The gasoline consumption in gallons per hour of a certain vehicle is known to be

Question

The gasoline consumption in gallons per hour of a certain vehicle is known to be the following function of velocity:

What is the optimal velocity which minimizes the fuel consumption of the vehicle in gallons PER MILE?
To solve this problem, we need to minimize the following function of v:
g(v)=
Hint for the above: Assume the vehicle is moving at constant velocity v. How long will it take to travel 1 mile? How much gas will it use during that time?
We find that this function has one critical number at v= .
To verify that g(v) has a minimum at this critical number we compute the second derivative g??(x) and find that its value at the critical number is , a positive number.

Explanation / Answer

we have

corresponding to 1 hr = f(v) gallons, where v is the velocity

time taken to travel one mile when velocity is v miles/ hour = 1/v

gas used during that time = f(v) / v

therefore

g(v) = f(v) / v = (v^2 - 90v + 4800)/150000

differentiating, we have

g'(v)= (2v - 90)/150000

for critical point, we equate it to 0

=> (2v - 90)/150000 = 0

=> v = 45miles/hour

we compute the second dertivative at v = 45miles/hour

g ''(v) = 2/150000 = 1/75000 = +ve value {we have the second derivative independent of value of v}

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