Find the absolute maximum and minimum values off on the set D. f\left(x,

Step 1
For the second step, we need to calculate the minima and maxima values along each boundary of the domain -i.e., along the three edges of the triangle.
Start by solving for the equation of the line segment that connects each point.
${\displaystyle \overline{AB}:y=2-x,0\le x\le 2}$
${\displaystyle \overline{BC}:y=x-2,0\le x\le 2}$
${\displaystyle \overline{AC}:x=0,-2\le y\le 2}$
Step 2
Solve for the values of the function along the boundary ${\displaystyle \overline{AB}}$ in terms of x by substituting ${\displaystyle y=2-x}$ in the original equation and simplifying
Since the result i a quadratic, find the x coordinate of the inflection point along this line by taking the first derivative and solving for 0, then find the corresponding y coordinate.
${\displaystyle f(x,2-x)=x^2+{(2-x)}^2-2x=2x^2-6x+4}$
${\displaystyle f^{\prime }(x,2-x)=4x-6=0\to x=\frac{3}{2},y=\frac{1}{2}}$
Step 3
Calculate the values of ${\displaystyle f(x,y)}$ at the two end points of ${\displaystyle \overline{AB}}$ and at its inflection point.
${\displaystyle f(0,2)=4}$
${\displaystyle f(\frac{3}{2},\frac{1}{2})=-\frac{1}{2}}$
${\displaystyle f(2,0)=0}$
Step 4
Next solve for the values of the function along the boundary ${\displaystyle \overline{BC}}$ by subisifuling ${\displaystyle y=x-2}$ and simplifying (This results in the same equation as for ${\displaystyle \overline{AB}}$ above, since ${\displaystyle {(x-2)}^2={(2-x)}^2}$.) Then find the inflection point as described above.
${\displaystyle =(x,x-2)=x^2+{(x-2)}^2-2x=2x^2-6x+4}$
${\displaystyle f^{\prime }(x,x-2)=4x-6=0\to x=\frac{3}{2},y=-\frac{1}{2}}$
Step 5
Calculate the values of ${\displaystyle f(x,y)}$ at the two end points of ${\displaystyle \overline{BC}}$ and at its inflection point.
${\displaystyle f(0,-2)=4}$
${\displaystyle f(\frac{3}{2},-\frac{1}{2})=-\frac{1}{2}}$
${\displaystyle f(2,0)=0}$ (same as above)
Step 6
Finally, solve for the equation of the line segment ${\displaystyle \overline{AC}}$. To do this, substitute ${\displaystyle y=0}$ in the original equation. This has an inflection point (a minimum value) at ${\displaystyle (0,0)}$.
${\displaystyle f(0,y)=y^2}$
Step 7
Determine the values of function at the two end points along the boundary ${\displaystyle \overline{AC}}$ and at this inflection point. Since we've already calculated the values at the end points ${\displaystyle (0,2)}$ and ${\displaystyle (0,-2)}$ above, there's no need to do so again.