A solid cylinder of radius r cm and height h cm is to be carved out of a piece of solid wood in the form of a solid cone of base diameter 24 cm and height 32 cm. Calculate the value of r for which the volume of the cylinder carved out will be a maximum and find this maximum volume.

To solve this problem, we need to maximize the volume of a cylinder that can be carved out of a cone. Let's break down the solution step-by-step.

### Step 1: Understand the Given Dimensions

- The cone has a base diameter of 24 cm, so its radius \( R = \frac{24}{2} = 12 \) cm.
- The height of the cone \( H = 32 \) cm.
- The cylinder to be carved has a radius \( r \) cm and a height \( h \) cm.

### Step 2: Set Up the Volume Formulas

1. **Volume of the Cone:**

   The volume \( V_{\text{cone}} \) of the cone is given by:
   \[
   V_{\text{cone}} = \frac{1}{3} \pi R^2 H
   \]
   Plugging in the values:
   \[
   V_{\text{cone}} = \frac{1}{3} \pi (12)^2 (32) = \frac{1}{3} \pi \times 144 \times 32 = 1536\pi \text{ cm}^3
   \]

2. **Volume of the Cylinder:**

   The volume \( V_{\text{cylinder}} \) of the cylinder is:
   \[
   V_{\text{cylinder}} = \pi r^2 h
   \]

### Step 3: Relate the Dimensions of the Cylinder and the Cone

The cylinder is inscribed in the cone, so its dimensions must fit within the cone. If the cylinder's radius is \( r \), and its height is \( h \), these dimensions are related to the cone's geometry.

Using similar triangles, the ratio of corresponding sides gives us:
\[
\frac{r}{R} = \frac{H - h}{H}
\]
\[
\frac{r}{12} = \frac{32 - h}{32}
\]
Solving for \( h \):
\[
r = 12 \left( \frac{32 - h}{32} \right)
\]
\[
r = 12 - \frac{12h}{32}
\]
\[
r = 12 - \frac{3h}{8}
\]

Now, solve for \( h \) in terms of \( r \):
\[
\frac{3h}{8} = 12 - r
\]
\[
h = \frac{8(12 - r)}{3}
\]
\[
h = 32 - \frac{8r}{3}
\]

### Step 4: Express the Volume of the Cylinder as a Function of \( r \)

Substitute \( h \) into the volume formula for the cylinder:
\[
V_{\text{cylinder}} = \pi r^2 \left(32 - \frac{8r}{3}\right)
\]
Expand and simplify:
\[
V_{\text{cylinder}} = \pi r^2 \left(32 - \frac{8r}{3}\right) = \pi \left(32r^2 - \frac{8r^3}{3}\right)
\]
\[
V_{\text{cylinder}} = \pi \left(32r^2 - \frac{8}{3}r^3\right)
\]

### Step 5: Find the Value of \( r \) for Maximum Volume

To find the maximum volume, we need to differentiate \( V_{\text{cylinder}} \) with respect to \( r \) and set the derivative equal to zero.

\[
\frac{dV_{\text{cylinder}}}{dr} = \pi \left(64r - 8r^2\right) = 0
\]

Simplify:
\[
64r - 8r^2 = 0
\]
\[
8r(8 - r) = 0
\]

So, \( r = 0 \) or \( r = 8 \).

Since \( r = 0 \) would imply no cylinder, we choose \( r = 8 \).

### Step 6: Calculate the Maximum Volume

Substitute \( r = 8 \) into the volume formula:
\[
h = 32 - \frac{8 \times 8}{3} = 32 - \frac{64}{3} = \frac{96 - 64}{3} = \frac{32}{3}
\]

Now, calculate \( V_{\text{cylinder}} \):
\[
V_{\text{cylinder}} = \pi r^2 h = \pi (8^2) \left(\frac{32}{3}\right)
\]
\[
V_{\text{cylinder}} = \pi (64) \left(\frac{32}{3}\right) = \frac{2048\pi}{3} \text{ cm}^3
\]

### Conclusion

- The value of \( r \) for which the volume of the cylinder is maximized is \( r = 8 \) cm.
- The maximum volume of the cylinder is \(\frac{2048\pi}{3}\) cm\(^3\).

Would you like any further details or have any questions?

### Related Questions

1. What is the surface area of the cylinder when \( r = 8 \) cm and \( h = \frac{32}{3} \) cm?
2. How does the volume of the cylinder change if the cone's height is halved?
3. What would be the maximum volume of a cylinder carved from a cone with a different base radius?
4. How does the volume change if the radius \( r \) is 6 cm instead of 8 cm?
5. What is the relationship between the radius and height of the cylinder for it to have maximum volume?

### Tip

When dealing with optimization problems, always start by defining the relationships between the variables and constraints to simplify the process of finding maxima or minima.

Learn how to maximize the volume of a cylinder carved from a cone with detailed steps and calculations. Discover the optimal radius 'r' for maximum volume and find out the maximum volume achieved. Suitable for high school students learning about geometric optimization.

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