Key Concepts and Formulas

• 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.
• Curved Surface Area of a cone: $A = \pi r l$, where $r$ is the radius of the base and $l$ is the slant height. The slant height is related to the radius and height by $l = \sqrt{r^2 + h^2}$.
• Optimization using Derivatives: To find the maximum or minimum of a function $f(x)$, we find the critical points by solving $f'(x) = 0$. We can then use the second derivative test to determine if the critical point is a maximum or minimum. If $f''(x) < 0$, then $x$ is a local maximum, and if $f''(x) > 0$, then $x$ is a local minimum.

Step-by-Step Solution

Step 1: Setting up the Geometry and Variables

Let $R$ be the radius of the sphere, which is given as $R = 3$ cm. Let $r$ be the radius of the base of the inscribed cone, and let $h$ be the height of the cone. We want to express $h$ in terms of $R$ and $r$. Let the center of the sphere be the origin. The height of the cone can be written as $h = R + x$, where $x$ is the distance from the center of the sphere to the base of the cone. Then, by the Pythagorean theorem, we have $r^2 + x^2 = R^2$, so $x = \sqrt{R^2 - r^2}$. Therefore, $h = R + \sqrt{R^2 - r^2}$. However, a simpler approach is to write $h = R + x$. Then $x = h - R$, so $r^2 + (h-R)^2 = R^2$, which gives $r^2 = R^2 - (h-R)^2 = R^2 - (h^2 - 2hR + R^2) = 2hR - h^2$.

Step 2: Expressing the Volume of the Cone

The volume of the cone is given by $V = \frac{1}{3}\pi r^2 h$. Substituting $r^2 = 2hR - h^2$, we get $V = \frac{1}{3}\pi (2hR - h^2)h = \frac{1}{3}\pi (2h^2R - h^3)$ Since $R = 3$, we have $V = \frac{1}{3}\pi (6h^2 - h^3)$

Step 3: Finding the Critical Points

To maximize the volume, we take the derivative of $V$ with respect to $h$ and set it equal to zero. $\frac{dV}{dh} = \frac{1}{3}\pi (12h - 3h^2) = \pi (4h - h^2) = \pi h(4-h)$ Setting $\frac{dV}{dh} = 0$, we get $h = 0$ or $h = 4$. Since $h=0$ corresponds to a cone with zero volume, we take $h = 4$.

Step 4: Verifying Maximum Volume

To confirm that $h=4$ gives a maximum volume, we take the second derivative of $V$ with respect to $h$. $\frac{d^2V}{dh^2} = \pi (4 - 2h)$ At $h = 4$, we have $\frac{d^2V}{dh^2} = \pi (4 - 2(4)) = -4\pi < 0$, which indicates that $h = 4$ corresponds to a maximum volume.

Step 5: Finding the Radius and Slant Height

We have $h = 4$ and $R = 3$. Then $r^2 = 2hR - h^2 = 2(4)(3) - 4^2 = 24 - 16 = 8$, so $r = \sqrt{8} = 2\sqrt{2}$. The slant height $l$ is given by $l = \sqrt{r^2 + h^2} = \sqrt{8 + 16} = \sqrt{24} = 2\sqrt{6}$.

Step 6: Calculating the Curved Surface Area

The curved surface area is $A = \pi r l = \pi (2\sqrt{2})(2\sqrt{6}) = 4\pi \sqrt{12} = 4\pi \sqrt{4 \cdot 3} = 4\pi (2\sqrt{3}) = 8\sqrt{3}\pi$. However, this doesn't match the answer.

Let's re-examine the height $h = R + x$. Then $x = h - R = 4 - 3 = 1$. $r^2 + x^2 = R^2$ so $r^2 + 1^2 = 3^2$ so $r^2 = 8$ and $r = 2\sqrt{2}$. $l = \sqrt{r^2 + h^2} = \sqrt{8 + 16} = \sqrt{24} = 2\sqrt{6}$. Then the curved surface area is $CSA = \pi r l = \pi (2\sqrt{2})(2\sqrt{6}) = 4\pi \sqrt{12} = 4\pi (2\sqrt{3}) = 8\sqrt{3}\pi$.

There must be a mistake in the given answer. The height should be $h=4$. Then $r^2 = 2hR - h^2 = 2(4)(3) - 4^2 = 24 - 16 = 8$, so $r = 2\sqrt{2}$. Then the slant height is $l = \sqrt{r^2 + h^2} = \sqrt{8+16} = \sqrt{24} = 2\sqrt{6}$. The curved surface area is $\pi r l = \pi (2\sqrt{2})(2\sqrt{6}) = 4\pi \sqrt{12} = 8\pi\sqrt{3}$.

Another approach: Let $h=R(1+\cos\theta)$ and $r=R\sin\theta$. Then $V = \frac{1}{3}\pi R^3 \sin^2\theta (1+\cos\theta)$. We want to maximize $f(\theta) = \sin^2\theta(1+\cos\theta) = (1-\cos^2\theta)(1+\cos\theta) = (1-\cos\theta)(1+\cos\theta)^2$. $f'(\theta) = \sin\theta (1+\cos\theta)^2 + (1-\cos\theta)2(1+\cos\theta)(-\sin\theta) = \sin\theta(1+\cos\theta)(1+\cos\theta - 2 + 2\cos\theta) = \sin\theta(1+\cos\theta)(3\cos\theta - 1)$. So $\cos\theta = 1/3$. $h = 3(1 + 1/3) = 4$. $r = 3\sin\theta = 3\sqrt{1-1/9} = 3\sqrt{8/9} = 3(2\sqrt{2}/3) = 2\sqrt{2}$. $l = \sqrt{r^2 + h^2} = \sqrt{8 + 16} = \sqrt{24} = 2\sqrt{6}$. $A = \pi r l = \pi(2\sqrt{2})(2\sqrt{6}) = 4\pi \sqrt{12} = 8\pi\sqrt{3}$.

Common Mistakes & Tips

• Careful with Geometry: Drawing a clear diagram is essential for relating the variables.
• Check Second Derivative: Always verify that the critical point you found is indeed a maximum (or minimum) using the second derivative test.
• Algebra Errors: Be meticulous with your algebraic manipulations, especially when taking derivatives and simplifying expressions.

Summary

We found the volume of a cone inscribed in a sphere of radius 3 cm as a function of its height. By taking the derivative of the volume with respect to the height and setting it to zero, we found the height that maximizes the volume. We then calculated the radius and slant height of the cone and finally found its curved surface area.

The final answer is \boxed{8\sqrt{3} \pi}, which corresponds to option (D).