A taxi is driving 2 0 0 kilometers on a highway at a uniform speed of x km/hr (speed rates of highway requires 4 0 ≤ x ≤ 7 0 ). The cost of fuel is $ 3 0 per litre and is consumed at the rate of ( 1 0 0 + 6 0 x 2 ) liter per hour. If the wage of the driver is $ 2 0 0 per hour then what will be the most economical speed to drive the taxi?
Details and Assumptions
Most economical refers to maximum profit or minimum loss. (As per the point of view of driver)
Profit or Loss = ∣ Wage earned - Price of fuel ∣
This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try
refreshing the page, (b) enabling javascript if it is disabled on your browser and,
finally, (c)
loading the
non-javascript version of this page
. We're sorry about the hassle.
bro, when you are riding a taxi, you give the money to the driver and driver fills petrol and not that we pay for both the petrol and his wage,
His wage is supposed to be more than petrol so he can buy petrol i think,
so doesnt it seem logically inadequate or it must be mentioned that we own the taxi and the driver as well and hence we must also bear the price of petrol
@pranjal i think you should mention that we own the taxi and not that we simply hire one
Log in to reply
In a public taxi, the money asked by the driver is the sum of petrol cost and his/her wage
w a g e = m i n w a g e + m u l t i p l i e r ∗ d i s t a n c e .
w a i t i n g c o s t = m u l t i p l i e r ∗ t i m e
t o t m o n e y = w a g e + p e t r o l c o s t + w a i t i n g c o s t + X
X is the unknown amount of money* that the driver asks above the actual fee(which people usually pay).
*Conditions Apply, read documents carefully before investing.
Log in to reply
oh, should have read the docs carefully then, :P
Still, it would have been better if mentioned :)
OMG! I agree with @Mvs Saketh , actually you are driver. And you are supposed to fill up the fuel on your own, and your client pays you the wage.
It seems that the wordings are not very clear. Can anyone suggest improvements? Also, I will be checking my calculations again as the answer I was getting was same as what @Ronak Agarwal got.
Problem Loading...
Note Loading...
Set Loading...
Money spent = Time × (Cost Of Fuel × Consumption Rate + Driver Wage)
Put the values to get the Money spent as :
Money spent M o n e y ( x ) = 1 0 0 ( x 6 4 0 0 + x )
Differentiate it with respect to x and equate it to zero to get x = 8 0
But 4 0 ≤ x ≤ 7 0
Hence the most economical speed is either 4 0 or 7 0
M o n e y ( 4 0 ) > M o n e y ( 7 0 )
Hence most economical speed is 70.