The fuel efficiency, E, in litres per 100 kilometres, for a car driven at speed v, in km/h, is E(v) = 1600v/(v^2+6400).
a) If the speed limit is 100 km/h, determine the legal speed that will maximize the fuel efficiency.
b) Repeat part a., using a speed limit of 50 km/h.
c) Determine the speed intervals, within the legal speed limit of 0 km/h to 100 km/h, in which the fuel efficiency is increasing.
d) Determine the speed intervals, within the legal speed limit of 0 km/h to 100 km/h, in which the fuel efficiency is decreasing.
a) To maximize the fuel efficiency within the legal speed limit (0 to 100 km/h), we need to find the critical points of E(v). By taking the derivative of E(v) with respect to v, we can find the places where the fuel efficiency is maximized or minimized.
The derivative E'(v) of E(v) can be found using the quotient rule for differentiation. For a function F(v) = g(v) / h(v), the derivative F'(v) is given by F'(v) = (g'(v) * h(v) - g(v) * h'(v)) / h^2(v). In our case, g(v) = 1600v, and h(v) = v^2 + 6400 with their respective derivatives, g'(v) = 1600, and h'(v) = 2v.
Applying the quotient rule:
E'(v) = (1600 * (v^2 + 6400) - 1600v * 2v) / (v^2 + 6400)^2
E'(v) = (1600v^2 + 10240000 - 3200v^2) / (v^2 + 6400)^2
E'(v) = (-1600v^2 + 10240000) / (v^2 + 6400)^2
To find critical points, set the derivative E'(v) to 0 and solve for v:
0 = (-1600v^2 + 10240000) / (v^2 + 6400)^2
Since the denominator is not equal to zero for all values of v within our domain (0 to 100), we can only focus on the numerator:
0 = -1600v^2 + 10240000
Solving for v:
v^2 = 10240000/1600 = 6400
v = ±sqrt(6400) ≈ ±80
Since we are only concerned with the legal speed limits (0 to 100 km/h), we can ignore the negative solution:
v = 80 km/h
Thus, the legal speed that will maximize fuel efficiency within the speed limit of 100 km/h is 80 km/h.
b) To find the legal speed that maximizes fuel efficiency within a speed limit of 50 km/h, we simply need to examine the behavior of the function E(v) within the range 0 to 50 km/h.
Since the critical point of E(v) is at 80 km/h, and the function E(v) will be increasing from 0 to 80 km/h, then the fuel efficiency will be maximized at the speed limit 50 km/h.
c) Since we found that the critical point for E(v) is at v = 80 km/h, we know that E(v) is increasing from 0 km/h to 80 km/h within the legal speed limit of 0 km/h to 100 km/h.
d) Since E(v) is increasing from 0 km/h to 80 km/h, then it must be decreasing from 80 km/h to the maximum speed limit 100 km/h.