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)If the speed limit is 50 km/h, determine the legal speed that will maximize the fuel efficiency.
Could someone explain to me why the legal speed is 50 km/h to maximize the fuel efficiency with calculations and explanations? My teacher it cannot be 50.
sorry - government regulations follow no logic which makes sense.
To find the legal speed that will maximize the fuel efficiency, we need to find the maximum of the function E(v). We can do this by taking the derivative of E(v) with respect to v and setting it equal to zero.
Given that E(v) = 1600v/(v^2 + 6400), let's differentiate E(v) with respect to v:
dE(v)/dv = (1600(v^2 + 6400) - 1600v(2v))/(v^2 + 6400)^2
= (1600v^2 + 10240000 - 3200v^2)/(v^2 + 6400)^2
= (1600v^2 - 3200v^2 + 10240000)/(v^2 + 6400)^2
= (-1600v^2 + 10240000)/(v^2 + 6400)^2
To find the critical points, we set dE(v)/dv = 0:
(-1600v^2 + 10240000)/(v^2 + 6400)^2 = 0
Since the numerator can never equal zero, the only possibility is when the denominator equals zero:
v^2 + 6400 = 0
Solving this equation for v, we have:
v^2 = -6400
This equation has no real solutions, which means there are no critical points.
To determine the maximum or minimum, we can examine the behavior of the function as v approaches positive or negative infinity.
As v approaches positive infinity, E(v) approaches zero.
Therefore, the maximum fuel efficiency cannot occur at v approaching positive infinity.
Similarly, as v approaches negative infinity, E(v) also approaches zero.
Hence, the maximum fuel efficiency cannot occur at v approaching negative infinity.
Since there are no critical points and E(v) approaches zero as v approaches positive or negative infinity, the maximum fuel efficiency of the car must occur at the boundary points of the speed limit.
Given that the speed limit is 50 km/h, the legal speed that will maximize the fuel efficiency is indeed 50 km/h.
To determine the legal speed that will maximize the fuel efficiency, let's first find the derivative of the fuel efficiency function E(v) with respect to the speed v. The derivative will help us determine where the function is increasing, decreasing, or reaches its maximum or minimum.
1. Find the derivative:
E'(v) = (1600*(v^2+6400) - 1600v(2v))/(v^2+6400)^2
= (1600v^2 + 10240000 - 3200v^2)/(v^2+6400)^2
= (1600v^2 - 3200v^2 + 10240000)/(v^2+6400)^2
= (-1600v^2 + 10240000)/(v^2+6400)^2
2. Set the derivative equal to zero to find critical points:
(-1600v^2 + 10240000)/(v^2+6400)^2 = 0
Since the numerator cannot equal zero (there is no value of v that satisfies -1600v^2 + 10240000 = 0), we can conclude that there are no critical points.
3. Determine the sign of the derivative on each side of the critical point(s) to determine if the function is increasing or decreasing:
If we substitute a value greater than zero (e.g., v = 100) and a value less than zero (e.g., v = -100) into the derivative E'(v), we find that E'(v) is negative for all values of v, meaning the function is decreasing for all values of v.
Since the function E(v) is always decreasing, the maximum fuel efficiency will occur at one of the endpoints of the speed range.
In this case, the legal speed limit is 50 km/h. To confirm that this is the maximum fuel efficiency, we can calculate the fuel efficiency at both endpoints (0 km/h and 50 km/h) and compare the values.
Let's calculate the fuel efficiency at v = 0 and v = 50:
E(0) = (1600*0)/((0^2)+6400) = 0/(0+6400) = 0
E(50) = (1600*50)/((50^2)+6400) = 80000/(2500+6400) ≈ 80000/8900 ≈ 8.9888
As we can see, the fuel efficiency at v = 0 is 0, and at v = 50, it is approximately 8.9888. Therefore, the maximum fuel efficiency occurs at the legal speed limit of 50 km/h.