Mathematics - Application of Derivative Question with Solution | TestHub

Let the speed of the train be $\mathrm{v}$ and distance to be covered be so that total time taken is $\mathrm{s}/\mathrm{v}$ hours. Cost of fuel per hour $={\mathrm{kv}}^2(\mathrm{k}$ is constant) Also $48=\mathrm{k}.16^2$ by given condition $\therefore \mathrm{k}=\frac{3}{16}$ $\therefore$ Cost to fuel per hour $\frac{3}{16}{\mathrm{v}}^2$. Other charges per hour are 300 . Total running cost, $$\begin{array}{r}C=\left(\frac{3}{16}{\mathrm{v}}^2+300\right)\frac{\mathrm{s}}{\mathrm{v}}=\frac{3\mathrm{s}}{16}\mathrm{v}+\frac{300\mathrm{s}}{\mathrm{v}}\\ \frac{\mathrm{dC}}{\mathrm{dv}}=\frac{3\mathrm{s}}{16}-\frac{300\mathrm{s}}{{\mathrm{v}}^2}=0\Rightarrow \mathrm{v}=40\end{array}$$ $\frac{{\mathrm{d}}^2\mathrm{C}}{{\mathrm{dv}}^2}=\frac{600\mathrm{s}}{{\mathrm{v}}^3}>0\therefore \mathrm{v}=40$ results in minimum running cost