area will be largest when r is largest. The maximum surface area is larger than 144*Pi, however, since r does not need to be an integer. Since r can take any value less than 9, the surface area could take any value less than 162*Pi.
If the question is to maximize volume, Sudhir has found the correct radius (6), though the height should be 3, not 2. I don't follow the approach, however. This is a problem that would normally be solved using calculus, though if you have answer choices, it is not difficult to test them here. If you know calculus (which you don't need for the GMAT!) you can solve as follows:
V = Pi*r^2*h = Pi*r^2*(9-r) = 9*Pi*r^2 - Pi*r^3
V' = 18*Pi*r - 3*Pi*r^2
Set V' = 0 to find maxima and minima:
0 = 18*Pi*r - 3*Pi*r^2
0 = r*(18*Pi - 3*Pi*r)
So we have a minimum value when r = 0, and a maximum value when r = 6. Plugging r = 6 in to find V, we see that V = 108*Pi.
If you don't know calculus, the above probably won't make much sense- it's definitely not something tested on the GMAT.
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