Prove that the volume of the largest cone that can be inscribed is a sphere of radius R is 8/27.
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Answer:
Let the centre of the sphere be O and radius be R. Let the height and radius of the variable cone inside the sphere be h and r respectively.
So, in the diagram, OA=OB=R,AD=h,BD=r
OD=AD−OA=h−R
Using Pythagoras Theorem in △OBD,
OB
2
=OD
2
+BD
2
⇒R
2
=(h−R)
2
+r
2
⇒R
2
=h
2
+R
2
−2hR+r
2
⇒r
2
=2hR−h
2
Volume of the cone V =
3
1
πr
2
h
=
3
1
π(2hR−h
2
)h
=
3
2πh
2
R
−
3
πh
3
For maximum volume,
dh
dV
=0
⇒0=
3
4πhR
−πh
2
⇒h=
3
4R
∴V =
3
2πR
9
16R
2
−
81
64πR
3
=
81
(96−64)πR
3
=
81
32πR
3
=
27
8
×
3
4
πR
3
We know that the volume of the sphere is V
s
=
3
4
πR
3
Therefore, V=
27
8
V
s
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