Sol. Let the angle AOB of the sectorial area removed be α . Then,

α =

⇒ AB = α

⇒ ACB = 2 π – α

⇒ Circumference of the base of the cone = 2 π – α

Let r be the radius of the base of the cone. Then, 2 π r = 2 π – α

⇒ r = 1 – ... (i)

Let V be the volume of the vessel. Then,

V = π r 2 × OD

⇒ V = π r 2

⇒ =

⇒ =

⇒ = –

For maximum or minimum values of V, we must have = 0

⇒ 2r –3r 3 = 0

⇒ r =

Now,

=–

Putting r = and = – , we get

= – × (2 –6) × < 0

Hence, V is maximum when r =

Putting r = in (i), we get α = 2 π –

Hence,

required sectorial area = . α

=

= π – π

= π