Channeling your inner Marty McFly, you travel one week back in time in an attempt to win the lottery. It’s worth $\$10$ million, and each ticket costs a dollar. Note that if you win, your ticket purchase is not refunded. All of this sounds pretty great.
The problem is, you’re not alone. There are 10 other time travelers who also know the winning numbers. You know for a fact that each of them will buy exactly one lottery ticket. Now, according to the lottery’s rules, the prize is evenly split among all the winning tickets (i.e., not evenly among winning people). How many tickets should you buy to maximize your profits?
Because you can depend on the kindness of your fellow time travelers and their close-minded purchasing behavior, you can game the system. If you buy $x$ tickets, then there will be $10 + x$ winning tickets, each of which would earn $\frac{10^7}{10 + x}$ and cost $1$. So the overall profit would be $$P(x) = x \left(\frac{10^7}{10 + x} - 1\right).$$
Differentiating we get $$P^\prime(x) = \frac{10^7}{10 + x} - 1 - \frac{10^7 x}{(10 + x)^2} = \frac{10^7 (10 + x) - (10+x)^2 - 10^7 x}{(10+x)^2} = \frac{10^8 - (10+x)^2}{(10+x)^2}.$$ So we see that the only positive critical point of the profit function is at $\hat{x} = 10^4 - 10 = 9990$ tickets. Since $P^\prime(x) \gt 0$ for $x \lt \hat{x}$ and $P^\prime(x) \lt 0$ for $x \gt \hat{x},$ we reason that this is indeed a local and global maximum of the profit function.
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