Answer
Maximum value of $\kappa= \dfrac{1}{2b}$
Work Step by Step
$\dfrac{d \kappa}{da}=\dfrac{d}{da}[\dfrac{a}{a^2+b^2}]=\dfrac{b^2-a^2}{(a^2+b^2)^2}$
Maximum value is: $ \dfrac{d \kappa}{da}=0$
or, $\dfrac{b^2-a^2}{(a^2+b^2)^2}=0$
or, $b^2-a^2 =0$
or, $ a=\pm \sqrt {b^2}$
Since, $a, b \geq 0$
So, $a=b$
Now, $\dfrac{d\kappa}{da} \lt 0$ when $a \gt b$ and $\dfrac{d\kappa}{da} \gt 0$ when $a \lt b$
Thus, $\kappa $ has a maximum value at $a=b$ .
$\implies \kappa (b)=\dfrac{b}{b^2+b^2}=\dfrac{1}{2b}$
Thus, the maximum value of $\kappa$ is $\dfrac{1}{2b}$.
