Set up the geometry

Let the sphere have radius $R=3$ cm and center at the origin.

Consider a right circular cone inscribed in the sphere with:

• vertex at the top point of the sphere,
• base in a plane at distance $x$ below the center.

Then:

• height of cone: $h = R + x$
• base radius: $r = \sqrt{R^2 - x^2}$

since the base circle lies on the sphere.

Write volume as a function of $x$

Volume of cone: $V = \frac{1}{3}\pi r^2 h$

Substitute $r^2 = R^2 - x^2$ and $h = R+x$: $V(x)=\frac{1}{3}\pi (R^2-x^2)(R+x)$

Factor: $V(x)=\frac{1}{3}\pi (R-x)(R+x)^2$

Differentiate to maximize volume

Let $f(x)=(R-x)(R+x)^2$

Differentiate: $f'(x)=-(R+x)^2 + (R-x)\cdot 2(R+x)$

Factor $(R+x)$: $f'(x)=(R+x)\left[-(R+x)+2(R-x)\right]$ $f'(x)=(R+x)(R-3x)$

For an interior maximum, $f'(x)=0 \implies R-3x=0 \implies x=\frac{R}{3}$

Since $R=3$, $x=1$

Find cone dimensions

Height: $h=R+x=3+1=4$

Base radius: $r=\sqrt{R^2-x^2}=\sqrt{9-1}=\sqrt{8}=2\sqrt{2}$

Slant height: $l=\sqrt{r^2+h^2}=\sqrt{8+16}=\sqrt{24}=2\sqrt{6}$

Compute curved surface area

Curved surface area of cone: $\text{CSA}=\pi r l$

Substitute $r=2\sqrt{2}$ and $l=2\sqrt{6}$: $\text{CSA}=\pi(2\sqrt{2})(2\sqrt{6})$ $\text{CSA}=4\pi\sqrt{12}$ $\text{CSA}=4\pi\cdot 2\sqrt{3}$ $\text{CSA}=8\sqrt{3}\pi$

Match with options

$8\sqrt{3}\pi$ corresponds to Option D.

Comparison with stored answer

Stored correct answer: D

Derived answer: D

Hence, the derived answer agrees with the stored answer.