A cone is made from a circular sheet of radius $\sqrt{3}$  by cutting out a sector and giving the cut edges of the remaining piece together. The maximum volume attainable for the cone is

First derivative zero and Verify second derivative

$\text{ radius of the sheet will be equal to slant height (1) of cone }$

$$\begin{gathered} therefore \\ l=\sqrt{3} \end{gathered}$$

$\text{ Volume of cone }\mathrm{V}=\frac{\pi {\mathrm{r}}^2\mathrm{h}}{3}=\frac{\pi \left(3-{\mathrm{h}}^2\right)\mathrm{h}}{3}$

$\text{ for maximum volume }\frac{\mathrm{dV}}{\mathrm{dh}}=0$

$$\begin{gathered} \begin{array}{r}\Rightarrow \frac{3\pi \left(1-{\mathrm{h}}^2\right)}{3}=0\\ \Rightarrow \mathrm{h}=1\end{array} \\ \text{ now, }\frac{{\mathrm{d}}^2\mathrm{V}}{{\mathrm{dh}}^2}=-2\mathrm{h} \\ \mathrm{h}=1\mathrm{V}<0 \end{gathered}$$

$\text{ Thus, }max\left(\mathrm{V}\right)=\frac{2\pi }{3}$