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

Question

The consumption of diesel per hour of a bus varies directly as square of its speed. When the bus is travelling at 40 kmph its consumption is 1 litre per hour. If each litre costs $40 and other expenses per hour is $40, then what would be the minimum expenditure required to cover a distance of 400 Km?

A. 600
B. 700
C. 800
D. 900
E. 1000

A. 600
B. 700
C. 800
D. 900
E. 1000

KarishmaB · Accepted Answer

We cannot assume that the bus must travel at 40 kmph only.

Given:
Consumption per hr/Speed^2 = k = 1/1600

C = (\frac{1}{1600})*Speed^2

Assume that time taken to cover 400 km is t hrs. Then Speed = 400/t

C = (\frac{1}{1600})*(\frac{400}{t})^2 
 C*t = \frac{100}{t}

C*t is the total fuel consumed.

We need to minimise 40*C*t + 40*t

40*(\frac{100}{t} + t)

Minimum value of (100/t + t) will be given when 100/t = t i.e. t = 10 (because the product of the two terms (100/t) * t = 100, a constant)

So the minimum value is 40*(10 + 10) = 800

Answer (C)

Archit3110 · Answer

so total hrs to cover 400 km ; 400/40 ; 10 hrs
and total fuel burned = 10*1; 10 ltrs 10*40 400 $
and 40*10 ; 400$ for other expenses
total expenditure ; 800$
IMO C

Lampard42 · Answer

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

chetan2u · Answer

Lampard42 · Answer

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

chetan2u · Answer

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 ....

ScottTargetTestPrep · Answer

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

Natansha · Answer

Bunuel  can you please share the official explanation of this ques?

Natansha · Answer

Can you please explain the logic or concept behind 
'' Minimum value of (100/t + t) will be given when 100/t = t i.e. t = 10 (because the product of the two terms (100/t) * t = 100, a constant)''

Krunaal · Answer

Because the product of the two terms (100/t) * t = 100, a constant , - we understand that there is an inverse relationship between 100/t and t => if t decreases, 100/t increases and if t increases, 100/t decreases. We will get the minimum value when 100/t = t (equality point), which is when t is 10. ( \frac{100}{t} = t => 100 = t^{2} => t =10 )

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

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

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615 to 715 in 15 Days (Ria's Strategies)

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