Let the distance of p be k, then the distance of q is 3k.

AC=a AB=√3a

① When 3k≤a, k≤a/3. At this time, Q is on AC and P is on AB.

Then the area is 1/2× 3k× k = 3/2k2. When k=a/3, the maximum area is 3/2a 2.

② when a≤3k≤3a and a/3≤k≤a, then q is on BC and p is on AB.

The height is (3a-3k)/2 based on AB.

So the area is1/2k (3a-3k)/2 =1/4 (3ak-3k2), the symmetry axis is k= 1/2a, and the opening is downward.

Therefore, the closer to the axis of symmetry, the greater the value, so the maximum area of k= 1/3a is 3/2a 2.

③ Because the speed ratio of PQ is 1:3.

The distance of meeting is a+2a+√3a p, and the distance of lines is (3a+√ 3a)/4.

So PQ will meet on AB.

Then when a < k < (3a+√ 3a)/4 area is 0.

In a word, the maximum area of APQ is 3/2a 2.