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.

Start with a heap $H$ with $n$ levels with all levels full. That's $2^{i - 1}$ nodes for each level $i$ for a total of $$|H| = 2^n - 1$$ nodes in the heap. Let $L$ denote the left sub-heap of the root and $R$ denote the right sub-heap. $L$ has a total of $$|L| = 2^{n - 1} - 1$$ nodes, as does $R$.  Since a binary heap is a complete binary tree, then new nodes must be added such that after heapification, nodes fill up the last level from left to right. So, let's add nodes to $L$ so that a new level is filled and let's denote this modified sub-heap as $L'$ and the modified whole heap as $H'$. This addition will require $2^{n - 1}$ nodes, bringing the total number of nodes in $L'$ to $$ |L'| = (2^{n - 1} - 1) + 2^{n - 1} = 2\cdot 2^{n - 1} - 1$$ and the total number of nodes in the entire heap to $$ |H'| = (2^{n} - 1) + 2^{n - 1} = 2^n + 2^{n - 1} - 1 $$
The amount of space $L'$ takes up out of the whole heap $H'$ is given by
$$ \frac{|L'|}{|H'|} = \frac{2\cdot 2^{n-1} - 1}{2^n + 2^{n - 1} - 1} =
\frac{2\cdot 2^{n-1} - 1}{2\cdot 2^{n - 1} + 2^{n - 1} - 1} =
\frac{2\cdot 2^{n-1} - 1}{3\cdot 2^{n - 1} - 1} $$
Taking the limit as $n \to \infty$, we get:
$$ \lim_{n\to\infty} { \frac{|L'|}{|H'|} } = \lim_{n\to\infty} { \frac{2\cdot 2^{n-1} - 1}{3\cdot 2^{n - 1} - 1} } = \frac{2}{3} $$
Long story short, $L'$ and $R$ make up effectively the entire heap. $|L'|$ has twice as many elements as $R$ so it makes up $\frac{2}{3}$ of the heap while $R$ makes up the other $\frac{1}{3}$.
This $\frac{2}{3}$ of the heap corresponds to having the last level of the heap half full from left to right. This is the most the heap can get imbalanced; adding another node will either begin to rebalance the heap (by filling out the other, right, half of the last level) or break the heap's shape property of being a complete binary tree. 
