Question

A right triangle whose hypotenuse is StartRoot 18 EndRoot18 m long is revolved about one of its legs to generate a right circular cone. Find the​ radius, height, and volume of the cone of greatest volume that can be made this way.

Answer 1

V =
`(1)/(3)`π(8 -
`(8)/(3)`)
`\sqrt{(8)/(3) }` = 9.11
`m^(3)`

r = 4
`(√(3) )/(3)`

h =
`\sqrt{(8)/(3) }`

Explanation:

Given that the right triangle whose hypotenuse is
`√(8)`

• Let r is the radius of the cone
• Let h is the height of the cone

We know that:

`r^(2) + h^(2) = 8`

<=>
`r^(2) = 8 - h^(2)`

The volume of the cone is:

V = π
`r^(2) (1)/(3) h`

<=> V = π
`(1)/(3)(8 - h^(2) )h`

Differentiate w.r.t h

`(dV)/(dh)` = π
`(1)/(3)` (8 -
`3h^(2)`)

For maximum/minimum:
`(dV)/(dh)` = 0

<=> π
`(1)/(3)` (8 -
`3h^(2)`) = 0

<=>
`h^(2)` =
`(8)/(3)`

<=> h =
`\sqrt{(8)/(3) }`

=>
`r^(2)` =
`(16)/(3)`

<=> r = 4
`(√(3) )/(3)`

So the volume of the cone is:

V =
`(1)/(3)`π(8 -
`(8)/(3)`)
`\sqrt{(8)/(3) }` = 9.11
`m^(3)`