# Finding the most economical speed of a train given the distance covered.

I got this question in my maths test:

"The fuel cost for running a train is proportional to the square of the speed generated in km per hour.

If the fuel costs Rs 48 per hour at speed 16 km per hour and the fixed charges amount to Rs 1200 per hour then find the most economical speed of train when total distance covered by train is 5 km."

My approach: I think we have to find the local minima of the Cost function. For that, if $v$ is the speed generated, we have Fuel cost, C = k ($v^2)$.

Thus when $v$ = 16, C = k (${16}^2$) = 48(x) + 1200 (x) = 1248(x) $\implies$ k = $(39x)/8$.

Now when distance covered, D = 5, then;

C = $(39x/8) v^2$ = $(39/8)(5/v) v^2$ = $(195v/8)$ .

But now for finding the the local minima we have to equate the derivative of C to be zero, but clearly that is not possible.

So I want to know what's wrong with my approach and if possible see some different ways to solve it.

• You would see why you were wrong after looking at my solution Jan 7 '18 at 9:24
• There is a similar problem solved at this site. The speed using the derivative method is 40. Check it out!! Jan 7 '18 at 10:22
• lofoya.com/aptitude/algebra/… Jan 7 '18 at 10:23
• @Satish, thanks for the suggestion. :) Jan 7 '18 at 11:18

Let C be the fuel cost

It is given that C is proportional to $v^2$

Thus $C = kv^2$

Also given is C = 48 when v = 16

$48 = k16^2\implies k = \frac{3}{16}$

Total Cost $$TC= \frac{3}{16}v^2.t + 1200. t$$

But $t = \frac{x}{v}$

$$TC= \frac{3}{16}v^2.\frac{x}{v} + 1200. \frac{x}{v}$$

Put x = 5

$$TC = \frac{15}{16}v + \frac{6000}{v}$$

Set $\frac{d TC}{dv} = 0$

$$\frac{d TC}{dv} = \frac{15}{16} - \frac{6000}{v^2}=0$$

$$\frac{15}{16} = \frac{6000}{v_{opt}^2}$$

$v_{opt} = 80$ km/hr

• >>"given is C = 48 when v = 16" - But you didn't include the fixed charge here. Jan 7 '18 at 9:55
• Fuel cost alone is proportional to speed not the fixed amount Jan 7 '18 at 9:58
• ..I see. I didn't read that carefully. Thanks for the answer, I understand it now! Jan 7 '18 at 10:01
• You are welcome!! Jan 7 '18 at 10:02