I'm doing the same problem as the one on pg. 5 of this link: http://www.math.lsa.umich.edu/~pwn/01~ch1_solutions.pdf
I got the same answer for my equation, although my approach was slightly different. I have the formula for cost in terms of the crew workers, but when I derive, I don't get the same result. Am I doing something wrong? This is my equation for cost:
$C=500 \frac{280}{k+1}+(18000+800 \frac{280}{k+1})(k)$
So,
$\frac{dC}{dn}= -\frac{140000}{(k+1)^2}+18000+\frac{224000}{(k+2)^2}$
Is that correct?
Note that the referenced page has two formulas, one for $k \lt 19$ and one for $k \ge 19$. The first one you have derived correctly (except for the $k+2$ term on the end, which should be $k+1$). The second is as follows:
$$C={500\cdot 280\over k+1}+18000k+{800\cdot 280\cdot k\over k+1}+{10000\cdot 280\over k+1}-140000$$
$$\iff C=18000k+800\cdot 280+{9700\cdot 280\over k+1}-140000$$
Then,
$${dC\over dk}=-{9700\cdot 280\over (k+1)^2}+18000$$
Since the first equation has an always-positive derivative, it is increasing for all $k$, and therefore it is minimized when $k$ takes on the smallest value within its allowed range ($k=19$). The second equation has the possibility of a zero-valued derivative within the allowed range for $k$, which will occur at
$$18000(k+1)^2=9700\cdot 280$$
or
$$k=\sqrt{9700\cdot 280\over 18000}-1$$
And this value, if it is within the range allowed for $k$ for this formula ($0 \le k\lt 19$), will be the minimum value for that portion.