The fuel charges for running a train are proportional to the square of the speed generated in miles per hour and costs $$Rs.48$$   per hour at $$16$$ miles per hour. The most economical speed if the fixed charges i.e. salaries etc. amount to $$Rs.300$$  per hour is :

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