Problem: The manager of a fish market speculates that the number of requests for salmon on any given day is a random variable $X$ with the probability function
$f(x)=\begin{cases} \frac14,&\text{if }x=0\\\\ \frac12,&\text{if }x=1\\\\ \frac14,&\text{if }x=2\;. \end{cases}$
There is a profit of $\$2$ on each salmon he sells and a loss of $\$1$ on each salmon he does not sell. Assuming that each salmon can be sold only on the day it is up for sale, that each request is for a single salmon, and that the managers's speculation is correct, find the number of salmons the market should have per day to maximize profit.
Attempt at a solution: So, what I think this is saying is that there is a $.25$ chance that no salmon will be requested, a $.50$ chance that $1$ salmon will be requested, and a $.25$ chance that $2$ salmon will be requested. This adds up to $1$, so it makes sense. In the second part, it states that each request is for a single salmon, but each the probability function of that only gives $.50$ (not sure if this is right)
Anyway, I attempt trial and error. So if we start with $10$ salmon, then he sells only $5$, and the other $5$ go to waste. So he loses $\$5$. If he starts with $8$ salmon, he loses $\$4$ ($8 *.5 *2 - 8*.5 * \$1$), gives us $4$... Continue this to $0$, he loses none, but has no salmon.
I'm lost
I think I need help modeling and solving , thanks in advance