The wall clock in the kitchen is 2 seconds slow every day, and the antique clock in the room is 15 seconds fast every day. On Sunday at noon, the wall clock shows 12:01, and the antique clock shows 11:59. When during the week will the sum of the squares of the differences between the times shown by the clocks and the actual time be the smallest?
Solution. Let the number of days elapsed since Sunday be $n$. Then the deviation of the wall clock from the actual time is: $60-2 n$, and that of the antique clock is $15 n-60$ seconds. The sum of the squares of the deviations is: $(60-2 n)^{2}+(15 n-60)^{2}=229 n^{2}-2040 n+7200$. This is a quadratic function, and since the coefficient of its square term is positive, the function has a minimum.

By transforming the quadratic function, we can read off the minimum value:

$$
f(n)+\left(n-\frac{1020}{229}\right)^{2}+\text { constant. }
$$

From this, $n_{\min }=\frac{1020}{229}$. The smallest deviation will occur on the fourth day from Sunday, that is, at 10:53:58 PM on Thursday.