Subtask 1

Note that when ~X \geq N~ or ~X \leq 0~, Eric's profit is non-positive. He would just make more money by not selling apples.

For this subtask, it suffices to try all values of ~X~ from 1 to ~N-1~, and take the first value of ~X~ that maximizes the profit.

Subtask 2

Note that Eric's revenue can be written as ~-X^2+Nx~. By completing the square, we get ~-(X-\frac{N}{2})^2+\frac{N^2}{4}~.

When ~N~ is even, then the profit is simply maximized if and only if ~X = \frac{N}{2}~. However, when ~N~ is odd, then the minimum value of this expression is ~\frac{N^2 - 1}{4}~, attained when ~|X-\frac{N}{2}| = \frac{1}{2}~. In other words, this occurs when ~X = \frac{N \pm 1}{2}~. Taking the smaller value, the solution is ~\frac{N-1}{2}~.