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BITSAT 2014
Exam
BITSAT
Year
2014
Subject
Mathematics
Chapter
Application of Derivatives

BITSAT 2014 · Mathematics · Application of Derivatives

The fuel charges for running a train are proportional to the square...

The fuel charges for running a train are proportional to the square of the speed generated in miles per hour and costs ' 48 per hour at 16 miles per hour. The most economical speed if the fixed charges i.e. salaries etc. amount to 300 per hour is

Options

  1. A10
  2. B20
  3. C30
  4. D40

Correct answer

D. 40

Step-by-step solution

Let the speed of the train be vanddistancetobecoveredbesothattotaltimetakenis\mathrm{v} and distance to be covered be so that total time taken is\mathrm{s} / \mathrm{v}hours.cos⁡toffuelperhourhours. \cos t of fuel per hour=\mathrm{kv}^{2}(\mathrm{k}isconstant)Alsois constant) Also48=\mathrm{k} .16^{2}bygivenconditionby given condition\therefore \mathrm{k}=\frac{3}{16}\thereforecos⁡ttofuelperhour\cos t to fuel per hour\frac{3}{16} \mathrmv^{2}.Otherchargesperhourare300.Totalrunningcos⁡t,. Other charges per hour are 300 . Total running \cos t,C=(316\mathrmv2+300)sv=3 s16v+300 svdCdv=3 s16−300 s\mathrmv2=0⇒v=40\begin{array}{r}C=\left(\frac{3}{16} \mathrmv^{2}+300\right) \frac{\mathrm{s}}{\mathrm{v}}=\frac{3 \mathrm{~s}}{16} \mathrm{v}+\frac{300 \mathrm{~s}}{\mathrm{v}}\\ \frac{\mathrm{dC}}{\mathrm{dv}}=\frac{3 \mathrm{~s}}{16}-\frac{300 \mathrm{~s}}{\mathrmv^{2}}=0 \Rightarrow \mathrm{v}=40 \end{array}\mathrmd2Cdv2=600 s\mathrmv3>0∴v=40\frac{\mathrmd^{2} \mathrm{C}}{\mathrm{dv}^{2}}=\frac{600 \mathrm{~s}}{\mathrmv^{3}} \gt 0 \therefore \mathrm{v}=40 results in minimum running \cos t
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