Find absolute maximum and minimum with domain Find absolute maximum and minimum of the function
$f(x,y)=3-x^2+y^2$
on the region
$R = \{(x,y):1≥x≥0, 2≥y≥0\}$
I found that the gradient is 
$∇f(x,y)=(2x,2y)$
and that the critical point inside the domain is (0,0) and it is a local minimum. I know I have to check the boundary, and I know how to do it for a fixed equation like $x^2+y^2=1$ but not for the domain I'm given. Can someone explain how this type of problem is to be solved? I don't seem to understand this problem logically or intuitively.
 A: Hint: border of $R$ has 4 parts, $[0,1]\times \{0\}$, $[0,1]\times \{2\}$, $\{0\}\times [0,2]$, $\{1\}\times [0,2]$.
Moreover you can solve this without gradient: note that $f$ is decreasing in $x$ and increasing in $y$ in your domain...
A: The boundary is:
$\{(x,0): 0\le x\le 1\}$ and $\{(x,1): 0\le x\le 1\}$; 
$\{(0,y): 0\le y\le 2\}$ and  $\{(1,y): 0\le y\le 2\}$

A: As Paul said, the boundary consists of four line segments:
$$
L_1: x=0,2\ge y\ge 0
$$
$$
L_2: x=1,2\ge y\ge 0
$$
$$
L_3: y=0,1\ge x\ge 0
$$
$$
L_4: y=2,1\ge x\ge 0
$$
Let's look at $L_1$, the left border. To find the maxima and minima of $f$ on $L_1$, you use the equation $x=0$ to think of $f(x,y)$ as a single variable function, $f(0,y)$. To find the critical points on $L_1$, take the derivative with respect to $y$:
$$
f'(0,y)=\frac{d}{dy}(3-x^2+y^2)=2y
$$
Since $2y$ is only zero when $y$ is zero, the only critical point is at $y=0$. The max and min of $f$ on $L_1$ are either at critical points or the endpoints $(0,0)$ and $(0,2)$, so max is $f(0,2)=7$ and the min is $f(0,0)=3$. 
Now, check for the max and min of $f$ on all other boundaries. The biggest of these maxes will be the global max, similarly for min. As Milly pointed out, there other clever observations you can make to simplify this, but this method is the most general.
A: It is not necessary to use derivatives ad all... just notice that your function
$f(x,y)$ apart from a constant and a sign, is the square distance from the origin. So you are looking for the points of your domain which are closer, farther from the origin...
