To solve the problem of finding the dimensions of the right circular cylinder of maximum volume inscribed in a cone, let's proceed step by step using optimization.

Given:

• The cone has radius $R$ and height $H$.
• The cylinder has radius $r$ and height $h$.
• We aim to maximize the volume $V$ of the cylinder.

Step 1: Relate the dimensions of the cylinder and cone

From the geometry of the cone, the slant sides of the cone provide the relationship between $r$, $h$, $R$, and $H$. Using similar triangles, we can express $h$ in terms of $r$:

$\frac{h}{H} = \frac{R - r}{R} \quad \Rightarrow \quad h = H\left(1 - \frac{r}{R}\right).$

Step 2: Write the volume of the cylinder

The volume $V$ of the cylinder is given by:

$V = \pi r^2 h.$

Substitute $h = H(1 - \frac{r}{R})$:

$V = \pi r^2 \cdot H\left(1 - \frac{r}{R}\right).$

Simplify:

$V = \pi H r^2 \left(1 - \frac{r}{R}\right),$ $V = \pi H \left(r^2 - \frac{r^3}{R}\right).$

Step 3: Maximize $V$

To maximize $V$, take the derivative of $V$ with respect to $r$ and set it to zero:

$\frac{dV}{dr} = \pi H \left(2r - \frac{3r^2}{R}\right).$

Set $\frac{dV}{dr} = 0$:

$2r - \frac{3r^2}{R} = 0.$

Factorize:

$r \left(2 - \frac{3r}{R}\right) = 0.$

This gives two solutions:

1. $r = 0$ (trivial, no cylinder),
2. $2 - \frac{3r}{R} = 0 \quad \Rightarrow \quad r = \frac{2R}{3}.$

Step 4: Verify maximum

To confirm this is a maximum, take the second derivative:

$\frac{d^2V}{dr^2} = \pi H \left(2 - \frac{6r}{R}\right).$

At $r = \frac{2R}{3}$:

$\frac{d^2V}{dr^2} = \pi H \left(2 - \frac{6 \cdot \frac{2R}{3}}{R}\right) = \pi H (2 - 4) = -2\pi H.$

Since $\frac{d^2V}{dr^2} < 0$, $r = \frac{2R}{3}$ is a maximum.

Final Answer:

The radius $r$ of the cylinder of maximum volume is:

$\boxed{r = \frac{2R}{3}}$

This corresponds to option (A).

Would you like me to verify this further or compute additional related values, such as the corresponding height $h$ of the cylinder?

Here are 5 follow-up questions for exploration:

1. What is the maximum volume of the inscribed cylinder?
2. How does the cylinder's height compare to the cone's height at maximum volume?
3. Can we generalize this result to cones of arbitrary dimensions?
4. What happens if the cone's height $H$ is doubled but the radius $R$ remains constant?
5. How does this result apply to real-world engineering problems?

Tip: Always verify if critical points give maxima or minima using the second derivative test for optimization problems.