, 2019

Solution:

With probability $\left(1-\frac{k}{60}\right)^{9}$, Dan reaches the last crosswalk without crossing at any previous site, in which case the expected value of his wait time is 30 seconds. Otherwise, with probability $1-\left(1-\frac{k}{60}\right)^{9}$, Dan crosses at an earlier crosswalk, in which case the expected value of his wait time is $\frac{k}{2}$. We want to find the $k$ that minimizes
$$30\left(1-\frac{k}{60}\right)^{9}+\frac{k}{2}\left(1-\left(1-\frac{k}{60}\right)^{9}\right)=30-\left(30-\frac{k}{2}\right)\left(1-\left(1-\frac{k}{60}\right)^{9}\right)$$
Letting $a=1-\frac{k}{60}$, we can use weighted AM-GM:
$$9^{\frac{1}{10}}\left(a\left(1-a^{9}\right)\right)^{\frac{9}{10}}=\left(9 a^{9}\right)^{\frac{1}{10}}\left(1-a^{9}\right)^{\frac{9}{10}} \leq \frac{9}{10}$$
where equality occurs when $9 a^{9}=1-a^{9}$, or $a=\left(\frac{1}{10}\right)^{\frac{1}{9}}$, meaning that $k=60\left(1-\left(\frac{1}{10}\right)^{\frac{1}{9}}\right)$. Because our original expression can be written as
$$30-30 a\left(1-a^{9}\right),$$
the minimum occurs at the same value, $k=60\left(1-\left(\frac{1}{10}\right)^{\frac{1}{9}}\right)$.