Boundary solution to a max-min problem Find the maximum and minimum values of $$f (x, y) = x − x^2 + y^2$$ on the rectangle $0 ≤ x ≤ 3, 0 ≤ y ≤ 2.$ 
I derived the determinant of the corresponding Hessian matrix and it turn out to be negative. If I'm not wrong that means the max-min values occur at the boundary. I'm not sure how to get there. All I can think of is substituting the corner point, but that is obviously not sufficient. So what do I do? Do I parameterise something? If yes, how?
 A: By a simple manipulation (by completing the square) you can see that
$$f(x,y)=y^2-\left(x-\frac12\right)^2+\frac14.$$
From this it is relatively easy to visualize the graph of function - we see that it is a hyperbolic paraboloid. 
And using this expression we can also find the extrema.
If we consider the function on $\mathbb R^2$ it does not have any extrema. 
If we consider it on the given rectangle:


*

*To maximize it we need $y^2$ to be as large as possible and $(x-1/2)^2$ as small as possible. So the maximum is attained at the point $(\frac12,2)$.

*To minimize it we need $y^2$ to be as small as possible and $(x-1/2)^2$ as large as possible. So the maximum is attained at the point $(3,0)$.



Back to your solution. You have found out that the function has only one critical point and determined that this point is not a local extremum. (In fact, it is a saddle point.)
So now you know that the function has no extrema, if you consider it as a function on $\mathbb R^2$. 
But you want to study this function only on the given rectangle, so you need to have a look also on the behavior of the function on the boundary.  Luckily, the boundary is in this case very easily parametrized. So you only have to study the function $f(x,0)$, $f(x,2)$, $f(0,y)$, $f(3,y)$. Each of these four functions is function of one variable on a compact interval. You find their maxima/minima. Then you choose the largest/smallest one from them. And that's all. 
