The maximum volume in cu.m of the right circular cone having slant height 3 m is:

For a right circular cone the three basic elements – radius $$r$$ of the base, vertical height $$h$$, and slant height $$l$$ – are joined by the Pythagoras relation

$$l^{2}=r^{2}+h^{2}.$$

Here the slant height is fixed at $$l=3\text{ m}$$, while $$r$$ and $$h$$ can vary subject to the above constraint.

The volume $$V$$ of a cone is given by the standard formula

$$V=\frac{1}{3}\pi r^{2}h.$$

Because $$h$$ must satisfy $$h^{2}=l^{2}-r^{2}$$, we first express $$h$$ in terms of $$r$$:

$$h=\sqrt{l^{2}-r^{2}}=\sqrt{3^{2}-r^{2}}=\sqrt{9-r^{2}}.$$

Substituting this expression for $$h$$ in the volume formula we obtain

$$V(r)=\frac{1}{3}\pi r^{2}\sqrt{9-r^{2}}.$$

Since $$l$$ is fixed, the only variable in this expression is $$r$$. We now find the value of $$r$$ that maximises $$V(r)$$ by differentiating with respect to $$r$$ and equating the derivative to zero.

Let us write

$$V(r)=\frac{\pi}{3}\;f(r),\qquad\text{where}\quad f(r)=r^{2}(9-r^{2})^{1/2}.$$

Differentiating $$f(r)$$ using the product rule:

$$\frac{d}{dr}\bigl[r^{2}(9-r^{2})^{1/2}\bigr] =2r(9-r^{2})^{1/2}+r^{2}\left(\frac{1}{2}\right)(9-r^{2})^{-1/2}(-2r).$$

Simplifying the second term first:

$$r^{2}\left(\frac{1}{2}\right)(9-r^{2})^{-1/2}(-2r)= -\,\frac{r^{3}}{(9-r^{2})^{1/2}}.$$

Hence

$$f'(r)=2r(9-r^{2})^{1/2}-\frac{r^{3}}{(9-r^{2})^{1/2}}.$$

For a critical point we set $$f'(r)=0$$. Multiplying through by $$(9-r^{2})^{1/2}$$ clears the denominator and keeps the sign of the equation unaltered because the square root is positive:

$$2r(9-r^{2})-r^{3}=0.$$

Taking the common factor $$r$$ out:

$$r\left[2(9-r^{2})-r^{2}\right]=0.$$

Since $$r=0$$ would give zero volume, we discard it and solve the bracketed equation:

$$2(9-r^{2})-r^{2}=0\quad\Longrightarrow\quad18-2r^{2}-r^{2}=0$$ $$\Longrightarrow\;18-3r^{2}=0$$ $$\Longrightarrow\;r^{2}=6\quad\Longrightarrow\quad r=\sqrt{6}\text{ m}.$$

With $$r=\sqrt{6}$$ we compute $$h$$ from $$h=\sqrt{9-r^{2}}$$:

$$h=\sqrt{9-6}=\sqrt{3}\text{ m}.$$

Now we substitute $$r$$ and $$h$$ back into the volume formula to get the maximum volume:

$$V_{\max}=\frac{1}{3}\pi r^{2}h =\frac{1}{3}\pi\,(6)\,(\sqrt{3}) =2\sqrt{3}\,\pi\;\text{cubic metres}.$$

(A check of endpoint values, $$r\to0$$ or $$r\to3$$, gives zero volume, confirming that the critical point indeed yields the maximum.)

Hence, the correct answer is Option 1.