Why is the formula for maximum speed in frictionless surface used to calculate optimum speed to avoid wear and tear of tires?

Question

In my school textbook , under the concept of friction I studied maximum possible velocity of a car while taking a turn in normal roads is given by: ${v}_{max}=\sqrt{{\mu}_{s}rg}$ , and on banked road it is: ${v}_{max}=\sqrt{rg\times \frac{{\mu}_{s}+\tan\left(\theta \right)}{1-{\mu}_{s}\tan\left(\theta \right)}}$ , where ${\mu}_{s}$ is the static coefficient of friction, $r$ is the radius and $g$ is acceleration due to gravity.

And then I studied that when we ignore friction, the formula turns out to be ${v}_{max}=\sqrt{rg\tan\left(\theta \right)}$, considering road can be banked too.

Now, I have seen when questions like these are posed:

What is the optimum speed of the car to avoid any wear and tear of tires? (Given some value of coefficient of friction)

The solutions use ${v}_{max}=\sqrt{rg\tan\left(\theta \right)}$ ? So here's my question:

1. When its mentioned that friction is present, why do we just assume it friction less?
2. I have also studied that both area of contact and velocity have no relation to friction (at least in case of static,rolling and sliding friction), then also it doesn't make sense to evaluate some "optimum" speed at some friction?

Answer 1

The optimal speed (for wear and tear) around a banked turn is the speed at which there is no friction force on the tire. Why? Because friction causes the most wear and tear. (Going slower than optimal would cause a lateral friction force too, due to gravity)

So

the optimal speed for wear and tear in the presence of friction

Equals

the maximum possible speed in the absence of friction.

Note that those are both different from “the maximum possible speed around a banked turn when there is friction”