To do this problem, I'm pretty sure you just have to find the maximum:
f(x)=11000-x^3+36x^2+600x
f'(x)=-3x^2+72x+600
Set f'(x) to be 0
0=-3x^2+72x+600
0=x^2-24x-200
The only positive x value is 30.54
Now we plug this into f(x)
f(30.54)=11000-30.54^3+36(30.54)^2+600(30.54)=34416.5
Thus we get (30.54,34416.5)
Of course, I'm not as confident about this one as the other ones, however, so you may need to double check my work!