= Solution
Use the <logic approximation for gene regulation>, zero basal production and initially vanishing X, Y and Z. Let $K=K_{xy}=K_{xz}=K_{yz}=K_{xx}/2$, $Y_\infty=Z_\infty=\beta_y/\alpha$, and assume the activated Q level exceeds the X activation threshold. The circuit combines a <coherent feed-forward loop> with <negative autoregulation>. While X is below its self-repression threshold it obeys $\dot X=\beta_x-\alpha X$, hence X first crosses $K$ at
$$
t_X=-\frac1\alpha\ln\left(1-\frac{\alpha K}{\beta_x}\right).
$$
Y production begins then. While X remains above $K$, $Y=Y_\infty[1-e^{-\alpha(t-t_X)}]$. The AND gate for Z first becomes active when Y reaches $K$, so
$$
\boxed{T_{\min}=t_X-\frac1\alpha\ln\left(1-\frac{\alpha K}{\beta_y}\right)
\simeq-\frac1\alpha\ln\left(1-\frac{\alpha K}{\beta_y}\right).}
$$
The approximation drops the short X build-up time because $\beta_x\gg\beta_y$. Strictly the input must last longer than this onset time to produce a nonzero amount of Z. This requires $\beta_y/\alpha>K$: at equality the threshold is approached only asymptotically, and below it Z never switches on. No numerical delay follows without the threshold-to-production ratio.
In the strong <negative autoregulation> limit, X is held close to $K_{xx}=2K$; a smooth repression law regularizes the rapid switching of the ideal Boolean rule. After the input disappears at $t_0$, production of X stops and $X\simeq2Ke^{-\alpha(t-t_0)}$. It takes $(\ln2)/\alpha$ for X to fall below $K$. Until then Y remains at $Y_\infty>K$ and the Z AND gate stays on, so Z stays at $Z_\infty$. After that Z decays by dilution, taking another $(\ln2)/\alpha$ to halve. Therefore
$$
\boxed{t_{Z,1/2}-t_0\simeq\frac{2\ln2}{\alpha}.}
$$
More generally replace the first $\ln2$ by $\ln[X(t_0)/K]$. This distinguishes disappearance of the external ligand from disappearance of the persistent X <transcription factor>.
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