91Ó°ÊÓ

Calculating the volume of a cylinder is a crucial step in solving optimization problems like this one. The volume of a cylinder can be calculated using the formula:\[ V = \pi x^2 y \]where:

• \( V \) is the volume of the cylinder,
• \( x \) is the radius of the base of the cylinder,
• \( y \) is the height of the cylinder.

Understanding this formula helps us translate problems about cylinders into mathematical statements that can be solved.
In this exercise, we relate the height and radius of an inscribed cylinder to maximize its volume within the constraints of a conical shape. This involves using geometry and algebra to express the volume in terms of just one variable, which simplifies the differentiation process in later steps.

An inscribed shape means that one shape is contained perfectly within another. Here, the cylinder is perfectly inscribed within a cone.
For a cylinder inscribed in a cone, the key geometric property is similarity of triangles, which allows us to establish a relationship between the radius of the cylinder and the height inside the cone.
By setting the ratio of the cylinder's dimensions equal to the ratio of the cone's dimensions, expressed as:\[ \frac{x}{r} = \frac{h-y}{h} \]we can solve for one variable in terms of another.

• \( x \) represents the radius of the cylinder,
• \( r \) represents the base radius of the cone,
• \( h \) is the total height of the cone,
• \( y \) is the height of the cylinder.

This relationship is pivotal in setting up the optimization equation, because it models how the cylinder expands or contracts inside a fixed-size cone.

Differentiation is a powerful tool in calculus used to find the maximum or minimum values of functions. In this problem, differentiation helps us find the height at which the cylinder reaches its maximum volume.
The process involves taking the derivative of the volume expression with respect to the cylinder's height, \( y \). We've simplified the volume expression to:\[ V = \pi r^2 \frac{(h-y)^2}{h^2} y \]Next, finding the derivative of \( V \), and setting it to zero, simplifies the task to finding critical points.
The critical point \( y = \frac{h}{3} \) is derived from solving the quadratic equation following the derivative. This value of \( y \) maximizes the volume, and ensures through the second derivative test (where the second derivative \( \frac{d^2V}{dy^2} \) is negative) that the volume expression indeed has a maximum at this point. This approach illustrates the practical use of differentiation in maximizing physical quantities constrained by geometrical shapes.