Arvind42 wrote:
Bunuel wrote:
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
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 fuelConsumption 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 maximaTo 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 \(\frac{16000}{s}+10s\)=\(\frac{16000}{40}+10*40=400+400=800\)....