Worst case analysis of MAX-HEAPIFY procedure . From CLRS book for MAX-HEAPIFY procedure :

The children's subtrees each have size at most 2n/3 - the worst case
  occurs when the last row of the tree is exactly half full

I fail to see this intuition for the worst case scenario . Can some one explain possibly with a diagram . Thanks .
P.S: I know Big O notation and also found this answer here  but still I have doubts.

Normally for a binary tree,
$ n \approx n/2 + n/2$
But in a binary-heap the left subtree has one more level that is filled completely.
What does this mean for the left subtree size?  Well, since, its a binary tree this is roughly the same as multiplying by 2 
(recall number of nodes in a binary tree of size $n = 2^h -1 $).
So the left subtree has 2x as many nodes as the right.
Now,
Let $a = $ size of left tree  
$b$ = size of right tree
Then we can express the size of the heap as
$$\begin{align}n &= (a + b) \\
&= (2b + b) \text{ (Since left has twice as many as right}) \\
&= 3b \\
&\implies b = \frac{n}{3}, a= \frac{2}{3}n \\
\end{align}
$$
