The problem asks us to use the digits 1 to 9, at most one time each, to fill in the boxes to make two fractions that have a quotient that is as close to 4/11 as possible.
The first step is to try to fill in the boxes so that the numerators of the two fractions are as close to each other as possible. This is because the quotient of two fractions is simply the ratio of their numerators to their denominators. So, if the numerators are close, then the quotient will also be close.
I started by filling in the first box with the largest digit, 9. This gave me a numerator of 9 for the first fraction. I then filled in the second box with the next largest digit, 8. This gave me a numerator of 8 for the second fraction.
Next, I filled in the third box with the next largest digit, 7. This gave me a denominator of 7 for both fractions. Finally, I filled in the fourth box with the only remaining digit, 6. This gave me a denominator of 6 for the first fraction and a denominator of 1 for the second fraction.
The two fractions I came up with are:
* Fraction 1: 9/7
* Fraction 2: 6/1
The quotient of the first fraction is 9/7 = 1.285714, and the quotient of the second fraction is 6/1 = 6. The average of these two quotients is 3.642857, which is very close to 4/11.
I believe that this is the closest quotient to 4/11 because the two fractions are as close to each other as possible. The difference between the two fractions is only 3, which is a relatively small difference. Additionally, the denominators of the two fractions are relatively close, which means that the error introduced by rounding the numerators is also relatively small.
Here are some other attempts I made before I arrived at the solution above:
* I tried filling in the boxes so that the denominators of the two fractions would be as close to each other as possible. However, this didn't work as well as filling in the boxes so that the numerators were close.
* I tried filling in the boxes so that the sum of the numerators and denominators of the two fractions would be as close to each other as possible. However, this didn't work as well as the other two methods.
I hope this explanation is more detailed and informative. Please let me know if you have any other questions.
Additional thoughts:
* The problem could also be solved by using a computer program to search for all possible combinations of digits in the boxes. However, this would be a very computationally expensive task.
* The problem could also be solved by using a mathematical model to represent the problem. This would allow us to find the optimal solution using a variety of optimization techniques.