Let $[-1,1] \times [-1,1]$ the unit square, Im thinking in compute the extreme values at the frontier of the unit circle. So I describe each face of the unit square as:
$$F_{1}= \lbrace (1,y) | -1 \leq y \leq 1 \rbrace\\ F_{2}= \lbrace (x,1) | -1 \leq x \leq 1 \rbrace\\ F_{3}= \lbrace (-1,y) | -1 \leq y \leq 1 \rbrace\\ F_{4}= \lbrace (x,-1) | -1 \leq x \leq 1 \rbrace\\$$
So first question,how I compute the candidates for maximum and minimum values at these face from the unit circle?
Also Im looking in compute the stationary points of $f(x,y)=x+y$. But $\frac{\partial f}{ \partial x}=1$ and $\frac{\partial f}{ \partial y}=1$ which give non information.
I would aprecciate any help to solve this problem.
$\endgroup$ 22 Answers
$\begingroup$Maximum
Fix $y$, then $f(x)=x+y \quad (-1\leq x \leq 1)$ is max at $x=1$.
Next, fix $x=1$ then $f(y)=1+y \quad (-1 \leq y \leq 1)$ is max at $y=1$
So, $f(x,y)=x+y \leq 2$.
Minimum
Fix $y$, then $f(x)=x+y \quad (-1\leq x \leq 1)$ is min at $x=-1$.
Next, fix $x=-1$ then $f(y)=-1+y \quad (-1 \leq y \leq 1)$ is min at $y=-1$
So, $f(x,y)=x+y \geq -2$.
$\endgroup$ $\begingroup$If $(x,y)$ is in the boundary of the unit square, in particular you have $-1\leq x\leq 1$ and $-1\leq y\leq 1$. So $-2\leq x+y\leq 2$. So, $$-2\leq f(x,y)\leq 2$$ And you actually have that $f(1,1)$ is maximum and $f(-1,-1)$ is minimum.
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