Shaggy has to learn the same 71 hiragana characters, and

I must admit the wording is pretty awkward. Anyway:

As "the number of characters he learns on any one day should be within 4 off the number he learns on any other day", then the number of characters he learns on any day must be between \(x\) and \(x+4\) characters .

A. To minimize # of characters per day he should learn \(x\) (so min possible) characters on one day and \(x+4\) (so max possible) characters on all other 6 days, so we would have: \(x+6(x+4)=71\) --> \(x_{min}=\frac{47}{7}=6.something\), as \(x\) must be an integer than round up to 7 (it can not be less than 7 as 6.something is minimum);

B. To maximize # of characters per day he should learn \(x\) (min possible) characters on 6 days and \(x\) characters on all other 6 days, so we would have: \(6x+(x+4)=71\) --> \(x_{max}=\frac{67}{7}=9.something\), as \(x\) must be an integer than round down to 9 (it can not be more than 9 as 9.something is maximum), so max value for \(x+4\) is 9+4=13.

Answer: 7 and 13.

General rule for such kind of problems:
to maximize one quantity, minimize the others;
to minimize one quantity, maximize the others.

Similar question to practice: a-certain-city-with-population-of-132-000-is-to-be-divided-76217.html

Hope it helps.