EAMCET Engineering 2010 Solved Paper

Let the height of the cone =h and the radius of the cone =r Given, radius of the sphere =R Now, In △OPB⇒‌R2‌=r2+(h−R)2
⇒‌r2‌=R2−(h−R)2
‌=(R+h−R)(R−h+R)
⇒‌r2‌=h(2R−h)
The volume of the cone is
‌V=‌

1

3

πr2h
⇒V‌=‌

1

3

πh(2R−h)h
⇒V‌=‌

π

3

(2Rh2−h3)
Differentiating with r to h
‌

dV

dh

=‌

π

3

(4Rh−3h2)
For maximum or minimum value of volume
‌‌

dV

dh

=0
⇒‌

π

3

(4Rh−3h2)=0
⇒h(4R−3h)=0
⇒h=0,h=‌

4R

3

‌‌‌ (Not possible) ‌
Now, ‌‌‌

d2V

dh2

=‌

π

3

(4R−6h)
‌(‌

d2V

dh2

)(‌at ‌h=‌

4R

3

)=‌

π

3

(4R−6⋅‌

4R

3

)
‌=‌

π

3

(4R−8R)=−‌

4π

3

R⇒‌ Negative ‌
ie., Maximum
Hence, the height of the cone of maximum volume is (‌

4R

3

).