Maximum for two-dimensional function (function of two variable) using calculus $P(q_1, q_2) = 1000(q_1+q_2) - q_1^2 -  q_2^2 - 2q_1\cdot q_2 - 100\cdot(q_1+q_2)$
Here's what I did:
Partial derivative of the function with respect to $q_1$; $$\frac{\partial P}{\partial q_1}(q_1, q_2)= 1000 - 2q_1 - 2q_2 - 100 = 0$$ 
Partial derivative of the function with respect to $q_2$; $$\frac{\partial P}{\partial q_2}(q_1, q_2) = 1000 - 2q_2 - 2q_1 - 100 = 0$$ 
The issue with this equation is that I have only one equation for two variables. How do I solve $q_1+q_2 = 450$?
Any help would be greatly appreciated.
Thanks
 A: It only means that you have an infinite number of relative minima/maxima at the points
$$r:(q_{1},450-q_{1})$$
The points sit on a straight line parallel to the bisectrix of the second quadrant of the $q_{1}q_{2}$ plane. If you substitute those values into the function, you get
$$P(r)=450^{2}$$
You can easily verify that those are maxima by noticing that $P(0,0)=0$. If the $r$'s were minima, there would need to be other maxima between $(0,0)$ and $r$, then you should have some other null-derivative points. Since you don't have any, you conclude that the $r$'s are relative maxima.
There is still the possibility that $P$ has a supremum at infinity. But as $r$ are relative maxima, if there were a supremum at infinity you should have at least a minima at any of the sides of $r$, after which the function should start to increase again. But since you don't have any minima, you can conclude that $r$ is a set of absolute maxima.
As for the minima, there aren't any. There is though an infimum at infinity. In fact, as $(q_{1},q_{2})\to\infty$, the quadratic terms outrun the others, and the function goes to $-\infty$ (this is, by the way, another proof of the fact that we don't have a supremum at infinity).
