To find the dimensions of the right-circular cylinder of greatest volume that can be inscribed in a cone with a radius of $R = 5 \, \text{cm}$ and a height $H = 12 \, \text{cm}$, we proceed as follows:

Step 1: Define variables and relationships

Let:

• $r$ be the radius of the cylinder,
• $h$ be the height of the cylinder.

The volume of the cylinder is given by: $V = \pi r^2 h$

The cone's shape imposes a geometric constraint:

• For any $r$, the height $h$ of the cylinder depends on $r$ because the cylinder is inscribed in the cone.

From the geometry of the cone:

• The slope of the cone is $-\frac{12}{5}$, so: $h = 12 - \frac{12}{5}r$

Step 2: Substitute $h$ in terms of $r$ into $V$

$V = \pi r^2 \left(12 - \frac{12}{5}r\right)$ Simplify: $V = \pi \left(12r^2 - \frac{12}{5}r^3\right)$ $V = 12\pi r^2 - \frac{12\pi}{5}r^3$

Step 3: Differentiate $V$ with respect to $r$

To maximize $V$, find $\frac{dV}{dr}$ and solve for when it equals zero: $\frac{dV}{dr} = 24\pi r - \frac{36\pi}{5}r^2$ Set $\frac{dV}{dr} = 0$: $24\pi r - \frac{36\pi}{5}r^2 = 0$ Factorize: $\pi r \left(24 - \frac{36}{5}r\right) = 0$

Step 4: Solve for $r$

• $\pi r = 0$ gives $r = 0$, which is not meaningful.
• Solve $24 - \frac{36}{5}r = 0$: $24 = \frac{36}{5}r$ $r = \frac{24 \times 5}{36} = \frac{120}{36} = \frac{10}{3} \, \text{cm}$

Step 5: Find $h$

Substitute $r = \frac{10}{3}$ into $h = 12 - \frac{12}{5}r$: $h = 12 - \frac{12}{5} \cdot \frac{10}{3}$ $h = 12 - \frac{120}{15}$ $h = 12 - 8 = 4 \, \text{cm}$

Final Answer:

• The radius of the cylinder is $r = \frac{10}{3} \, \text{cm}$,
• The height of the cylinder is $h = 4 \, \text{cm}$.

Would you like detailed visual explanations or calculations?

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Relative Questions:

1. How do constraints affect optimization problems in geometry?
2. Can you explain other uses of calculus in finding extrema for volumes?
3. How does the slope of the cone influence the cylinder's dimensions?
4. What are some real-world applications of such optimization problems?
5. How does the derivative test confirm the maximum in this case?

Tip: Always check the second derivative to ensure whether the critical point is a maximum or minimum.