AP EAMCET Engineering 22 Apr 2019 Shift 1 Paper

Consider the diagram,ln‌Δ‌OAM
r2+(

h

2

)2=32
r2+

h2

4

=9
r2=9−

h2

4

The volume of the cylinder is given by,
V=πr2h
=π(9−

h2

4

)2h
=π(9h−

h3

4

)
Differentiate above with respect to h

dV

dh

=(9−

3h2

4

)π
For maximum or minimum volume,

dV

dh

=0
9−

3h2

4

=0

3h2

4

=9
h2=12
h=±2√3
So,
r2=9−3
r2=6
The maximum volume of cyclinder is,
Vmax=π(6)(2√3)
=12√3π sq units ‌‌‌‌‌[∵

d2V

dh2

=−

3h

2

<0]