Estimate the dimension of the right-circular Cylinder of greatest volume that can be inscalel -in a right-circular cone with a radius of 5 cm and a height of 12cm. derivativs

To find the dimensions of the **right-circular cylinder** of greatest volume that can be inscribed in a cone with radius \( R = 5 \, \text{cm} \) and height \( H = 12 \, \text{cm} \), we need to use calculus and optimization.

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### Step 1: Setup the Problem

1. **Cone Geometry**: The equation of the cone can be related to the geometry:
   - The cone's slant height gives a linear relationship between the height \( y \) of a point on the cone and the corresponding radius \( r \) at that height.
   - Slope: \(\frac{5}{12}\), so:
     \[
     r = \frac{5}{12}y
     \]

2. **Cylinder Volume**: The cylinder's volume is:
   \[
   V = \pi r^2 h
   \]
   where \( r \) is the base radius of the cylinder and \( h \) is its height.

---

### Step 2: Express Variables in Terms of \( y \)

- Let the cylinder's base lie at height \( y \) from the apex of the cone.
- The cylinder's radius \( r \) is then \( r = \frac{5}{12}y \).
- The cylinder's height \( h \) is the remaining part of the cone: \( h = 12 - y \).

Substitute these expressions into the volume formula:
\[
V = \pi \left(\frac{5}{12}y\right)^2 (12 - y)
\]

Simplify:
\[
V = \pi \frac{25}{144} y^2 (12 - y)
\]
\[
V = \frac{25\pi}{144} y^2 (12 - y)
\]

---

### Step 3: Differentiate to Maximize \( V \)

1. Expand \( V \):
\[
V = \frac{25\pi}{144} \left(12y^2 - y^3\right)
\]
\[
V = \frac{25\pi}{144} \cdot 12y^2 - \frac{25\pi}{144} \cdot y^3
\]

2. Differentiate \( V \) with respect to \( y \):
\[
\frac{dV}{dy} = \frac{25\pi}{144} \left(24y - 3y^2\right)
\]
\[
\frac{dV}{dy} = \frac{25\pi}{144} \cdot 3y (8 - y)
\]

3. Set \( \frac{dV}{dy} = 0 \):
\[
\frac{25\pi}{144} \cdot 3y (8 - y) = 0
\]

This gives:
\[
y = 0 \quad \text{or} \quad y = 8
\]

---

### Step 4: Verify Maximum Volume

- When \( y = 0 \), the cylinder has no volume.
- When \( y = 8 \), we compute the cylinder's volume.

At \( y = 8 \):
- Radius \( r = \frac{5}{12}y = \frac{5}{12} \cdot 8 = \frac{40}{12} = \frac{10}{3} \, \text{cm} \),
- Height \( h = 12 - y = 12 - 8 = 4 \, \text{cm} \).

Volume:
\[
V = \pi r^2 h = \pi \left(\frac{10}{3}\right)^2 \cdot 4
\]
\[
V = \pi \frac{100}{9} \cdot 4 = \frac{400\pi}{9} \, \text{cm}^3
\]

---

### Final Answer:

The **dimensions** of the cylinder of greatest volume are:
- Radius: \( \frac{10}{3} \, \text{cm} \),
- Height: \( 4 \, \text{cm} \).

The maximum **volume** is:
\[
V = \frac{400\pi}{9} \, \text{cm}^3 \approx 139.63 \, \text{cm}^3
\]

---

Do you need further clarifications or a deeper breakdown? 

### Related Questions:
1. How do you derive the formula for volume in a general cone or cylinder?
2. Can this method be applied to other shapes, such as a sphere inside a cone?
3. What happens to the cylinder's volume if the cone's dimensions change?
4. Why does \( \frac{dV}{dy} = 0 \) represent a critical point?
5. How does scaling affect the results of the optimization?

### Tip:
Always verify maximum or minimum values by testing endpoints or using the second derivative test!

Estimate the dimension of the right-circular cylinder of greatest volume that can be inscribed in a right-circular cone with a radius of 5 cm and a height of 12 cm.

Explore how to estimate the dimensions of the cylinder with the greatest volume that can fit inside a cone with a radius of 5 cm and a height of 12 cm. Learn step-by-step optimization techniques using calculus and geometry. Suitable for advanced high school students.

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