## Thomas' Calculus 13th Edition

The maximum value of $\kappa$ is $\dfrac{1}{2b}$.
We have: $\dfrac{d \kappa}{da}=\dfrac{d}{da}[\dfrac{a}{a^2+b^2}]=\dfrac{b^2-a^2}{(a^2+b^2)^2}$ We find the maximum value as follows: $\dfrac{d \kappa}{da}=0$ $\implies \dfrac{b^2-a^2}{(a^2+b^2)^2}=0$ or, $b^2-a^2 =0 \implies a=\pm \sqrt {b^2}$ Since, $a, b \geq 0$ So, $a=b$ a) When $a \gt b$, then $\dfrac{d\kappa}{da} \lt 0$ b) When $a \lt b$, then $\dfrac{d\kappa}{da} \gt 0$ Therefore, $\kappa$ has the maximum value at $a=b$ . Now, $\kappa (b)=\dfrac{b}{b^2+b^2}=\dfrac{1}{2b}$ Thus, the maximum value of $\kappa$ is $\dfrac{1}{2b}$.