Question

The maximum volume of the right circular cone with slant height $6$ units is:

• A. $4\sqrt{3} \pi$ cubic units
• B. $16\sqrt{3} \pi$ cubic units
• C. $3\sqrt{3} \pi$ cubic units
• D. $6\sqrt{3} \pi$ cubic units

Answer 1

The Correct Option is B

1. Understand the problem:

We need to find the maximum volume of a right circular cone with a given slant height of 6 units.

2. Relate variables using the Pythagorean theorem:

For a cone with radius \( r \), height \( h \), and slant height \( l = 6 \):

\[ r^2 + h^2 = l^2 \implies r^2 + h^2 = 36 \implies r^2 = 36 - h^2 \]

3. Express volume in terms of \( h \):

The volume \( V \) of a cone is:

\[ V = \frac{1}{3} \pi r^2 h = \frac{1}{3} \pi (36 - h^2) h = \frac{1}{3} \pi (36h - h^3) \]

4. Find the critical points for maximum volume:

Differentiate \( V \) with respect to \( h \):

\[ \frac{dV}{dh} = \frac{1}{3} \pi (36 - 3h^2) \]

Set \( \frac{dV}{dh} = 0 \):

\[ 36 - 3h^2 = 0 \implies h^2 = 12 \implies h = 2\sqrt{3} \]

5. Verify it's a maximum:

Second derivative:

\[ \frac{d^2V}{dh^2} = \frac{1}{3} \pi (-6h) = -2\pi h \]

At \( h = 2\sqrt{3} \), \( \frac{d^2V}{dh^2} = -4\sqrt{3}\pi < 0 \), confirming a maximum.

6. Compute the maximum volume:

Substitute \( h = 2\sqrt{3} \) back into the volume formula:

\[ V = \frac{1}{3} \pi (36 \cdot 2\sqrt{3} - (2\sqrt{3})^3) = \frac{1}{3} \pi (72\sqrt{3} - 24\sqrt{3}) = \frac{1}{3} \pi (48\sqrt{3}) = 16\sqrt{3}\pi \]

Correct Answer: (B) \( 16\sqrt{3} \, \pi \) cubic units

Answer 2

1. Volume of a Cone:

$ V = \frac{1}{3} \pi r^2 h $, where $ r $ is the radius of the base and $ h $ is the height.

2. Relationship between Slant Height, Radius, and Height:

Given the slant height ($ l = 6 $), we have the relationship: $ r^2 + h^2 = l^2 = 6^2 = 36 $.

3. Express Volume in terms of a Single Variable:

From the slant height relationship, we can express $ r^2 $ as $ r^2 = 36 - h^2 $. Substituting this into the volume formula:

\[ V = \frac{1}{3} \pi (36 - h^2) h = \frac{\pi}{3} (36h - h^3) \]

4. Maximize the Volume:

To find the maximum volume, we need to find the critical points by taking the derivative of $ V $ with respect to $ h $ and setting it to zero:

\[ \frac{dV}{dh} = \frac{\pi}{3} (36 - 3h^2) \]

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

\[ \frac{\pi}{3} (36 - 3h^2) = 0 \] \[ 36 - 3h^2 = 0 \] \[ 3h^2 = 36 \] \[ h^2 = 12 \] \[ h = \sqrt{12} = 2\sqrt{3} \quad \text{(We take the positive root since height cannot be negative)} \]

\[ \frac{d^2V}{dh^2} = \frac{\pi}{3} (-6h) = -2\pi h \]

Since $ h $ is positive, $ \frac{d^2V}{dh^2} $ is negative, indicating a maximum.

\[ r^2 = 36 - h^2 = 36 - 12 = 24 \] \[ r = \sqrt{24} = 2\sqrt{6} \]

\[ V = \frac{1}{3} \pi r^2 h = \frac{1}{3} \pi (24)(2\sqrt{3}) = \frac{1}{3} \pi (48\sqrt{3}) = 16\sqrt{3}\pi \]

Therefore, the maximum volume of the right circular cone is $ 16\sqrt{3}\pi $ cubic units.