The consumption of diesel per hour of a bus varies directly as square

If the bus is traveling at 40 kmph, to travel a distance of 400 km, it will take 10 hours. Since at this speed, the consumption is 1 liter per hour, we need 10 liters of diesel fuel, and thus it will cost 40 x 10 = $400 for the fuel and another $400 for other expenses. So the total cost would be $800.

Of course, in the above, we assume the speed is 40 kmph. The question is this: can we lower the cost if we change the speed or can we show that the cost is minimized when the bus is traveling at 40 kmph? Let f = the diesel fuel, in liters per hour, and r = the speed of the bus, in kmph. Thus we can create the equation where k is a positive constant:

f = kr^2

1 = k(40)^2

k = 1/1600

Therefore, we see that f = 1/1600 x r^2 = r^2/1600. We need to determine the value of r that minimizes the total cost, which consists of the cost of the fuel and the cost of other expenses.

The cost of the fuel = Time in hours x Fuel/Hr x Unit cost of fuel = 400/r x r^2/1600 x 40 = 10r

The cost of other expenses = Time in hours x Unit cost of other expenses = 400/r x 40 = 16000/r

Therefore, the total cost is 10/r + 16000/r. To minimize this sum, we set the two addends equal to each other:

10r = 16000/r

10r^2 = 16000

r^2 = 1600

r = 40

We see that if the speed is 40 kmph, the total cost will be minimized. Since we have already calculated it as $800, $800 is the minimum total cost.

Answer: C

Bunuel chetan2u request you to share the official solution/Expert reply

chetan2u was confused by the relation only. Thought we are required to minimize the cost using the relation. Thanks for the reply.

Hi,

I misread the question completely in a hurry, and took 40 as constant speed, so the answer would be as under and one such elegant way has been given by VeritasKarishma above. So as not to confuse others, I will delete the initial response

Now, the consumption of diesel per hour of a bus varies directly as square of its speed.----\(c=k*(speed)^2\)
When the bus is travelling at 40 kmph its consumption is 1 litre per hour. ---- \(1=k*40^2.....k=\frac{1}{40^2}\)

If s is the speed at which we get the minimum expenditure, the hours spent = \(\frac{400}{s}\)

Cost of other expenses @40ph=\(\frac{400}{s}*40\)

Cost of fuel
Consumption of fuel per liter as per the relation above.. = \(k*s^2=\frac{1}{40^2}*s^2\)
Cost @40 per liter and for \(\frac{400}{s}\) hr = \(40*\frac{400}{s}*\frac{1}{40^2}*s^2=10s\)

Total cost = \(\frac{16000}{s}+10s\)..

Now, differentiation is a very simple method to find minima and maxima
To find the minimum value, we can differentiate it as - \(\frac{d}{ds}(\frac{16000}{s}+10s)=0\)
Now \(\frac{d}{dx}(x)=1\), \(\frac{d}{dx}(x^2)=2x\), and \(\frac{d}{dx}(\frac{1}{x})=-\frac{1}{x^2}\)
So, \(\frac{d}{ds}(\frac{16000}{s}+10s)=0......10-\frac{16000}{s^2}=0.......s^2=1600...s=40\)

Cost at s=40 is \(\frac{16000}{s}+10s\)=\(\frac{16000}{40}+10*40=400+400=800\)....

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