4. Traffic Lights (from 9th grade. 2 points). Long Highway intersects with Narrow Street and Quiet Street (see fig.). There are traffic lights at both intersections. The first traffic light allows traffic on the highway for $x$ seconds, and for half a minute on Narrow St. The second traffic light allows traffic on the highway for two minutes, and for $x$ seconds on Quiet St. The traffic lights operate independently of each other. For what value of $x$ will the probability of driving through both intersections on Long Highway without stopping at the traffic lights be the greatest? What is this maximum probability?

![](https://cdn.mathpix.com/cropped/2024_05_06_1584a5c3bdc918dddbecg-09.jpg?height=483&width=919&top_left_y=604&top_left_x=614)
Solution. First method. We will measure time in seconds. The probability of passing the intersection with Narrow St. without stopping is $\frac{x}{x+30}$. The probability of passing the intersection with Quiet St. without stopping is $\frac{120}{x+120}$. Since the traffic lights operate independently of each other, the probability of passing both intersections without stopping is

$$
p(x)=\frac{120 x}{(x+30)(x+120)}=120 \cdot \frac{x}{x^{2}+150 x+3600}
$$

We need to find the value of $x$ for which the function

$$
f(x)=\frac{120}{p(x)}=\frac{x^{2}+150 x+3600}{x}=x+\frac{3600}{x}+150
$$

has the smallest value on the ray $(0 ;+\infty)$. By the Cauchy inequality for means $^{4}$

$$
f(x) \geq 150+2 \sqrt{x \cdot \frac{3600}{x}}=150+2 \cdot 60=270
$$

therefore, for all $x$

$$
p(x)=\frac{120}{f(x)} \leq \frac{120}{270}=\frac{4}{9}
$$

and equality is achieved if $x=\frac{3600}{x}$, from which $x=60$.[^3]

Second method. The point of maximum of the function $p(x)$ can be found using the derivative.

Answer: 60 seconds. The probability is $4 / 9$.