Question

The fuel economy of a car, measured in miles per gallon, is modeled by the function ƒ(s) = –0.009s2 + 0.699s + 12 where s represents the average speed of the car, measured in miles per hour. What's the maximum fuel economy of the car?

Answer 1

24.43225 miles per gallon

Explanation:

We have a function to know the fuel economy of a car, f(s)=-0.009s²+0.669s+12.

We need to find the first derivative to get the maximums of the function.

Derivative (formulas on the attached file):

f'(s)=-0.018s + 0.669 +0

Having the first derivative we can make the function equal to 0 to get the maimum number of s, the maximum is when the slope of the tangent is 0 (attached in as a graphic).

-0.018s+0.669=0

We solve:

-0.018s = -0.669

s= -0.669/-0.018

s= 223/6 miles/hour

Now we are going to substitute the optimal s in the original function to get the maximum fuel economy of the car:

f(s)= -0.009(223/6)² + 0.669(223/6) + 12

f(s)= -0.009(1381.3611) + 24.8645 + 12

f(s)= -12.432249 + 36.8645

f(s)= 24.43225 miles per gallon.

Answer 2

The maximum fuel economy of the car is 25.57225

Explanation:

we are given

`f(s)=-0.009s^2+0.699s+12`

where

where s represents the average speed of the car, measured in miles per hour

So, we have to maximize f(s)

So, firstly we will find derivative

`f'(s)=-0.009*2s+0.699*1+0`

`f'(s)=-0.018s+0.699`

now, we can set it to 0

and then we can solve for s

`f'(s)=-0.018s+0.699=0`

`s=(233)/(6)`

To find maximum fuel economy , we can plug s into f(s)

`f((233)/(6))=-0.009((233)/(6))^2+0.699((233)/(6))+12`

`f((233)/(6))=25.57225`

So,

The maximum fuel economy of the car is 25.57225