Problem A grocer stock pear. On any given day, demand for one, two or three pears is 0.2, 0.5, 0.3, respectively. He buys the fruits at 60 cents each and sells them at one dollar each. Left overs will be thrown away by the end of the day. How many pears should the grocer purchase to maximize his daily profit?
I think I should obtain an expression for the mean profit in terms of number of fruits stocked, let's say n.
The profit function in general is given by $$ p(q)q-C(q) \text{ or, less often as } q(p)p-C(p)$$ Daily profit will be a random variable: Y=100-60=40X
Therefore: $$E[Y]=40\sum_{k=0}^{n}kP(X=k)=40\left(0\cdot0.2+1\cdot0.5+2\cdot0.3...\right)$$
This is what have so far, and I do not know how to continue from there....
It's clear that he should stock at least $1$ pear and at most $3$ pears.
Let $n$ be the number of pears stocked.
Take cases, based on the value of $n$.
If $n=1$, he sells $1$, hence the profit is
$$(1)(1) - (1)(.6) = .4$$
If $n=2$, the expected number sold is
$$(.2)(1) + (.8)(2) = 1.8$$
hence the expected profit is
$$(1.8)(1) - (2)(.6) = .6$$
If $n=3$, the expected number sold is
$$(.2)(1) + (.5)(2) + (.3)(3) = 2.1$$
hence the expected profit is
$$(2.1)(1) - (3)(.6) = .3$$
Since the expected profit is maximized for $n=2$, he should stock $2$ pears.