Key Concepts and Formulas

• Volume of a right circular cone: $V = \frac{1}{3}\pi r^2 h$, where $r$ is the radius and $h$ is the height.
• Relationship between slant height ($l$), radius ($r$), and height ($h$): $l^2 = r^2 + h^2$.
• Application of Derivatives: Finding maxima/minima by setting the first derivative to zero and using the second derivative test.

Step-by-Step Solution

Step 1: Define variables and establish the relationship

Let $r$ be the radius, $h$ be the height, and $l$ be the slant height of the cone. We are given that $l = 3$ m. The relationship between $r$, $h$, and $l$ is given by the Pythagorean theorem: $r^2 + h^2 = l^2$ Substituting $l = 3$, we get: $r^2 + h^2 = 3^2 = 9$

Step 2: Express the volume in terms of a single variable

The volume of the cone is given by: $V = \frac{1}{3}\pi r^2 h$ From the relationship $r^2 + h^2 = 9$, we can express $r^2$ in terms of $h$: $r^2 = 9 - h^2$ Substitute this into the volume equation: $V = \frac{1}{3}\pi (9 - h^2)h = \frac{1}{3}\pi (9h - h^3)$

Step 3: Find the critical points by taking the first derivative and setting it to zero

To find the maximum volume, we need to find the critical points of the volume function. We differentiate $V$ with respect to $h$: $\frac{dV}{dh} = \frac{1}{3}\pi (9 - 3h^2)$ Set $\frac{dV}{dh} = 0$: $\frac{1}{3}\pi (9 - 3h^2) = 0$ $9 - 3h^2 = 0$ $3h^2 = 9$ $h^2 = 3$ $h = \sqrt{3}$ Since $h$ represents the height, we only consider the positive root.

Step 4: Verify that the critical point corresponds to a maximum using the second derivative test

Now, we find the second derivative of $V$ with respect to $h$: $\frac{d^2V}{dh^2} = \frac{1}{3}\pi (-6h) = -2\pi h$ Evaluate the second derivative at $h = \sqrt{3}$: $\frac{d^2V}{dh^2}\Big|_{h=\sqrt{3}} = -2\pi (\sqrt{3}) = -2\sqrt{3}\pi$ Since $\frac{d^2V}{dh^2} < 0$, the volume is maximized at $h = \sqrt{3}$.

Step 5: Calculate the radius and the maximum volume

We have $h = \sqrt{3}$. We can find the corresponding radius: $r^2 = 9 - h^2 = 9 - (\sqrt{3})^2 = 9 - 3 = 6$ $r = \sqrt{6}$ Now, we can calculate the maximum volume: $V_{max} = \frac{1}{3}\pi r^2 h = \frac{1}{3}\pi (6)(\sqrt{3}) = 2\sqrt{3}\pi$

Common Mistakes & Tips

• Always remember the formula for the volume of a cone and the relationship between $r$, $h$, and $l$.
• Don't forget to use the second derivative test to confirm that you have found a maximum (not a minimum or an inflection point).
• Pay attention to units. In this case, the volume is in cubic meters.

Summary

We found the maximum volume of a right circular cone with slant height 3 m by expressing the volume in terms of a single variable (height), finding the critical points using the first derivative, and confirming that it was a maximum using the second derivative test. We then calculated the maximum volume.

Final Answer

The final answer is $\boxed{2\sqrt{3}\pi}$, which corresponds to option (A).