Answer
Absolute minimum: $f(0,0)=0$
Absolute maximum: $f(4,0)=4$
Work Step by Step
To calculate the critical points, we have to first put $f_x(x,y), f_y(x,y)$ equal to $0$.
Thus, $f_x=1-y,f_y=1-x$
This yields, $x=1,y=1$ and $f(1,1)=1$
Thus, $f_{xx}=f_{yy}=0$
This yields, $x=0,y=0$
From the vertices of the triangle, after simplifications, we have
$f(x,y)=x+y-xy$
$=x+(-\dfrac{1}{2}x+2)-x(-\dfrac{1}{2}x+2)$
$=x-\dfrac{1}{2}x+2+\dfrac{1}{2}x^2-2x$
$=\dfrac{1}{2}x^2-\dfrac{3}{2}x+2$
Therefore, it has a minimum and maximum value at $x=0,4$
Hence,
Absolute minimum: $f(0,0)=0$
Absolute maximum: $f(4,0)=4$
