Question

The fuel charges for running a train are proportional to the square of the speed generated in km per hour, and the cost is Rs $48$ at $16$km per hour. If the fixed charges amount to Rs.$300$ per hour, the most economical speed is

• A. $60$ kmph
• B. $40$ kmph
• C. $48$ kmph
• D. $36$ kmph

Answer 1

The correct option is B $40$ kmph

$\because$ charges $\propto v^2.$
Let $F$ represents fuel charges
$\Rightarrow F\propto v^2\Rightarrow F=k{v}^2$
Given that $F$=Rs.48 per hour, $v=16$km per hour
$\Rightarrow 48=k(16{)}^2\Rightarrow k=\frac{3}{16}$
$\Rightarrow$ $F=\frac{3v^2}{16}$
Let the train covers $\lambda$ km in $t$ hours
$\Rightarrow \lambda =vt$ or $t=\frac{\lambda }{v}$
$\Rightarrow$ Fuel charges for time $t$,
$t=\frac{3}{16}{v}^2\times \frac{\lambda }{v}=\frac{3v\lambda }{16}$
$\Rightarrow$ Total cost for running the train,
$C=\frac{3v\lambda }{16}+300\times \frac{\lambda }{v}$
$\Rightarrow \frac{dC}{dv}=\frac{3\lambda }{16}-\frac{300\lambda }{v^2}$ and $\frac{d^{2}C}{d{v}^2}=\frac{600\lambda }{v^3}$
For the maximum or minimum value of
$C,\frac{dC}{dv}=0$
$\Rightarrow v=40\text{km/hr}$
Also,${\left(\frac{d^{2}C}{d{v}^2}\right)}_{v=40}=\frac{600\lambda }{(40{)}^3}>0(\because \lambda >0)$
$\Rightarrow$ $C$ is mimimum when $v=40$km/hr.