Question

Pleases solve the case study.​

Answer 1

$\large\underline{\sf{Solution-}}$

Given that,

Speed of the train is v km per hour.

Let assume that fuel cost be Rs x per hour.

Further it is given that, Cost of fuel per hour is directly proportional to square of the speed in km per hour.

$\rm\implies \:x \: \alpha \: {v}^{2}$

$\rm\implies \:x \: = k \: {v}^{2}$

Now, given that x = 48 and v = 16, so on substituting the values, we get

$\rm \: 48 = k {(16)}^{2}$

$\rm\implies \:k = \frac{48}{16 \times 16} = \frac{3}{16}$

So, option (d) is correct.

Now, Further given that Fixed charges to run the train per hour is ₹ 1200

So, Total cost of running the train per hour is

$\rm \: Cost \: per \: hour, \: c \: = \: 1200 + {kv}^{2}$

Now, its a cost per hour to travel v km.

So, cost to travel per km is

$\rm \: Cost \: per \: km\: \: = \: \frac{1200}{v} + kv$

Now,

$\rm \: Cost \: for \: 500 \: km,\:x \: = \: \frac{600000}{v} + 500kv$

On substituting the value of k, we get

$\rm \: Cost \: for \: 500 \: km,\:x \: = \: \frac{600000}{v} + 500 \times \frac{3}{16} \times v$

$\rm \: Cost \: for \: 500 \: km,\:x \: = \: \frac{600000}{v} + 125 \times \frac{3}{4} \times v$

$\rm \: Cost \: for \: 500 \: km,\:x \: = \: \frac{600000}{v} + \frac{375v}{4}$

So, option b is correct.

Now, we have

$\rm \: Cost \: per \: km,\: y \: = \: \frac{1200}{v} + kv$

$\rm \: Cost \: per \: km,\: y \: = \: \frac{1200}{v} + \frac{3}{16} v$

On differentiating both sides w. r. t. v, we get

$\rm \: \frac{d}{dv} \:y \: = \: \frac{d}{dv}\bigg(\frac{1200}{v} + \frac{3v}{16}\bigg)$

$\rm \: \frac{dy}{dv} = - \frac{1200}{ {v}^{2} } + \frac{3}{16}$

For maxima or minima,

$\rm \: \frac{dy}{dv} = 0$

$\rm \: - \frac{1200}{ {v}^{2} } + \frac{3}{16} = 0$

$\rm \: - \frac{1200}{ {v}^{2} } = - \: \frac{3}{16}$

$\rm \: \frac{400}{ {v}^{2} } = \: \frac{1}{16}$

$\rm \: {v}^{2} = 6400$

$\rm\implies \:v = 80 \: km \: per \: hour$

So, option (c) is correct

Now, the fuel cost to travel 500 km for the most economical speed is

$\rm \: = \: 500 \times \frac{3v}{16}$

$\rm \: = \: 500 \times \frac{3}{16} \times 80$

$\rm \: = \: 500 \times 15$

$\rm \: = \: 7500$

So, option (c) is correct.

Now, Total cost of the train to travel 500 km with most economical speed is

$\rm \: Cost \: for \: 500 \: km,\:x \: = \: \frac{600000}{v} + 500kv$

$\rm \: Cost \: for \: 500 \: km,\:x \: = \: \frac{600000}{80} + 500 \times \frac{3}{16} \times 80$

$\rm \: Cost \: for \: 500 \: km,\:x \: = \: 15000$

So, option (d) is correct.

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Additional Information :-

$\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \dfrac{d}{dx}f(x) \\ \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf 0 \\ \\ \sf sinx & \sf cosx \\ \\ \sf cosx & \sf - \: sinx \\ \\ \sf tanx & \sf {sec}^{2}x \\ \\ \sf cotx & \sf - {cosec}^{2}x \\ \\ \sf secx & \sf secx \: tanx\\ \\ \sf cosecx & \sf - \: cosecx \: cotx\\ \\ \sf \sqrt{x} & \sf \dfrac{1}{2 \sqrt{x} } \\ \\ \sf logx & \sf \dfrac{1}{x}\\ \\ \sf {e}^{x} & \sf {e}^{x} \end{array}} \\ \end{gathered}$

Answer 2

Question:-

CASE STUDY

The fuel cost per hour for running a train is proportional to the square of the speed it generates in km per hour. If the fuel costs ₹48 per hour at speed 16km per hour and the fixed charges to run the train amount to ₹1,200 per hour. Assume the speed of the train as v km/h.

