Question 822130
You are definitely on the right track.  This is an optimization problem, and we are trying to maximize the enclosed volume.  
{{{V = pi*r^2h}}}, and the volume will be maximized where dV/dr = 0.  You have already expressed h as a function of r using the information provided.
{{{h = (6-4pi*r)/6 = 1 - (2/3)pi*r}}}
So we make the substitution before taking the derivative:
{{{V = pi*r^2*(1-((2pi*r)/3))}}}
Taking the derivative and setting equal to zero you will end up with an expression like {{{2pi*r(1-pi*r) = 0}}}
Since r=0 is not a solution, we are left with {{{1-pi*r = 0}}} or {{{r = 1/pi}}}
Now solve for h:
{{{h = 1 - (2/3)pi/pi = 1/3}}}
The graph of the volume as a function of r is shown below.  We can see the volume maximizes at r = {{{1/pi}}}, a little over 0.3.
{{{graph(300,300,-1,2,-1,1,pi*x^2*(1-(2/3)*pi*x))}}}