= Solution
The given <facilitated target search> time contains a rising term $L\lambda/D_1$ and a falling term $V/(D_3\lambda)$. Longer sliding explores more neighbouring <DNA> without another bulk encounter, reducing the cost of three-dimensional relocation; but it also spends longer revisiting local contour. Roughly $L/\lambda$ local scans, each taking time of order $\lambda^2/D_1$, give the first term. The second represents the bulk encounter cost for an effective capturable contour range of order $\lambda$, with a <Smoluchowski association rate> scaling as $D_3\lambda$. Numerical geometrical factors are already absorbed in the stipulated scaling formula.
Differentiating at fixed positive $L,V,D_1,D_3$ gives
$$
\frac{d\tau_{1+3}}{d\lambda}=\frac L{D_1}-\frac V{D_3\lambda^2},\qquad
\boxed{\lambda_* =\sqrt{\frac{VD_1}{LD_3}},\qquad
\tau_{\min}=2\sqrt{\frac{LV}{D_1D_3}}.}
$$
The second derivative $2V/(D_3\lambda^3)>0$ proves the minimum is unique. The two contributions are equal there. The formal curve diverges at both zero and infinite $\lambda$; real sliding lengths are bounded, for example by the available contour, so an inaccessible optimum must be replaced by the nearest physical boundary. In dimensionless form, $\tau/\tau_{\min}=(u+u^{-1})/2$ with $u=\lambda/\lambda_*$. The right panel of the original illustration shows its minimum and the two contributing terms. The supplied expression does not include a factor of 6000; it should not silently be identified with the preceding earliest-encounter estimate.
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