A cylinder is inscribed in a right circular cone of height 8 and radius (at the base) equal to 2. What are the dimensions of such a cylinder which has maximum volume?
Radius?
Height?
did you make a sketch??
let the radius of the cylinder be r
let the height of the cylinder by h
by similarity I can set up the ratio h/(2-r) = 8/2
---> h = 8-4r
volume of cylinder
= pi(r^2)(8-4r)
= 8pi(r^2) - 4pi(r^3)
differentiate, set equal to zero, solve for r, etc
(I got r = 4/3)
To find the maximum volume of a cylinder that is inscribed in a cone, we can use the method of optimization.
Let's start by finding the relationship between the dimensions of the cylinder and cone.
The cylinder is inscribed in the cone such that its base lies in the base of the cone and its height is equal to the height of the cone.
Let's denote the radius of the cylinder as r and the height of the cylinder as h.
Given:
Height of the cone (and cylinder) = 8 units
Radius of the cone (at the base) = 2 units
To find the dimensions of the cylinder with maximum volume, we need to find the values of r and h that maximize the volume of the cylinder.
The volume of a cylinder can be calculated using the formula: V = πr^2h
To proceed, we need to express h in terms of r.
Taking a cross-section of the cone, we can use similar triangles to establish a relationship between the height of the cone (8 units), the radius of the cone (2 units), and the height of the cylinder (h units):
In the right triangle formed by the height of the cone (8 units), the radius of the cone (2 units), and the slant height of the cone (which is the hypotenuse), we can use the Pythagorean theorem:
(2)^2 + (8)^2 = r^2 + h^2
4 + 64 = r^2 + h^2
68 = r^2 + h^2
Since the cylinder is inscribed in the cone, the cylinder's radius is equal to the cone's radius at the base. So, r = 2.
Plugging in r = 2 into the equation 68 = r^2 + h^2, we can solve for h:
68 = 2^2 + h^2
68 = 4 + h^2
64 = h^2
h = 8
Therefore, the dimensions of the cylinder with maximum volume are:
Radius = 2 units
Height = 8 units.