Non-strict extremum of $z = (x-y+1)^2$ What's the proper way to find the minima and maxima of this function:
$$z = (x-y+1)^2$$
Well, in Russian books it is claimed that such a function has non-strict extrema, but how should I handle it? Should I rewrite it like 
$$z=(x-y+1)(x-y+1)$$
and then get rid of brackets and then use
$$AC-B^2$$ 
finding appropriate $A$ $B$ and $C$ before
or easier way?
 A: You don't really need partial derivatives.
One clearly has $z=\left( x-y+1\right)^2 \ge 0$, with the equality (and thus minimal value):
$$\left( x-y+1\right)^2 = 0 \iff x-y+1 = 0$$
This is the equation of a line in the $xy$-plane; for all points $(x,y)$ on this line, you have $f(x,y) = 0$. In all other points, you have $f(x,y) > 0$. Note that there is no maximum since you can make $x-y+1$, and thus also its square, as large as you want.

Alternatively, consider the function $f(t) = t^2$ of one real variable, which has no maximum since $t^2$ becomes arbitrarily large if $t$ is arbitrarily large and with a minimum in $t=0$, then $f(0)=0$.
In your case, you have $z=f(x,y) = \left( x-y+1\right)^2$ or, with $t=x-y+1$, also $z=f(t)$, so:


*

*no maximum since you can make $t=x-y+1$ arbitrarily large;

*a minimum where $t = 0 \iff x-y+1=0$.



If you would prefer using partial derivatives, you're looking for the critical points $(x,y)$ where both partial derivatives are equal to $0$, so with $f(x,y) = \left( x-y+1\right)^2$, you (again) find:
$$\left\{\begin{array}{l}
\displaystyle \frac{\partial f}{\partial x}(x,y)=0 \\[4pt]
\displaystyle \frac{\partial f}{\partial y}(x,y)=0
\end{array}\right. \iff
\left\{\begin{array}{l}
\displaystyle 2\left( x-y+1\right)=0 \\[2pt]
\displaystyle -2\left( x-y+1\right)=0
\end{array}\right. \iff x-y+1=0$$
The Hessian is $0$ but the nature of the critical points simply follows from arguments as above.
