The fuel efficiency for a certain midsize car is given by

E(v) = −0.018v^2 + 1.440v + 3.9 where E(v)is the fuel efficiency in miles per gallon for a car traveling v miles per hour.
(a) What speed will yield the maximum fuel efficiency? Round to the nearest mile per hour.
1 mph

(b) What is the maximum fuel efficiency for this car? Round to the nearest mile per gallon.
2 mpg

find the derivative, set it equal to zero and solve.

To find the speed that yields the maximum fuel efficiency, we need to find the vertex of the parabolic function E(v) = -0.018v^2 + 1.440v + 3.9. The vertex represents the maximum point on the graph.

(a) To find the speed that yields the maximum fuel efficiency, we can use the formula for the x-coordinate of the vertex: v = -b/2a.
In this case, a = -0.018 and b = 1.440.

Substituting these values into the formula:
v = -1.440 / (2 * -0.018)
v = 1.440 / 0.036
v ≈ 40

Therefore, the speed that yields the maximum fuel efficiency is approximately 40 miles per hour.

(b) To find the maximum fuel efficiency for this car, we substitute the speed (v) that yields the maximum fuel efficiency into the fuel efficiency function.

Substituting v = 40 into E(v):
E(40) = -0.018 * (40)^2 + 1.440 * 40 + 3.9
E(40) = -0.018 * 1600 + 57.6 + 3.9
E(40) = -28.8 + 57.6 + 3.9
E(40) ≈ 32.7

Therefore, the maximum fuel efficiency for this car is approximately 32.7 miles per gallon.