The cost of fuel for running a train is proportional to the square of the speed generated in km/h. It costs ₹48 per hour when the train is moving at the speed of 16 km/h. What is the most economical speed if the fixed charges are ₹300 per hour, over and above the running cost?

40 km/h

The correct answer is Option (2) → 40 km/h

Let the speed of the train be $v$ km/h,

then the cost of fuel = $₹kv^2$ per hour.

Given when $v = 16\, km/h$, cost = ₹48

$⇒48 = k(16)^2⇒ k=\frac{3}{16}$.

∴ The cost of fuel for running the train = $₹\frac{3}{16}v^2$ per hour.

Let the total distance travelled by the train be $d$ km, then

the time taken to cover the distance = $\frac{d}{v}$ hours.

Let the total cost of the journey be $₹C$, then

$C=\frac{3}{16}v^2.\frac{d}{v}+300.\frac{d}{v}=\frac{3}{16}d.v+\frac{300d}{v}$   ...(i)

Differentiating (i) w.r.t. v, we get

$\frac{dC}{dv}=\frac{3}{16}d.1+300d.(-\frac{1}{v^2})$ and 

$\frac{d^2C}{dv^2}=0-300d.(-\frac{2}{v^3})=\frac{600d}{v^3}$

Now, $\frac{dC}{dv}=0⇒\frac{3}{16}d-\frac{300d}{v^2}=0⇒\frac{1}{16}=\frac{100d}{v^2}$

$⇒v^2=1600⇒v=40$

Also, $\left[\frac{d^2C}{dv^2}\right]_{v=40}=\frac{600d}{(40)^3}>0$ ⇒ C is least when $v=40$.

Hence, the most economical speed is 40 km/h.