The fuel efficiency for a certain midsize car is given by
E(v) = −0.017v^ + 1.462v + 3.5 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. answer 43 mph
(b) What is the maximum fuel efficiency for this car? Round to the nearest mile per gallon. ____ mpg (unsure of this answer)
I assume the first term mean to have v squared.
find the zeroes....
0=−0.017v^2 + 1.462v + 3.5
use the quadratic equation to find the two v soutions. Then max will be half way between the two v valuse.
b. put that max v into the fomrula, and copute E(v)
The vertex of a quadratic equation
y = a x ^ 2 + b x + c
is either a maximum or a minimum of the function.
x = - b / 2 a
gives the x - value (or the value of the function) at its vertex.
y = f ( - b / 2 a )
gives the y - value (or the value of the function) at its vertex.
If the value of a is positive, you're going to get the minimum value because as such the parabola opens upwards (the vertex is the lowest the graph can get.
If the value of a is negative, you're going to find the maximum value because as such the parabola opens downward (the vertex is the highest point the graph can get.
In this case a is negative, so a quadratic equation
y = a x ^ 2 + b x + c
have maximum value.
a = − 0.017
b = 1.462
c = 3.5
x = - b / 2 a = - 1. 462 / [ ( 2 * ( - 0.017 ) ] = - 1.462 / - 0.034 = 43
y = f ( - b / 2 a ) =
- 0.017 * 43 ^ 2 + 1.462 * 43 + 3.5 =
-0.017 * 1849 + 62.866 + 3.5 =
-31.433 + 62.866 + 3.5 =
34.933 =
35 mpg
Round to the nearest mile per gallon.
Answers :
a = 43 mph
b = 35 mpg
To find the speed that yields the maximum fuel efficiency and the corresponding maximum fuel efficiency for the given equation E(v) = -0.017v^2 + 1.462v + 3.5, we can use calculus. Here's how:
(a) To find the speed that yields the maximum fuel efficiency, we need to find the critical points of the function E(v) by taking the derivative E'(v) and setting it equal to zero.
Let's differentiate E(v) with respect to v:
E'(v) = d/dv(-0.017v^2 + 1.462v + 3.5)
E'(v) = -0.034v + 1.462
Now, set E'(v) = 0 and solve for v:
-0.034v + 1.462 = 0
-0.034v = -1.462
v = -1.462 / -0.034
v ≈ 43 (rounded to the nearest mile per hour)
Therefore, the speed that yields the maximum fuel efficiency is approximately 43 mph.
(b) To find the maximum fuel efficiency, substitute the value obtained in (a) into the original function E(v).
E(43) = -0.017(43)^2 + 1.462(43) + 3.5
E(43) ≈ 36.1
Therefore, the maximum fuel efficiency for this car is approximately 36.1 mpg.
Note: It's always good practice to double-check the answers obtained through calculations.