Maximum value of simulation I'm writing a simulation for a newspaper stand in Scala. Before I wrote the simulation, I tried to come up with the max value for the simulation analytically (to no avail). I should be able to calculate this value with ease, but it's been a few years since I've done any math. 
Anyway, the simulation has the following constraints: sales are randomly generated by a function call to gen(1,10) * 100 where gen is a scala function from a simulation library assumed to be actually random. The orders must be constant(cannot vary day to day). Left over orders (as in orders in excess of sales on a given day), are sold at ten cents a paper. The formula is I came up with for calculating profit: 
profit = $1.00 * sales -$0.50 * orders + $0.1(orders-sales)

From here though, I'm stumped. I tagged it as hw, but it's not for a grade or anything. HW is the most viewed tag.
 A: TL;DR: You should order $60$ copies of the newspaper.
Let $P$ be the price of ordering a newspaper, $S$ be the price of selling it, and $L$ be the price at which you sell leftovers.  In your case, $P = \$0.50, S = \$1.00, L = \$0.10$.
Also, let $x$ be the number of orders, and let $s$ be the number of sales.  The formula you wrote, using these variables, is
$$ \text{Profit}(x,s) = S s - Px + L(x - s)$$
But your profit formula needs to take into account the scenario in which you run out of papers before the sales are complete.  Therefore, you should modify your formula:
$$
\text{Profit}(x,s) =
\left\{
\begin{array}{ll}
Ss - Px + L(x-s) &\text{ if } s \le x \\
Sx - Px &\text{ if } s \ge x
\end{array}
\right.
$$
Now, $s$ is uniformly random from $10 \cdot 1$ to $10 \cdot 10$, let us say from $a$ to $b$.
So
$$
\text{Expected Profit} = \frac{1}{b-a} \int_a^b \text{Profit}(x, s) \; ds
$$
It would never be good to order more papers than you know you are going to sell; nor would it ever be good to order less papers than you know are going to sell.  Therefore, we set $x$ somewhere between $a$ and $b$.  So
\begin{align*}
\int_a^b \text{Profit}(x, s) \; ds
&=
\int_a^x [Ss - Px + L(x-s)] \; ds
+
\int_x^b [Sx - Px] \; ds \\
&=
\int_a^b (-Px ) \; ds
+ \int_a^x [Ss + L(x-s)] \; ds
+ \int_x^b [Sx] \; ds \\
&=
-(b - a) Px + \left[\frac12 S s^2 - \frac12 L (x-s)^2 \right]_a^x
+
(b - x) Sx \\
&=
(-Pb + Pa)x + \frac{S}{2} (x^2 - a^2)
+ \frac{L}{2} (x - a)^2 + bSx - Sx^2
\end{align*}
Ignoring constants, the important terms are
$$
Pax - Pbx + \frac12 S x^2 + \frac12 L x^2 - La x + bS x - Sx^2
$$
$$
= Pax - Pbx + \left( bSx - \frac12 S x^2 \right) + \left(\frac12 L x^2 - Lax \right)
$$
At this point we may as well plug in actual values for $P, S, L, a, b$.
I'm assuming that your gen function makes $a = 10, b = 100$.
\begin{align*}
&\quad (.5) (10) x - (.5) (100) x + (100)(1)(x) - \frac12 (1) x^2 + \frac12 (.1) x^2 - (.1) (10) x
\\
&= 5x - 50x + 100x - \frac12 x^2 + \frac{1}{20} x^2 - x \\
&= 54x - \frac{9}{20} x^2 \\
\end{align*}
This function is quadratic and its maximum is at
$$
x = \frac{-54}{2(-9/20)} = \frac{540}{9} = \boxed{60}.
$$
