Answer: The maximum possible weight of the lightest box would be 3.
Explanation:
Since we have given that
Average of three boxes = 7
Let the weights of three boxes be 'x', 'y', 'z'.
So, we get
![(x+y+z)/(3)=7\\\\x+y+z=7* 3=21](https://img.qammunity.org/2020/formulas/mathematics/high-school/g4xu31cy7koswhrw6hujz24uf6usx33hua.png)
Median = 9
Since it has odd number of boxes i.e. 3
So, median would be the middle term i.e. b = 9
So, it becomes
a,9,c
Suppose c at its highest i.e. c= 9
Then it becomes,
![a+9+9=21\\\\a+18=21\\\\a=21-18\\\\a=3](https://img.qammunity.org/2020/formulas/mathematics/high-school/lcg1rlxnlkpfxdfxuei5hspko5gamwvvba.png)
Hence, the maximum possible weight of the lightest box would be 3.