Posted by Isaac on Tuesday, August 9, 2011 at 11:25am.
This is a continuation of the question I answered for you earlier.
Since you titled it "Calculus" you should be able to finish it.
1. expand and simplify the expression for V
2. differentiate, you will have a quadratic
3. solve that quadratic by setting it equal to zero for a max of V
4. sub the value you found in 3. into the original volume equation
for the restriction, all you have to do is look at the equation for V
V of course has to be positive.
so x > 0 and x < 3 or else the base side values make no sense.
restriction on x : 0 < x < 3
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