Chapter 3 - Applications of Differentiation - 3.7 Optimization Problems - 3.7 Exercises - Page 264: 9

$\frac{1}{2}k$

We need to find the maximum $Y$ for $N\geq 0$. $Y(N)$ = $\frac{kN}{1+N^{2}}$ $Y'(N)$ = $\frac{(1+N^{2})k-kN(2N)}{(1+N^{2})^{2}}$ = $\frac{k(1+N)(1-N)}{(1+N^{2})^{2}}$ $Y'(N)$ $\gt$ $0$ for $0$ $\lt$ $N$ $\lt$ $1$ $Y'(N)$ $\lt$ $0$ for $N$ $\gt$ $1$ so $Y$ has an absolute maximum of $Y(1)$ = $\frac{1}{2}k$ at $N$ = $1$.