Based on the given information, answer the following questions.

46. Given that the fuel cost per hour is k times the square of the speed the train generates in km/h, the value of k is:

Options:

(a) 16/3. b) 1/3. c) 3. b) 3/16.

________________________________________

47. If the train has travelled a distance of 500km, then the total cost of running the train is given by function:

Options:

(a) 15/16 v + 6,00,000/v. (b) 375/4 v + 6,00,000/v.

(c) 5/16 v² + 1,50,000/v (d) 3/16 v + 6,000/v.

________________________________________

48. The most economical speed to run the train is:

Options:

(a) 18km/h. (b) 5km/h. (c) 80km/h. (d) 40km/h.

________________________________________

49. The fuel cost for the train to travel 500km at the most Economical speed is:

Options:

(a) ₹3,750. (b) ₹750. (c) ₹7,500. (d) ₹75,000.

_________________________________________

50. The total cost of the train to travel 500km at the most Economical speed is:

Options:

(a) ₹3,750. (b) 75,000. (c) ₹7,500. (d) ₹ 15,000.

_________________________________________

Given:-

• Speed of the train is v km/h.

Solution:-

46. Fuel cost = k (speed)²

⇒ 48 = k . 16²

${ \boxed{ \sf \large \color{red}⇒ k = \frac{3}{16} .}} \\ { \boxed{ \sf \large \blue{Option(d) \: is \: correct. }}}$

_________________________________________

47.

$\sf \large Total \: cost \: of \: running \: train \: (Let, C) = \frac{3}{16} v {}^{2} t + 1200t.$

$\sf \large Distance \: covered = 500km \: \\ \sf \large ⇒time = \frac{500}{v} hrs$

$\sf \large Total \: cost \: of \: running \: train \: 500km = \frac{3}{16} v {}^{2} ( \frac{500}{v} ) + 1200( \frac{500}{v} )$

${ \boxed{ \sf \large \red{⇒C = \frac{375}{4} v + \frac{6,00,000}{v} .}}} \\ { \boxed{ \sf \large \blue{Option(b) \: is \: correct. }}}$

________________________________________

48.

$\sf \large \frac{dc}{dv} = \frac{375}{4} - \frac{6,00,000}{v}$

$\sf \large Let, \frac{dc}{dv} = 0.$

${ \boxed{ \sf \large \red{⇒v = 80 \frac{km}{h}. }}} \\ { \boxed{ \sf \large \blue{ Option(c) \: is \: correct. }}}$

________________________________________

49.

$\sf \large Fuel \: cost \: for \: running \: 500km = \frac{3}{16} v = \frac{375}{4} \times 80 = 7,500.$

${ \boxed{ \sf \large \red{ =₹7,500. }}} \\ { \boxed{ \sf \large \blue{ Option(c) \: is \: correct. }}}$

________________________________________

50.

$\sf \large The \: total \: cost \: for \: running \: 500km = \frac{375}{4} v + \frac{6,00,000}{v} = \frac{375 \times 80}{4} + \frac{6,00,000}{80} = 15,000$

${ \boxed{ \sf \large \red{= ₹15,000.}}} \\ { \boxed{ \sf \large \blue{Option(d) \: is \: correct. }}}$

Answers:-

46. Option(d) 3/16. ✔

47. Option(b) 375/4 v + 6,00,000/v. ✔

48. Option(c) 80km/h. ✔

49. Option(c) ₹7,500. ✔

50. Option(d) ₹15,000. ✔

Hope you have satisfied. ⚘