The fuel efficiency, E, in litres per 100 km, for a car driven at speed v, in km/h is E(v)=1600v/(v^2+6400).
a) Determine the legal speed speed that will maximize the fuel efficiency if the speed limit is 50 km/h.
The answer is 50 km/h but my teacher said it can't be that. Any explain please.
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b) Determine the speed intervals, within the legal speed limit of 0 km/h to 100 km/h, in which the fuel efficiency is increasing.
To find the value of v that maximizes the fuel efficiency, we need to find the derivative of the fuel efficiency equation and set it equal to zero. This is because the maximum or minimum of a function occurs when its derivative is zero.
Let's start with part (a):
a) To find the maximum fuel efficiency for a given speed limit of 50 km/h, we need to determine the speed that maximizes the fuel efficiency equation E(v) = 1600v / (v^2 + 6400).
1. Find the derivative of E(v) with respect to v:
E'(v) = dE(v) / dv
Hint: To simplify the process, we can write the equation in terms of the square power: E(v) = 1600v / (v^2 + 6400) = 1600v / (v^2 + 80^2).
Applying the quotient rule of differentiation, where g(v) = v and h(v) = v^2 + 80^2, the derivative is:
E'(v) = [g'(v) * h(v) - g(v) * h'(v)] / h(v)^2
E'(v) = [(1) * (v^2 + 80^2) - (v) * (2v)] / (v^2 + 80^2)^2
2. Set E'(v) equal to zero and solve for v:
[(v^2 + 80^2) - 2v^2] / (v^2 + 80^2)^2 = 0
[v^2 + 6400 - 2v^2] / (v^2 + 80^2)^2 = 0
-v^2 + 6400 = 0
v^2 = 6400
v = ±80
Since we are considering the speed limit, which cannot be negative, we discard the negative solution, leaving us with v = 80 km/h.
Therefore, the legal speed that maximizes the fuel efficiency is 80 km/h, not 50 km/h as stated by your teacher. You may want to verify this with your teacher or double-check the calculations.
Now let's move on to part (b):
b) To determine the speed intervals within the legal speed limit of 0 km/h to 100 km/h in which the fuel efficiency is increasing, we need to analyze the behavior of the derivative E'(v).
1. Determine where E'(v) is positive.
We know that when E'(v) > 0, the fuel efficiency is increasing.
E'(v) = [(v^2 + 80^2) - 2v^2] / (v^2 + 80^2)^2 > 0
-v^2 + 6400 > 0
v^2 < 6400
-80 < v < 80
Therefore, the fuel efficiency is increasing for speeds between -80 km/h and 80 km/h.
2. Exclude the speed range beyond the legal limit.
Since the speed limit is between 0 km/h and 100 km/h, we need to exclude the negative speed range.
Therefore, the fuel efficiency is increasing for speeds between 0 km/h and 80 km/h.
In summary, the fuel efficiency is increasing in the speed interval of 0 km/h to 80 km/h, within the legal speed limit.