Answer
$v_{min}=16.4\frac{m}{s}=59.0\frac{km}{h}$
$v_{max}=31.7\frac{m}{s}=114\frac{km}{h}$
Work Step by Step
The angle at which the road is banked can be calculated by the formula $\theta=\arctan\Big(\frac{v^2}{rg}\Big)$
$85\frac{km}{h}\times \frac{1000m}{1km}\times \frac{1h}{3600s}=23.6\frac{m}{s}$
$\theta=\arctan\Big(\frac{(23.6\frac{m}{s})^2}{78m\times 9.8\frac{m}{s^2}}\Big)=36.1^o$
When driving at higher speed with the same radius, you require a greater centripetal acceleration than the normal force. Therefore, the force of friction would act downward along the incline.
$\sum F_y=F_N\cos(\theta)-F_g-F_f\sin(\theta)=0N$
because there is no acceleration in the y direction
Rearranging, $F_N=\frac{mg}{\cos(\theta)-\mu_s\sin(\theta)}$
$\sum F_x=F_N\sin(\theta)+F_f\cos(\theta)=\frac{mv^2}{r}$
$F_N=\frac{mv^2}{r(\sin(\theta)+\mu_s\cos(\theta))}$
$\frac{mg}{\cos(\theta)-\mu_s\sin(\theta)}=\frac{mv^2}{r(\sin(\theta)+\mu_s\cos(\theta))}$
Rearranging and solving for $v_{max}$
$v_{max}=\sqrt{(78m)(9.8\frac{m}{s^2})\frac{\sin(36.1)+0.3\cos(36.1)}{\cos(36.1)-0.3\sin(36.1)}}=31.7\frac{m}{s}$
For minimum velocity, the force of friction is acting in the opposite direction so the sign is negative.
$v_{min}=\sqrt{(78m)(9.8\frac{m}{s^2})\frac{\sin(36.1)-0.3\cos(36.1)}{\cos(36.1)+0.3\sin(36.1)}}=16.4\frac{m}{s}$
