The answer is: #r=4/3sqrt2#, #h=8/3#

We can imagine a vertical section of the figure, that would appear:

Let #r# be the radius of the cone, #R# be the radius of the sphere and #h# be the height of the cone.

Let's put #DhatOB=x# with limitations #0<=x<=pi#.

In the right-angled triangle #DOB#:

#r=DB=Rsinx#, #OD=Rcosx#, than

#h=CD=OC+OD=R+Rcosx=R(1+cosx)#.

So the volume of the cone is:

#V=1/3pir^2hrArrV=1/3pi(Rsinx)^2*R(1+cosx)rArr#

#V=1/3piR^3sin^2x(1+cosx)#

#V'=1/3piR^3*[2sinx*cosx*(1+cosx)+sin^2x(-sinx)]=#

#=1/3piR^3sinx[2cosx(1+cosx)-sin^2x]=#

#=1/3piR^3sinx(2cosx+2cos^2x-sin^2x)=#

#=1/3piR^3sinx(2cosx+2cos^2x-1+cos^2x)=#

#=1/3piR^3sinx(3cos^2x+2cosx-1)#

Now let's find the signum of the derivative, since #sinx>=0# for every #x# in the limitations, than:

#V'>=0rArr3cos^2x+2cosx-1>=0rArr#

#Delta/4=(b/2)^2-ac=1+3=4#

#cosx=((-b/2)+-sqrt(Delta/4))/a=(-1+-2)/3#,

So:

#cosx<=-1vvcosx>=1/3#

The first one has only the solution:

#x=pi#,

the second:

#-arccos(1/3)<=x<=arccos(1/3)#, but for the limitations:

#0<=x<=arccos(1/3)#.

The function growths from zero to #arccos(1/3)#, than it decreases.

So #x=arccos(1/3)# is the maximum requested.

Let's find now #r# and #h#:

#r=Rsinx=Rsqrt(1-cos^2x)=2sqrt(1-(1/3)^2)=2sqrt(1-1/9)=#

#=2sqrt((9-1)/9)=2sqrt(8/9)=2*2sqrt2/3=4/3sqrt2#

#h=R(1+cosx)=2(1+1/3)=2((3+1)/3)=8/3#.