The fuel efficiency for a certain midsize car is given by E(v)= -0.016v^2+1.408v+3.8, where E(v)is the fuel efficiency in miles per gallon for a car traveling v miles per hour.
What speed will yield the maximum fuel efficiency? _______mph.
What is the maximum fuel efficiency for this car? _____mpg
If you are taking Calculus ....
E ' (v) = -.032v + 1.408
= 0 for max of E
.032v = 1.408
v = 44
E(44) = ....just sub in
Cuan capitulo en el libro de Leviticos habla sobre la comida que no podemos comer?
To find the speed that yields the maximum fuel efficiency, we can take the derivative of the given function with respect to velocity and set it equal to zero. Then we can solve for the velocity. The maximum fuel efficiency will occur at this velocity.
1. Take the derivative of the given function E(v) with respect to v:
E'(v) = -0.032v + 1.408
(Taking the derivative of each term separately and applying the power rule)
2. Set the derivative equal to zero and solve for v:
-0.032v + 1.408 = 0
-0.032v = -1.408
v = -1.408 / (-0.032) ≈ 43.75 mph
Therefore, the speed that yields the maximum fuel efficiency is approximately 43.75 mph.
To find the maximum fuel efficiency, substitute the found velocity back into the original equation.
3. Plug the value of v into the function E(v):
E(43.75) = -0.016(43.75)^2 + 1.408(43.75) + 3.8
Simplify the equation to find the maximum fuel efficiency.
4. Calculate the result:
E(43.75) ≈ -0.016(1914.06) + 61.75 + 3.8
E(43.75) ≈ -30.624 + 61.75 + 3.8
E(43.75) ≈ 35.925 mpg
Therefore, the maximum fuel efficiency for this car is approximately 35.925 mpg.