optimal speed of a truck

Question

A truck gets 800/x

 miles per gallon (mpg) when driven at a constant speed of x mph, where 40 ≤ x ≤ 80.

 If the price of fuel is $2.80/gal and the driver is paid $15/hr, at what speed is it most economical for the trucker to drive? (Round your answer to two decimal places.)

Answer 1

Let us define the distance that needs to be traveled as d miles, where d is an unknown constant. Since we are traveling at x mph, it will take d/x hours to travel the distance. Since the driver is paid $15/hr, they are paid 15d/x for the trip.

At 800/x miles per gallon, d miles can be traveled with d/(800/x) gallons, or dx/800. Fuel costs $2.80/gal, so the fuel costs 2.80dx/800 for the trip.

Therefore, the trucker makes 15d/x and has to pay 2.80dx/800 for the trip, for a total profit of f(x) = 15d/x - 2.80dx/800. To find the speed at which it is most economical for the trucker to drive, we have to maximize the above equation. We do this by taking the derivative of f(x) = 15d/x - 2.80dx/800 with respect to x.

Using the power rule, we get the derivative is f'(x) = -15d/x² - 2.80d/800. We then have to find the critical points, we are values of x where the above function is equal to 0. Since x² is always positive, we find that this equation has no critical points.

To find the maximum value of the function, we evaluate it at the endpoints of the interval (x = 40, 80) and the critical points of the equation (none).

f(40) = 0.235d.

f(80) = -0.0925d.

So, the optimal speed for the trucker to drive is 40 mph, and they will earn 0.235d dollars for the trip, or 23.5 cents per mile driven.