Find the maximum value of the expression

$$
x \sqrt{1-y^{2}}+y \sqrt{1-x^{2}}
$$
Answer: The maximum value of the expression is 1. Since the expression $\sqrt{1-x^{2}}$ is defined, $|x| \leq 1$, which means there exists an angle $\varphi$ such that $x=\cos \varphi, \sqrt{1-x^{2}}=\sin \varphi$. Similarly, $y=\cos \psi, \sqrt{1-y^{2}}=\sin \psi$.

Therefore,

$$
x \sqrt{1-y^{2}}+y \sqrt{1-x^{2}}=\cos \varphi \sin \psi+\sin \varphi \cos \psi=\sin (\varphi+\psi) \leq 1
$$

On the other hand, when $x=1, y=0$, the value 1 is achieved.

There are balance scales, any weights, and ten bags of coins. All coins in all bags look the same, but in one of the bags, all the coins are counterfeit and each weighs 15 g, while in the other nine bags, all the coins are genuine and each weighs 20 g. How can you determine with one weighing which bag contains the counterfeit coins?

## Hint

Try taking 1 coin from the first bag, 2 from the second, 3 from the third, ..., 10 from the last and weigh them.

## Solution

Take 1 coin from the first bag, 2 from the second, 3 from the third, ..., 10 from the last. In total, there will be $1+2+3+\ldots+10=55$ coins. Weigh them. If all of them were genuine, they would weigh (55×20) = 1100 g, but in our case, they will weigh less. If there is one counterfeit coin, there will be a shortage of 5 g, if two 10 g, ... if ten counterfeit coins, there will be a shortage of 50 g. Thus, knowing how much is missing from 1100 g, we can immediately determine the number of counterfeit coins. The number of counterfeit coins, in turn, will show us the number of the bag in which they are located.

## Answer

Take 1 coin from the first bag, 2 from the second, 3 from the third, ..., 10 from the last. Weigh them. If the counterfeit coin is in the first bag, there will be a shortage of 5 g, if in the second 10 g, ... if in the last 50 g.