Final answer:
To prove that the inscribed cylinder with the greatest volume has a height that is one third of the cone's height, we use optimization and calculus. This results in an optimal cylinder height of h/3 and the cylinder's volume being 4/9 of the cone's volume.
Step-by-step explanation:
To show that the height of the right circular cylinder of greatest volume that can be inscribed in a right circular cone of height h and radius r is one third of the height of the cone, we use the concept of optimization and calculus. We first express the volume of the cylinder, V = πr²h_cylinder, in terms of the dimensions of the cone. Through differentiation and applying the condition for maximum volume, we can establish the relationship between the cylinder's height and the cone's height
Considering the proportions of similar triangles formed by the cone and the inscribed cylinder, the ratio of the cylinder's radius to its height is the same as the ratio of the cone's radius to its height. Letting x represent the height of the cylinder, the radius of the cylinder at a given height will be r(x/h). By substituting these values into the volume formula and finding the derivative, we set it to zero to find the critical point that gives the maximum volum
The critical point indicates that the optimal height of the cylinder, x_max, is indeed h/3, which is one third of the height of the cone. Furthermore, after calculating the maximum volume of this cylinder and comparing it with the volume of the cone, V_cone = (1/3)πr²h, we can confirm that the volume of the optimal cylinder is 4/9 times the volume of the cone.