Expert: Ahmed Salami - 4/11/2010

Question
The question is:

The fuel efficiency, E, (in litres per 100 kilometres) of a car driven at speed v (in km/h) is E(v) = 1600v/v^2+6400. If the speed limit is 100 km/h, find the legal speed that will maximize the fuel efficiency.  

I tried finding the derivative of E(v) and setting it to 0 and solving for v but I think I'm doing something wrong. Could you please show me the steps to arrive to the answer? Thanks for the help.

Answer
Hi Sarah,
E(v) = 1600v/(v^2 + 6400)
E'(v) = [(v^2 + 6400)(1600) - (1600v)(2v)] / (v^2 + 6400)^2
     = [(1600v^2 + 10240000) - (3200v^2)] / (v^2 + 6400)^2
     = (10240000 - 1600v^2) / (v^2 + 6400)^2
Equating to zero, we have
(10240000) - 1600v^2) / (v^2 + 6400)^2 = 0
10240000 - 1600v^2 = 0
10240000 = 1600v^2
v^2 = 6400
v = 80 km/h

You can always get back to me.

Regards