3. (15 points) In a box, there are 22 red and 25 blue balls. They are distributed into two boxes: the first box should have 24 balls, and the second box should have 23. After the distribution, the percentage of blue balls in each box was calculated and the obtained numbers were added together. What should be the distribution of blue balls between the boxes to maximize the sum of these percentages?

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# Solution.

Solution 1. Instead of the total percentage, we will calculate the total share of blue balls, which are obviously 100 times different and reach their maximum simultaneously. Each blue ball in a box of 24 balls constitutes $1 / 24$ of the total number of balls in that box, and in a box of 23 balls, it constitutes $1 / 23$ of the total number of balls. Therefore, if we swap blue balls from the box with more balls with red balls from the box with fewer balls, the total percentage of blue balls in the boxes will increase. Thus, the maximum is achieved when all such swaps are made, that is, when the box with fewer balls consists entirely of blue balls, and the box with more balls contains 2 blue balls and 22 red balls.

Solution 2.

|  | Total number of balls | Blue balls | Share in each box |  |  |
| :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- |
| 1st box | 24 pieces | $x$ | $x / 24$ |  |  |
| 2nd box | 23 pieces | $25-x$ | $(25-x) / 23$ |  |  |
| Therefore, the total share of blue balls in the two boxes is equal to |  |  |  |  |  |

$\frac{x}{24}+\frac{25-x}{23}=-\frac{x}{24 \cdot 23}+\frac{25}{23}=-\frac{x}{552}+\frac{25}{23}$ and represents a linear function with a negative slope. Therefore, this function reaches its maximum value at the left end of the interval [2; 24], that is, when $x=2$. Thus, in the box with fewer balls, there are only blue balls, and in the box with more balls, there are 2 blue balls and 22 red balls.

Answer. In the second box - 23 blue balls, in the first box - 2 blue balls and 22 red balls.

| Content of the criterion | Points |
| :--- | :---: |
| The correct answer is obtained with justification. | 15 |
| An arithmetic error is made in the correct solution process. | 10 |
| The problem is solved by selecting values of the variable that satisfy the condition of the problem. | 5 |
| There are individual advances in the solution. | 2 |
| The solution does not meet any of the criteria listed above | 0 |
| Maximum score | 15 |