Michael drives x miles due north at arrives at Point A. He then heads

At the risk of being that guy, I believe the answer is E. Here's why:

We already know that, if we start at a place on the equator and walk 40,000 km (the approximate circumference of Earth) due east, we will end up at the same place we started.
If we start at a place further north (say Los Angeles) and walk due east, we will return to our starting place in less than 40,000
In fact, the further north we move our starting point, the less the distance one must walk due east to return to the starting point.

So, there must exist a point (very close to the North Pole) where, if we walk due east, we will return to our starting place in 5 miles.
Let's call this point Point Q.
To reiterate, if we start at Point Q and walk due east for 5 miles, we end up at the exact point we started (Point Q).

So, if we start at a point that is 6 miles due south of Point Q, then Michael's journey goes like this:
Michael drives 6 miles due north at arrives at Point A (aka Point Q). He then heads due east for 5 miles (at which point, he arrives back at Point Q) . Finally, he drives 6 miles in a straight line (due south) until he reaches his starting point.

So, the length of the 3 legs of his journey are: 5 miles, 6 miles and 6 miles (the shortest leg being 5 miles)
So, the total trip was 17 miles.

Of course, there's also the option where the total trip is 30 miles.

Since we cannot definitively answer the question, the correct answer must be E

Cheers,
Brent