A) 10
B) 20
C) 30
D) 40
Correct Answer: D
Solution :
[d] Let the speed of the train be v and distance to be covered be s so that total time taken is 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, \[C=\left( \frac{3}{16}{{v}^{2}}+300 \right)\frac{s}{v}=\frac{3s}{16}v+\frac{300s}{v}\] \[\frac{dC}{dv}=\frac{3s}{16}-\frac{300s}{{{v}^{2}}}=0\Rightarrow v=40\] \[\frac{{{d}^{2}}C}{d{{v}^{2}}}=\frac{600s}{{{v}^{3}}}>0\therefore v=40\] results in minimum running cost.
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