The most economical speed of a truck

Cost per km.
$$c(v)=\frac{18+0.005v^2}{v}=\frac{18}v+0.005v,\qquad v>0.$$

Derivative. $c'(v)=-\dfrac{18}{v^2}+0.005$, which vanishes when $v^2=3600$, i.e. $v=60$. It is negative on $(0,60)$ and positive on $(60,\infty)$, so $v=60$ is the global minimum on $(0,\infty)$.

Answer. $60$ km/h, at a cost of $c(60)=0.30+0.30=0.60$ €/km.

Comparison. At $80$ km/h: $c(80)=0.225+0.40=0.625$ €/km, and at $40$ km/h: $c(40)=0.45+0.20=0.65$ €/km.

Remark. At the optimum the two parts of the cost are equal ($0.30$ each). This is typical of $\frac av+bv$: the minimum is where the two terms balance.