Question 1122184: Find the volume of the largest cylinder that can be inscribed in a right circular cone of radius 3 inches and whose height is 10 inches.
Answer by solver91311(24713)   (Show Source):
You can put this solution on YOUR website!
See diagram. The isosceles triangle is a representation of a cross-section of the given cone. The rectangle is an arbitrary cylinder inscribed in the cone. The cross-section of the cone is placed on a coordinate grid such that the center of the base of the cone is at the origin and the apex of the cone is at the point (0,10). That means the intersection of a generator of the cone and a radius of the base of the cone is at the point (3,0).
Using the two points, (0,10) and (3,0), and designating the coordinate axes as  and ) , we can write an equation of the right-hand cone generator as \ =\ -\frac{10}{3}(r\ -\ 3)) . From the diagram, we can see that represents the radius of the cylinder and represents the height of the cylinder as a function of the radius. Note the domain of this function is the open interval (0,3) because the endpoints would represent zero volume cylinders.
The volume of a cylinder is given by the area of the base times the height. Using the information developed so far, we can write a function that represents the volume of the cylinder as a function of the radius of its base.
\ =\ \pi{r^2}\(-\frac{10}{3}\(r\ -\ 3\)\)\ =\ 10\pi\(-\frac{r^3}{3}\ +\ r^2\))
Then
)
Note  is not in the domain of either the cone generator function or the cylinder volume function. Clearly, there is an extremum at  . We could go through the process of checking the second derivative to ensure that this is a maximum, but intuitively it is pretty clear that this has to be a maximum since the volume near the endpoints of the function domain approaches zero and we have a substantial volume at and near
\ =\ 10\pi\(-\frac{2^3}{3}\ +\ 2^2\))
\ =\ 10\pi\(4\ -\ \frac{8}{3}\)\ =\ \frac{40\pi}{3}\text{ in^3})
John
My calculator said it, I believe it, that settles it
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