If a store owner defines total daily profit as total revenue from newspaper sales, less total cost of newspapers ordered, less goodwill loss from unsatisfied demand, how many copies per day should be order to maximize expected profit?

Cost to storeowner: $.70
Price charged: $.90
Any copies left over are destroyed.
Any requests that can't be met are considered a $.05 loss.

Probability distribution of number of requests for paper in a day:
Number of Requests | Probability
0 | .12
1 | .16
2 | .18
3 | .32
4 | .14
5 | .08


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So, I need to find the maximized profit. I don't really know where to start even. I've found E(X) = 2.44 = 0*.12+1*.16+2*.18+3*.32+4*.14+5*.08 for the Expected number of requests.

To maximize the profit, I was thinking it would be right to find the expected profit for each number requested:

So you'd do it for 0-5 requests, but for 5 requests, the expected profit it would be: revenue - cost - unused
2.44*$.9 - 5*$.7 - (5-2.44)*$.05 = 2.196 - 3.5 - .128 = $-1.432

Doing it for all of them, the largest value will be the maximum profit?