Answer
A nitrogen level of $~~N=1~~$ gives the best yield ($Y(1)=\frac{1}{2}k$).
Work Step by Step
$Y = \frac{kN}{1+N^2}$
We can assume that $N \geq 0$ since $N$ is the nitrogen level.
We can find the point where $Y'(N) = 0$:
$Y'(N) = \frac{k(1+N^2)-2N(kN)}{(1+N^2)^2} = \frac{k(1-N^2)}{(1+N^2)^2} = 0$
$k(1-N^2) = 0$
$N^2 = 1$
$N = 1$
When $0 \leq N \lt 1$, then $Y'(N) \gt 0$
When $N \gt 1$, then $Y'(N) \lt 0$
Thus, $N=1$ must be the point where $Y$ is a maximum.
A nitrogen level of $N=1$ gives the best yield.
