Question

A truck gets 10 miles per gallon (mpg) of diesel fuel traveling along an interstate highway at 50 mph. This mileage decreases by 0.15 mpg for each mile per hour increase above 50 mph. If the truck driver is paid $31/h and diesel fuel costs P = $3/gal, which speed v between 50 and 70 mph will minimize the cost of a trip along the highway? Notice that the actual cost depends on the length of the trip, but the optimal speed does not. (Round your answer to four decimal places.)

Answer 1

Answer:

$V \approx 64.2 \text{mph}$

Step-by-step explanation:

Let $d$ be the distance measured in miles that the track travels at a speed of $v$ miles per hour. Then, the required time equals

                                         $t = \dfrac{d}{V}$

Therefore, the amount paid to the driver equals

                                         $30 \cdot \dfrac{d}{V}$

Since the mileage decreases by 0.15 mpg for each mile per hour increase above 50 mph and the diesel fuel costs $3\gal, the cost of fuel is

                           $\dfrac{3d}{10 - 0.15(V-50)} = \dfrac{3d}{17.5 - 0.15V}$

So, the total cost is

                          $F(V) = 30 \cdot \frac{d}{V} + \dfrac{3d}{17.5 - 0.15V}$

Let's find the minima of this function. To do so, first, find its derivative.

              $F'(V) = -30 \cdot \dfrac{d}{V^2} - \dfrac{3d}{(17.5 - 0.15V)^2} \cdot (17.5 - 0.15V)'\\\\\phantom{F'(V)} = -30 \cdot \dfrac{d}{V^2} + \dfrac{0.45d}{(17.5 - 0.15V)^2}$

Next, solve  

                                          $F'(V) = 0$

for $V$.

                        $-30 \cdot \dfrac{d}{V^2} + \dfrac{0.45d}{(17.5 - 0.15V)^2} = 0$

                        $d \left( \dfrac{-30}{V^2} + \dfrac{0.45}{(17.5 - 0.15V)^2} \right)= 0 \quad \Big/ : d \neq 0$

                                $\dfrac{-30}{V^2} + \dfrac{0.45}{(17.5 - 0.15V)^2} = 0$

                                           $\dfrac{0.45}{(17.5 - 0.15V)^2} = \dfrac{30}{V^2}$

                                        $30 (17.5 - 0.15V)^2 = 0.45V^2$

                   $30(306.25 - 5.25V + 0.0225V^2) = 0.45V^2$

                         $9187.5 - 157.5V+ 0.675V^2 = 0.45V^2$

                         $0.225V^2 - 157.5V+ 9187.5 = 0$

Solving the quadratic equation yields

                                       $V \approx 64.2 \text{mph}$

which is the speed between 50 and 70 mph that  will minimize the cost of the trip along the highway.