Answer
The maximum value of $\kappa$ is $\dfrac{1}{2b}$.
Work Step by Step
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 $\dfrac{1}{2b}$.
