If the petrol burnt per hour in driving a motor boat varies as the cube of its velocity when going against a current of ‘C’ kmph, the most economical speed Is (in kmph)

$$\begin{gathered} \text{ Let the petrol burnt per hour be ' }y\text{ ' units} \\ a\text{nd }V \mathrm{kmph}\text{ be the velocity of the } \\ \text{ Since, }y\propto V^3 \\ \Rightarrow \mathrm{y}={\mathrm{kV}}^3\text{ where ' }\mathrm{k}\text{ 'is constant. } \\ \text{ Velocity of boat against current }=(\mathrm{V}-\mathrm{C})\mathrm{kmph}\text{. } \\ \text{ Let }\mathrm{S}\text{ be the distance travelled by the boat in time }t\text{. } \\ \text{ Then petrol burnt }=\frac{\mathrm{S}}{\mathrm{V}-\mathrm{C}}\times {\mathrm{kV}}^3 \\ \text{ Let }f\left(V\right)=\frac{S}{V-C}\times k{V}^3 \\ {f}^{\mathrm{\prime }}\left(V\right)=\frac{2V^3-3V^{2}C}{(V-C{)}^2} \\ \begin{array}{r}{\mathrm{f}}^{\mathrm{\prime }}\left(\mathrm{V}\right)=0\\ \Rightarrow \mathrm{V}=0,\frac{3\mathrm{C}}{2}\end{array} \\ \text{ Clearly }{f}^{\prime \prime }\left(\mathrm{V}\right)>0\text{ for }\mathrm{V}=\frac{3\mathrm{C}}{2} \\ \text{ Hence, most economical speed is }\frac{3\mathrm{C}}{2} \end{gathered}$$