Minimizing total trip cost as a function of speed

The total cost per hour of travel is fuel cost plus fixed cost: $k v^3 + 800$. Since the trip covers $500$ miles at speed $v$, the travel time is $t = 500/v$ hours. Multiplying, the total cost is $C(v) = \left(kv^3 + 800\right)\cdot \frac{500}{v} = 500kv^2 + \frac{400000}{v}.$ With $k = 0.02$, this becomes $C(v) = 10v^2 + \dfrac{400000}{v}$, defined for $v > 0$.

Differentiate: $C'(v) = 20v - \dfrac{400000}{v^2}$. Setting $C'(v) = 0$ gives $20v = \dfrac{400000}{v^2}$, so $v^3 = 20000$, giving $v = 20000^{1/3} \approx 27.14$ mph. Since $C''(v) = 20 + \dfrac{800000}{v^3} > 0$ for all $v>0$, this critical point is a minimum, and it is the only one, so it minimizes total cost.