Respuesta :
Answer:
[tex]V \approx 64.2 \text{mph}[/tex]
Step-by-step explanation:
Let [tex]d[/tex] be the distance measured in miles that the track travels at a speed of [tex]v[/tex] miles per hour. Then, the required time equals
                     [tex]t = \dfrac{d}{V}[/tex]
Therefore, the amount paid to the driver equals
                     [tex]30 \cdot \dfrac{d}{V}[/tex]
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
              [tex]\dfrac{3d}{10 - 0.15(V-50)} = \dfrac{3d}{17.5 - 0.15V}[/tex]
So, the total cost is
             [tex]F(V) = 30 \cdot \frac{d}{V} + \dfrac{3d}{17.5 - 0.15V}[/tex]
Let's find the minima of this function. To do so, first, find its derivative.
       [tex]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}[/tex]
Next, solve Â
                     [tex]F'(V) = 0[/tex]
for [tex]V[/tex].
            [tex]-30 \cdot \dfrac{d}{V^2} + \dfrac{0.45d}{(17.5 - 0.15V)^2} = 0[/tex]
            [tex]d \left( \dfrac{-30}{V^2} + \dfrac{0.45}{(17.5 - 0.15V)^2} \right)= 0 \quad \Big/ : d \neq 0[/tex]
                [tex]\dfrac{-30}{V^2} + \dfrac{0.45}{(17.5 - 0.15V)^2} = 0[/tex]
                      [tex]\dfrac{0.45}{(17.5 - 0.15V)^2} = \dfrac{30}{V^2}[/tex]
                    [tex]30 (17.5 - 0.15V)^2 = 0.45V^2[/tex]
          [tex]30(306.25 - 5.25V + 0.0225V^2) = 0.45V^2[/tex]
             [tex]9187.5 - 157.5V+ 0.675V^2 = 0.45V^2[/tex]
             [tex]0.225V^2 - 157.5V+ 9187.5 = 0[/tex]
Solving the quadratic equation yields
                    [tex]V \approx 64.2 \text{mph}[/tex]
which is the speed between 50 and 70 mph that  will minimize the cost of the trip along the highway.
