BITSAT 2014 Paper

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 = $kv^2$ (k is constant) Also 48 = k. $16^{2}$ by given condition ∴ k = $\frac{3}{16}$ ∴ Cost to fuel per hour $\frac{3}{16} v^{2}$ Other charges per hour are 300. Total running cost C= $\left(\frac{3}{16}v62+300\right)\frac{s}{v}$ = $\frac{3s}{16}v + \frac{300s}{v}$ $\frac{dC}{dv}$ = $\frac{3s}{16} - \frac{300s}{v^{2}}$ = 0 ⇒ v = 40 $\frac{d^{2}C}{dv^{2}}$ = $\frac{600s}{v^{3}}$ > 0 ∴ v = 40 results in minimum running cost