Finding extrema of $z(x,y)$ given some conditions

Question

I am given the following exercise and I am not sure I understand what it asks:

Find the extrema for $z=2x^2-y^2$, given $x+y=2$.

This part "given $x+y=2$" is confusing. I know that if any point is extremum, then partial derivatives should be zero. In this case: $$\begin{align} \partial_x z &= 4x\\ \partial_y z &= -2y \end{align}$$

This gives $x = 0$ and $y = 0$. So, what is the point of the condition given?

Answer 1

The purpose of given constraint is to scope down the range of $(x,y)$. In your case, instead of finding an optimum in $\mathbb{R}^2$, you should find an optimum of $z$ along the line $x + y = 2$.

Normally, I would not do any partial derivative to this problem. We know that $x + y = 2$, hence the problem can be simplified as

$$z = 2x^2 - (2 - x)^2$$

Answer 2

Solution $x=0$ and $y=0$ does not belong to the set {(x,y): x+y = 2} so it it not solution of your exercise.
What is wanted from you is to find $\min_{(x,y) \in M} f(x,y)$ and $\max_{(x,y) \in M} f(x,y)$ where $M= \left\{(x,y): x+y = 2\right\}$ and $f(x,y)=2x^2-y$.

Hint: Given your condition for all $(x,y) \in M$ $$f(x,y)=f(x,2-x).$$