The dimensions of the radius is 4/3[tex]\sqrt{2}[/tex] and that of height is 8/3.
Let r be the radius of the cone, be the radius of the sphere and h be the height of the cone.
Let's put DˆOB=x with limitations 0≤x≤π.
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=1/3πr²h ⇒V=1/3π(Rsinx)²⋅R(1+cosx)
V=1/3πR³sin²x(1+cosx)
V'=1/3πR³⋅[2sinx⋅cosx⋅(1+cosx)+sin2x(−sinx)]
=1/3πR³sinx[2cosx(1+cosx)−sin2x]
=1/3πR³sinx(2cosx+2cos2x−sin2x)
=1/3πR³sinx(2cosx+2cos2x−1+cos2x)
=1/3πR³sinx(3cos2x+2cosx−1)
Now let's find the signum of the derivative, since sinx≥0 for every x in the limitations, then:
V'≥0⇒3cos²x+2cosx−1≥0
⇒Δ/4=(b/2)²−ac =1+3=4
cosx=(−b/2)±√Δ/4 / a
=−1±2/3,
So:
cosx≤−1∨cosx≥1/3
The first one has only the solution:
x=π
the second:
−arccos(1/3)≤x≤arccos(1/3), but for the limitations:
0≤x≤arccos(1/3).
The function gros from zero to arccos(1/3), and then it decreases.
So x=arccos(1/3) is the maximum requested.
Let's find now r and h:
r=Rsinx
=R√1−cos2x
= 2√1−(1/3)²
= 2√1−1/9
=2√9−1/9
=2√8/9
=2⋅2√2/3
=4/3√2
h=R(1+cosx)=2(1+1/3)
=2(3+1/3)
=8/3.
Therefore the radius is 4/3√2 and the height is 8/3.
To know more about the cone refer to the link given below:
https://brainly.com/question/1082469
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