Show that a right-circular cylinder of greatest volume that can be inscribed in a right-circular cone that has a volume that is 4/9 the volume of the cone.

Please, help out, this is for a study guide for a quiz. :) Thank you.

Amy, take a look at the solution I gave to the first of the "Related Questions" below.

It is the same kind of question except it has specific values.
Instead of a height of 12 and a radius of 4 for the cone, in your question you would have a height of H and a radius of R, where H and R are constants, that is, they are fixed values.
See if you can follow the same steps, then take the volume of the cone and the volume of the cylinder.
They should be in the ratio of 4:9

Get back to me if you run into problems, but give it a try.

To find the right-circular cylinder of greatest volume inscribed in a right-circular cone, we need to set up an optimization problem.

Let's denote the radius of the cone's base as "R" and the height of the cone as "H". We also need to find the radius and height of the cylinder.

The volume of the cone is given as V_cone = (1/3) * π * R^2 * H.
The volume of the cylinder is given as V_cylinder = π * r^2 * h, where r is the radius of the cylinder and h is the height of the cylinder.

We are given that the volume of the cylinder is 4/9 the volume of the cone, so we can write:

V_cylinder = (4/9) * V_cone.

Substituting the formulas for V_cone and V_cylinder:

π * r^2 * h = (4/9) * ((1/3) * π * R^2 * H).

We can simplify this equation by canceling out the common terms:

r^2 * h = (4/9) * (1/3) * R^2 * H.

Now, the goal is to maximize the volume of the cylinder, subject to the constraint of being inscribed in the cone.

To proceed, we need to introduce another constraint. Since the cylinder is inscribed in the cone, its height (h) will be equal to the height of the cone (H). Therefore, we can substitute h = H in the equation above:

r^2 * H = (4/9) * (1/3) * R^2 * H.

Simplifying further:

r^2 = (4/9) * (1/3) * R^2.

r^2 = (4/27) * R^2.

To maximize the volume of the cylinder, we can take the derivative of both sides with respect to r:

d/dr (r^2) = d/dr ((4/27) * R^2).

2r = (4/27) * 2R^2 * dR/dr.

Simplifying further:

r = (4/27) * R^2 * (dR/dr).

Since dr/dR represents the rate of change of r with respect to R, we can see that for a right-circular cylinder inscribed in a right-circular cone, the ratio of r to R is constant.

Therefore, to maximize the volume of the cylinder, we need to maximize the ratio of r to R. The maximum value of this ratio is achieved when r is equal to R.

Therefore, the right-circular cylinder of greatest volume that can be inscribed in a right-circular cone is a cylinder whose base radius (r) is equal to the cone's base radius (R).