Answer:
The maximum fuel economy of the car is 25.57225
Step-by-step explanation:
we are given
[tex]f(s)=-0.009s^2+0.699s+12[/tex]
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
[tex]f'(s)=-0.009*2s+0.699*1+0[/tex]
[tex]f'(s)=-0.018s+0.699[/tex]
now, we can set it to 0
and then we can solve for s
[tex]f'(s)=-0.018s+0.699=0[/tex]
[tex]s=\frac{233}{6}[/tex]
To find maximum fuel economy , we can plug s into f(s)
[tex]f(\frac{233}{6})=-0.009(\frac{233}{6})^2+0.699(\frac{233}{6})+12[/tex]
[tex]f(\frac{233}{6})=25.57225[/tex]
So,
The maximum fuel economy of the car is 25.57225
Answer:
24.43225 miles per gallon
Step-by-step 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.
