Find min/max values of $f(x,y)=x^3-y^3+3xy$ in the set $K=[0,4]\times[-4,0]$ Find the biggest and the smallest values of the function $f(x,y)=x^3-y^3+3xy$ in the set $K=[0,4]\times[-4,0]$.
So using  partial derivatives we find that the critical points are $(0,0)$ and $(1,-1)$. After that we find that $(1,-1)$ is local minimum and $(0,0)$ is saddle point.
What generally is the procedure of finding the min and max values in the set $K$? As far as I know we should check for min/max values on the bounds of K. Does that mean that we should check the points $(0,0)$, $(0,-4)$, $(4,-4)$ and $(4,0)$ in this case, or what?
 A: I will write about the function
$$f(x,y)=x^3+y^3-3xy$$
for $x,y\in[0,4]$.
This is basically the same problem, it seems a bit more symmetric after the change of variables.

Local extrema:
You correctly found
\begin{align*}
f_x &= 3x^2-3y\\
f_y &=-3x+3y^2
\end{align*}
and identified critical points as $(0,0)$ and $(1,1)$.
We can calculate also the second derivatives: $f_{xx}=6x$, $f_{xy}=-3$, $f_{xy}=6y$. We have
\begin{align*}
D_1&=6x\\
D_2&=\begin{vmatrix}6x&-3\\-3&6y\end{vmatrix}=36xy-9
\end{align*}
We see that for $x=y=1$ we get $D_1>0$ and $D_2>0$. So the corresponding matrix is
positive definite
and this is a
local minimum.
For $x=y=0$ we have $D_1=0$, so we cannot say from this whether it is maximum or minimum. But it is relatively easy to see that it is neither of them, since we can find close to $(0,0)$ points with positive values and points with negative values. (It suffices to check $f(\varepsilon,0)=\varepsilon^3$ for small $\varepsilon$.)
We can also ask Wolfram Alpha to check for local extrema of x^3+y^3-3xy to verify whether our calculations are correct.

Extrema on $K$
But we are interested in the extrema on the set $K$, not in the local extrema of the function $f$.
Since this set is closed and bounded and the function $f$ is continuous, it is guaranteed that the function $f$ attains maximum and minimum on $K$.
If the extrema are in the interior, then they are local extrema. So one candidate is $(1,1)$. We have $f(1,1)=-1$.
Now we check what happens on the boundary. We have $f(0,0)=0$, $f(4,0)=f(0,4)=4^3=64$ and $f(4,4)=4^3+4^3-3\cdot4^2=4^2(4+4-3)=80$.
What happens on the sides of the square?


*

*For $y=0$ we have $f(x,0)=x^3$. This function is clearly increasing for $x\in[0,4]$. So on this side we have minimum for $x=0$ and maximum for $x=4$.

*For $y=4$ we have $f(x,4)=x^3-12x+64$. We can check that this function is decreasing for $x\in[0,2]$ and increasing for $x\in[2,4]$. So the values interesting for us are $f(0,4)=64$, $f(2,4)=48$ and $f(4,4)=80$.

*The other two sides behave symetrically.


So we see that the values attained on the boundary are between $f(0,0)=0$ and $f(4,4)=80$.
The conclusion is that the maximum on $K$ is $f(4,4)=80$ and the minimum is $f(1,1)=-1$.
Again we can check WA: extrema x^3+y^3-3xy for 0<=x<=4, 0<=y<=4.
