The Prisoner's Dilemma

Consider the case $n=1$ in which you precisely fit one cookie in a 2×2 square, leaving $4-\pi$ square units of cookie dough, which are “wasted” in that you cannot make any further cookies. This case shows two things:

(a) you must always waste at least $4-\pi$ square units of dough, and

(b) you can never waste 4 or more square units of dough.

Therefore, the process described is equivalent to making unit-circle cookies (each with area π) one at a time until the area of dough left lies in the interval $[4-\pi ,4)$ (it just takes fewer steps when there is at least 16 square units of dough). If you start with an area $A$ of dough, that process produces $\lfloor \frac{A+\pi -4}{\pi }\rfloor$ cookies. So the general answer for Fill the Jar with an $n\times n$ grid is

$\lfloor \frac{4n^2+\pi -4}{\pi }\rfloor ,$

which comes out to 20 cookies in the case of starting with a 4-by-4 grid.