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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

60 kmph
40 kmph
48 kmph
36 kmph

The correct answer is: 40 kmph

$because blank$ Fuel charges $proportional to v to the power of 2 end exponent$ . Let $F$ represents fuel charges
$rightwards double arrow F proportional to v to the power of n end exponent rightwards double arrow F equals k v to the power of 2 end exponent$ (1)
Given that $F equals$ Rs. 48 per hour, $v equals 16$ km per hour
$rightwards double arrow 48 equals k open parentheses 16 close parentheses to the power of 2 end exponent rightwards double arrow k equals fraction numerator 3 over denominator 16 end fraction$
From (1), $F equals fraction numerator 3 v to the power of 2 end exponent over denominator 16 end fraction$
Let the train covers $lambda$ km in $t$ hours
$rightwards double arrow lambda equals v t$ or $t equals fraction numerator lambda over denominator v end fraction$
$rightwards double arrow$ Fuel charges in time $t equals fraction numerator 3 over denominator 16 end fraction v to the power of 2 end exponent cross times fraction numerator lambda over denominator v end fraction equals fraction numerator 3 v lambda over denominator 16 end fraction$
$rightwards double arrow$ Total cost for running the train,
$C equals fraction numerator 3 v lambda over denominator 16 end fraction plus 300 cross times fraction numerator lambda over denominator v end fraction$
$rightwards double arrow fraction numerator d C over denominator d v end fraction equals fraction numerator 3 lambda over denominator 16 end fraction minus fraction numerator 300 lambda over denominator v to the power of 2 end exponent end fraction$ and $fraction numerator d to the power of 2 end exponent C over denominator d v to the power of 2 end exponent end fraction equals fraction numerator 600 lambda over denominator v to the power of 3 end exponent end fraction$
For the maximum or minimum value of $C comma fraction numerator d C over denominator d v end fraction equals 0$
$rightwards double arrow v equals 40 blank k m divided by h r$ . Also, $open fraction numerator d to the power of 2 end exponent C over denominator d v to the power of 2 end exponent end fraction close vertical bar subscript v equals 40 end subscript equals fraction numerator 60 lambda over denominator open parentheses 40 close parentheses to the power of 3 end exponent end fraction greater than 0 blank left parenthesis because blank lambda greater than 0 right parenthesis$
$rightwards double arrow C$ is minimum when $v equals 40 blank k m divided by h r$

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