If a right circular cone having maximum volume, is inscribed in a sphere of radius 3 cm, then the curved surface area (in cm$$^2$$) of this cone is:

Let the sphere have centre $$O$$ and radius $$R=3\text{ cm}$$. We place the origin at $$O$$ and take the $$z$$-axis along the line joining the centre to the vertex of the cone. Thus the vertex (apex) of the cone is at the “north-pole’’ of the sphere, i.e. at the point $$A(0,0,R)$$.

Denote by $$c$$ the $$z$$-coordinate of the centre of the base circle of the cone. Because the base plane is perpendicular to the axis, every point on the base has the same $$z$$-coordinate $$c$$, with $$-R\le c\le R$$ inside the sphere.

The height of the cone, measured from the vertex down to the base plane, is therefore $$h = R - c.$$

The base radius $$r$$ is obtained from the equation of the sphere $$x^{2}+y^{2}+z^{2}=R^{2}.$$ For the base points we have $$z=c$$, so $$r^{2}=R^{2}-c^{2} \; \Longrightarrow \; r=\sqrt{R^{2}-c^{2}}.$$

The volume of a right circular cone is given by the formula $$V=\dfrac{1}{3}\pi r^{2}h.$$ Substituting the expressions of $$r$$ and $$h$$ we get $$V(c)=\dfrac{1}{3}\pi\left(R^{2}-c^{2}\right)\left(R-c\right).$$

To obtain the cone of maximum volume we differentiate $$V(c)$$ with respect to $$c$$ and set the derivative equal to zero. Since the constant $$\tfrac{1}{3}\pi$$ does not affect the location of the maximum, we differentiate the function $$f(c)=\left(R^{2}-c^{2}\right)\left(R-c\right).$$

Using the product rule, $$\dfrac{df}{dc}=(-2c)(R-c)+(R^{2}-c^{2})(-1).$$ Expanding each product, $$\dfrac{df}{dc}=-2cR+2c^{2}-R^{2}+c^{2}=-2cR+3c^{2}-R^{2}.$$

Setting $$\dfrac{df}{dc}=0$$ gives the quadratic equation $$3c^{2}-2Rc-R^{2}=0.$$ Solving, $$c=\dfrac{2R\pm\sqrt{(2R)^{2}+12R^{2}}}{6}= \dfrac{2R\pm4R}{6}.$$ Thus $$c_{1}=R \quad\text{and}\quad c_{2}=-\dfrac{R}{3}.$$

The value $$c=R$$ corresponds to zero height (no cone), so the maximum volume occurs at $$c=-\dfrac{R}{3}.$$

Hence $$h = R - c = R -\!\left(-\dfrac{R}{3}\right)=R+\dfrac{R}{3}=\dfrac{4R}{3},$$ and $$r = \sqrt{R^{2}-c^{2}}=\sqrt{R^{2}-\left(\dfrac{R}{3}\right)^{2}} =\sqrt{R^{2}-\dfrac{R^{2}}{9}}=\sqrt{\dfrac{8R^{2}}{9}}=\dfrac{2\sqrt{2}\,R}{3}.$$

Substituting the numerical value $$R=3\text{ cm}$$, we obtain $$h=\dfrac{4\times3}{3}=4\text{ cm},\qquad r=\dfrac{2\sqrt{2}\times3}{3}=2\sqrt{2}\text{ cm}.$$

The slant height $$l$$ of the cone is found from the Pythagoras theorem: $$l=\sqrt{r^{2}+h^{2}}=\sqrt{(2\sqrt{2})^{2}+4^{2}}=\sqrt{8+16}=\sqrt{24}=2\sqrt{6}\text{ cm}.$$

The curved surface area (lateral surface area) of a cone is given by $$\text{CSA}= \pi r l.$$ Substituting the values of $$r$$ and $$l$$, $$\text{CSA}= \pi\,(2\sqrt{2})\,(2\sqrt{6}) =\pi\;4\sqrt{12} =\pi\;4\,(2\sqrt{3}) =8\sqrt{3}\,\pi\text{ cm}^{2}.$$

Hence, the correct answer is Option A.