Maximise the smallest piece of grid Given a big rectangular chocolate bar that consists of n × m unit squares. We wants to cut this bar exactly k times. Each cut must meet the following requirements:
each cut should be straight (horizontal or vertical);
each cut should go along edges of unit squares (it is prohibited to divide any unit chocolate square with cut);
each cut should go inside the whole chocolate bar, and all cuts must be distinct.

Imagine we have made k cuts and the big chocolate is splitted into several pieces. Consider the smallest (by area) piece of the chocolate, We want this piece to be as large as possible.
What is the maximum possible area of smallest piece we can get with exactly k cuts? 
Note : The area of a chocolate piece is the number of unit squares in it.
Example : If we have grid of 3*4 and k=1 then here answer is 6

 A: Suppose that each cut is either vertical across the "width" of the chocolate bar, or horizontal across the "length" of the chocolate bar, so that $k=v+h$ where $v$ denotes the number of vertical cuts and $h$ the number of horizontal cuts.
The notation of the problem gives $n$ rows and $m$ columns, so that $0 \le v \lt m$ and $0 \le h \lt n$.  For any choice of $v,h$ the size of the smallest piece of the "grid" is the product of the smallest vertical increment in columns and the smallest horizontal increment in rows.  To maximize these sizes we make the division of rows and columns as equal as possible.  That is, we have the smallest column increment $\lfloor \frac{m}{v+1} \rfloor$ and the smallest row increment $\lfloor \frac{n}{h+1} \rfloor$.  Thus the objective is to maximize:
$$ \lfloor \frac{m}{v+1} \rfloor \times \lfloor \frac{n}{h+1} \rfloor $$
subject to the constraints $v+h=k$ and $0 \le v \lt m,\; 0 \le h \lt n$.
This reduces the discovery of the exact maximum smallest grid piece to a one dimensional search, substituting (say) $v = 0,1,\ldots,m-1$ if $m \le n$, and choosing the largest of the values above so obtained.
However the search really comes down to checking the endpoints of the range of values.  We can motivate this by considering a continuous approximation of the discrete function above.  Let $v = x-1$ and accordingly $h = k+1-x$ where $x \in [1,k+1]$, and omit the floor from the calculation.  Then the product $mn$ is merely a constant multiplier, and we have as an approximation to located the maximum of:
$$ y = (x(k+2-x))^{-1} \; \text{ on } \; [1,k+1] $$
This function is concave up, so the only critical point $x = \frac{k+2}{2}$ corresponds to a local (global) minimum, and the maximum occurs at an endpoint.
This is actually not deep, but rather tedious to argue for the discrete function directly.  The maximum smallest piece of the grid is achieved by applying as many of the cuts as possible in one direction and the remaining cuts in the other.  Checking the two endpoints in this way suffices to give the exact answer.
