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 that will maximize the fuel efficiency using speed limit of 50 km/h

If E is measured in litres per 100 km, that is the reciprocal of efficiency. You do not want to maximize that number. You want to minimize it. The question does not make sense to me.

There is a relative maximum of E at v = 80 km/h, but that is a MINIMUM in fuel efficiency. The best efficiency is obtained as v-> 0, well below the speed limit.

To determine the legal speed that will maximize the fuel efficiency using a speed limit of 50 km/h, we need to find the maximum value of the fuel efficiency equation E(v) = 1600v / (v^2 + 6400) when v is less than or equal to 50 km/h.

Since the speed limit is 50 km/h, we can substitute v with 50 in the fuel efficiency equation:
E(50) = 1600 * 50 / (50^2 + 6400)
= 80000 / (2500 + 6400)
= 80000 / 8900
≈ 8.9888 liters per 100 km

Therefore, the fuel efficiency will be approximately 8.9888 liters per 100 km when the car is driven at the legal speed limit of 50 km/h.

To determine the legal speed that will maximize the fuel efficiency using a speed limit of 50 km/h, we need to find the maximum value of the fuel efficiency function, E(v), within the given speed limit.

First, let's substitute the speed limit of 50 km/h into the fuel efficiency formula:

E(v) = 1600v / (v^2 + 6400)

Replacing v with 50:

E(50) = 1600 * 50 / (50^2 + 6400)
E(50) = 80000 / (2500 + 6400)
E(50) = 80000 / 8900
E(50) ≈ 8.99 liters per 100 km

Now, we need to find the maximum value of E(v) within the speed limit by calculating the derivative of the function and solving for v.

To find the derivative, we can differentiate E(v) with respect to v:

dE(v)/dv = (1600(v^2 + 6400) - 1600v * 2v) / (v^2 + 6400)^2
dE(v)/dv = (1600v^2 + 6400 * 1600 - 3200v^2) / (v^2 + 6400)^2
dE(v)/dv = (4800v^2 + 6400 * 1600) / (v^2 + 6400)^2

Next, set the derivative equal to zero to find the critical points:

(4800v^2 + 6400 * 1600) / (v^2 + 6400)^2 = 0

Since the numerator cannot be equal to zero, we set it equal to zero:

4800v^2 + 6400 * 1600 = 0
4800v^2 = -6400 * 1600
v^2 = (-6400 * 1600) / 4800
v^2 = -canceled out-
v^2 = 10666.67
v ≈ √10666.67
v ≈ 103.28 km/h

Therefore, the critical point is at v ≈ 103.28 km/h.

However, since the speed limit is 50 km/h, the maximum fuel efficiency within the legal limit would be at v = 50 km/h.

So, the legal speed that maximizes the fuel efficiency using a speed limit of 50 km/h is 50 km/h.