= Solution
For the dynamically regulated interpretation, take $X(0)=S$, $Y(0)=Z(0)=0$, and let X production respond to Y repression after the <ligand> appears. Define $h=K/S\in(0,1)$. Initially active X drives both Y and Z, giving $Y=Z=S(1-e^{-\alpha t})$. Y reaches $K$ at $t_1=-\ln(1-h)/\alpha$ and switches X production off. X then decays from $S$ to $K$, reaching it at
$$
t_2=t_1+\frac{\ln(1/h)}\alpha=\frac1\alpha\ln\frac1{h(1-h)}.
$$
Until $t_2$, X remains active and Y keeps being produced. Thus $Y_2=S(1-h+h^2)$. At $t_2$, Y production stops, but Y still activates Z. Its subsequent decay crosses $K$ at
$$
\boxed{t_{\max1}=t_3=t_2+\frac1\alpha\ln\frac{Y_2}{K}
=\frac1\alpha\ln\frac{1-h+h^2}{h^2(1-h)}.}
$$
Immediately afterwards both inputs to the Z OR gate are below threshold, so Z production stops and its <derivative> changes from positive to negative. Since Z has been continuously produced from zero until $t_3$,
$$
\boxed{Z_{\max1}=S(1-e^{-\alpha t_3})
=S\left[1-\frac{h^2(1-h)}{1-h+h^2}\right].}
$$
For $h=1/2$, $t_{\max1}=\ln6/\alpha$ and $Z_{\max1}=5S/6$. Y repression is lifted at $t_3$, but X must first rebuild from a subthreshold value, so Z has a genuine first local maximum before any later response.
\b[Wording qualification:] if “X is always present at its <steady state> <concentration>” is read as an external clamp maintaining $X=S$ \i[after] <ligand> arrival, it overrides Y's repression and changes the problem. Then $Z=S(1-e^{-\alpha t})$ rises monotonically, with no finite first maximum. The finite-peak result uses the consistent initial-steady-state interpretation in which the stated <negative feedback> can act.
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