Is there a method to calculate the maximum gain of a biquad?
in this previous answer, the squared-magnitude of a biquad frequency response is shown to be:

$$\begin{align} f(\phi) \triangleq |H(e^{j\omega})|^2 
&amp; = \left| \frac{b_0 + b_1 e^{-j\omega} + b_2 e^{-j2\omega}}{a_0 + a_1 e^{-j\omega} + a_2 e^{-j2\omega}} \right|^2 \\ 
&amp; = \frac{B_0 + B_1 \phi + B_2 \phi^2}{A_0 + A_1 \phi + A_2 \phi^2} \\ \end{align}$$

where $ \phi \triangleq \sin^2\left(\frac{\omega}{2} \right) $

and

$\begin{align} 
B_0 &amp; = \left(\frac{b_0+b_1+b_2}{2}\right)^2 \\ 
B_1 &amp; = - (4b_0b_2 + b_1(b_0+b_2)) \\
B_2 &amp; = 4b_0b_2 \\
A_0 &amp; = \left(\frac{a_0+a_1+a_2}{2}\right)^2 \\ 
A_1 &amp; = - (4a_0a_2 + a_1(a_0+a_2)) \\
A_2 &amp; = 4a_0a_2 \\
\end{align}$

maxima and minima occur when the derivative is zero.

the derivative of the above is zero when

$$ \phi = \frac{A_2B_0 - A_0B_2 \pm \sqrt{A_0^2B_2^2 + A_2^2B_0^2 - 2A_0A_2B_0B_2 - A_0A_1B_1B_2 \\ + A_0A_2B_1^2 + A_1^2B_0B_2 - A_1A_2B_0B_1}}{A_1B_2 - A_2B_1} $$

then evaluating $f(\phi)$ at those two values of $\phi$ is another laborious pain in ass.  and it won't be my ass.