The answer is: r=4/3sqrt2r=43√2, h=8/3h=83
We can imagine a vertical section of the figure, that would appear:
Google image
Let rr be the radius of the cone, RR be the radius of the sphere and hh be the height of the cone.
Let's put DhatOB=xDˆOB=x with limitations 0<=x<=pi0≤x≤π.
In the right-angled triangle DOBDOB:
r=DB=Rsinxr=DB=Rsinx, OD=RcosxOD=Rcosx, than
h=CD=OC+OD=R+Rcosx=R(1+cosx)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)rArrV=13πr2h⇒V=13π(Rsinx)2⋅R(1+cosx)⇒
V=1/3piR^3sin^2x(1+cosx)V=13πR3sin2x(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.
