Question

What is the volume of the largest circular cylinder that can be

inscribed in a circular cone of radius 5″ and height 8″?

Answer 1

The volume of the largest circular cylinder that can be inscribed in a circular cone with radius 5 inches and height 8 inches is
`\( (320\pi)/(3) \)` cubic inches.

Step-by-step explanation:

Let's denote:

-
`\( r \)` as the radius of the cylinder,

-
`\( h \)` as the height of the cylinder.

The cylinder is inscribed in the cone, so its dimensions must satisfy the conditions of the cone. In particular, the radius of the cylinder
`(\( r \))` must be equal to the radius of the cone's base (5 inches).

The height of the cylinder
`(\( h \))` plus the height of the remaining frustum of the cone must be equal to the height of the cone (8 inches).

Let's express the height of the frustum
`(\( h' \))` in terms of
`\( r \)` using similar triangles:

`\[ (h')/(r) = (8 - h)/(5) \]`

Solving for
`\( h' \):`

`\[ h' = (r(8 - h))/(5) \]`

The volume
`(\( V \))` of the cylinder is given by:

`\[ V = \pi r^2 h \]`

Substituting the expressions for
`\( r \)` and
`\( h' \):`

`\[ V = \pi \left(5^2\right) \left((r(8 - h))/(5)\right) \]`

Simplifying:

`\[ V = (\pi)/(5) r(8 - h)r^2 \]`

We know that
`\( r = 5 \)` (radius of the cone), and
`\( h + h' = 8 \)` (height of the cone). Substituting these values:

`\[ V = (\pi)/(5) \cdot 5(8 - h) \cdot 5^2 \]`

Simplifying further:

`\[ V = (320\pi)/(3) \]`

Therefore, the volume of the largest circular cylinder that can be inscribed in the given circular cone is
`\( (320\pi)/(3) \)\\` cubic inches.