As it turns out, the solution to this problem is known. Say that your profit on each cup of soup you sell is $\pi$ -- in your case, 4 dollars -- and your loss if you make a cup of soup but don't sell it is $\lambda$ -- in you case, 1 dollar. Let $Y$ be the distribution of demand.
Then in order to maximize your expected profit, you should make $y$ cups, where $y$ is the least integer such that $P(Y \le y) \ge \pi/(\lambda + \pi)$. You can obtain this by analyzing the equation Ross Millikan gives in his comment. In particular if $\pi > \lambda$, as in your case, you want to stock more than the median demand, since unsold stock is relatively cheap. In your case $\pi/(\lambda + \pi) = 4/(1+4) = 0.8$, so you want to stock an amount that will meet demand 80 percent of the time.
Proving that this is the right formula is a bit tedious. But here's an explanation that may help. Let's ask the question: is it better to stock to meet the demand 80 percent of the time, or to stock $x$ more and therefore have enough to meet the demand 81 percent? Well, (100-81)% of the time the demand is above what you stock in the second case, and so you sell $x$ more cups for an extra profit of $4x$. But 80% of the time this just results in your overstock being slightly bigger, and so you lose $x$. So on average, slightly increasing your stock costs you $4x(.19)-x(.80) = -0.04x$. The same is true if you compare stocking at the 80th percentile to stocking at the 79th percentile.
I don't know if this goes under some standard name -- I recently learned this from a problem in Probability by Jim Pitman (specifically, this is problem 3.2.15(b)).
So in your case you want the 80th percentile of the distribution; for a normal distribution this will be about $0.85$ standard deviations above the mean, or about $175$ with the numbers you gave. Of course there is no reason to expect the distribution to be normal, or for the standard deviation to be 30. But a useful rule of thumb here could be that you want to stock in such a way that you sell out one time out of every five, and you can find that point by experimenting.