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Question

The altitude of the right circular cone of maximum volume that can be inscribed in a sphere of radius r is

A
r2
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B
r3
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C
3r4
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D
4r3
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Solution

The correct option is D 4r3
Let R be the radius and h be the height of cone.
OA=hr
In ΔOAB,
r2=R2+(hr)2
r2=R2+h2+r22rh
R2=2rhh2
The volume V of the cone is given by
V=13πR2h
=13πh(2rhh2)=13π(2rh2h3)
On differentiating with respect to h, we get
dVdh=13π(4rh3h2)
For maximum and minimum, put dVdh=0
4rh=3h2
4r=3h
h=4r3 .....(h0)
Now, d2Vdh2=13π(4r6h)
AT h=4r3,
(d2Vdh2)h=4r3=13π(4r6×4r3)
=π3(4r8r)
=4rπ3<0
V is maximum when h=4r3.
Hence, volume of the cone is maximum when h=4r3, which is the altitude of cone.
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