Given a right circular cone, you put an upside-down cone inside it so that its vertex is at the center of the base of the larger cone, and its base is parallel to the base of the larger cone. If you choose the upside-down cone to have the largest possible volume, what fraction of the volume of the larger cone does it occupy?
To find the fraction of the volume occupied by the upside-down cone, we need to compare the volume of the smaller cone to the volume of the larger cone.
First, let's label the dimensions of the cones. Let the radius of the larger cone's base be R and the height of the larger cone be H. Similarly, let the radius of the smaller cone's base be r and the height of the smaller cone be h.
We need to determine the values of r and h that would maximize the volume of the smaller cone.
Since the vertex of the smaller cone coincides with the center of the base of the larger cone, the smaller cone is similar to the larger cone. This means that the ratio of corresponding side lengths is constant:
r / R = h / H
Rearranging this equation, we get:
r = (R / H) * h
To maximize the volume of the smaller cone, we need to find the maximum value of r. Since r is directly proportional to h, we can choose the largest value for h, which would be the height of the larger cone H.
Therefore, the maximum value of r is:
r = (R / H) * H = R
So, the radius of the smaller cone's base is equal to the radius of the larger cone's base, and they share the same height.
The volume of a cone is given by the formula:
V = (1/3) * π * (r^2) * h
Substituting the values of r and h into the volume formula:
V_small = (1/3) * π * (R^2) * H
V_large = (1/3) * π * (R^2) * H
Notice that the volume of the smaller cone is equal to the volume of the larger cone.
Therefore, the fraction of the volume of the larger cone occupied by the upside-down cone is 1 (or 100%).