2008 AMC 10B Problem 4

The team has a fixed amount of money. Every dollar paid to someone else is a dollar the star cannot get, so pay everyone else as little as the rules allow.

1. Besides the star there are $21 - 1 = 20$ players.
2. Each must get at least $15{,}000$. Paying each of them exactly that costs
$$20 \times 15{,}000 = 300{,}000.$$
3. The total cannot be more than $700{,}000$, so the star can get at most
$$700{,}000 - 300{,}000 = 400{,}000.$$
4. This really happens: twenty players at $15{,}000$ and one at $400{,}000$ add to exactly $700{,}000$, and everyone gets at least the minimum.

The answer is $\boxed{\textbf{(C)}\ 400,000}$.

When a total has a ceiling and each piece has a floor, one piece is biggest when all the other pieces sit at their floor. So "make one as large as possible" becomes "make all the others as small as possible", and the problem is one subtraction. Be careful to count the others, not everyone: $20$ players, not $21$.

The team's payroll is at most $700{,}000$. To leave as much as possible for one player, every one of the other $20$ players should earn the minimum, $15{,}000$:
$$20 \cdot 15{,}000 = 300{,}000.$$
The remaining money goes to the highest-paid player:
$$700{,}000 - 300{,}000 = 400{,}000.$$
This is attainable (twenty players at $15{,}000$ and one at $400{,}000$ totals exactly $700{,}000$), so it is the maximum.

The answer is $\boxed{\textbf{(C)}\ 400,000}$.