Question

A hollow cone with a base radius of a cm and’ height of b cm is placed on a table. Show that) the volume of the largest cylinder that can be hidden underneath is `4/9` times the volume of the cone

Answer 1

Cone

Heigthof the cone = b cm

Base radius = a cm

Cylinder

Let the base radius be ‘r’ cm

Height be ‘h’ cm

From the figure, 

`"h"/("a" - "r") = "b"/"a"`

Using similar triangles property

⇒ h = `"b"/"a" ("a" - "r")`

= `"b" - "b"/"a"  "r"`

Volume of cylinder V = πr²h

= `pi"r"^2 ("b" - "b"/"a"  "r")`

V = `"r"("br"^2 - "b"/"a'  "r"^3)`

`"dV"/"dr" = pi(2"br" - 3  "br"^2/"a")`

= `pi/"a" (2"abr" - 3"br"^2)`

For maximum or minimum,

`"dV"/"dr"` = 0

⇒ br(2a – 3r) = 0

r = 0 and r = `(2"a")/3`

r = 0 is not possible

∴ r = `(2"a")/3`

`("d"^2"V")/("dr"^2) = "r"(2"b" - (6"br")/"a")`

At r = `(2"a")/3`

`("d"^2"V")/("dr"^2) = 2pi"b" - (6pi"b")/"a"((2"a")/3)`

= – 2πb < 0

∴ Volume is maximum when r = `(2"a")/3`

h = `"b" - "b"/"a" ((2"a")/3) = "b"/3`

Now, Volume of cylinder = `pi"r"^2"h"`

= `pi((2"a")/3)^2("b"/3)`

= `4/9 (1/3 pi"a"^2"b")`

= `4/9` ......(Volume of cone)

Hence proved.