Looks like
.
![f_x=2x-2xy=0\implies2x(1-y)=0\implies x=0\text{ or }y=1](https://tex.z-dn.net/?f=f_x%3D2x-2xy%3D0%5Cimplies2x%281-y%29%3D0%5Cimplies%20x%3D0%5Ctext%7B%20or%20%7Dy%3D1)
![f_y=2y-x^2=0\implies2y=x^2](https://tex.z-dn.net/?f=f_y%3D2y-x%5E2%3D0%5Cimplies2y%3Dx%5E2)
- If
, then
- critical point at (0, 0). - If
, then
- two critical points at
and ![(\sqrt2,1)](https://tex.z-dn.net/?f=%28%5Csqrt2%2C1%29)
The latter two critical points occur outside of
since
so we ignore those points.
The Hessian matrix for this function is
![H(x,y)=\begin{bmatrix}f_{xx}&f_{xy}\\f_{yx}&f_{yy}\end{bmatrix}=\begin{bmatrix}2-2y&-2x\\-2x&2\end{bmatrix}](https://tex.z-dn.net/?f=H%28x%2Cy%29%3D%5Cbegin%7Bbmatrix%7Df_%7Bxx%7D%26f_%7Bxy%7D%5C%5Cf_%7Byx%7D%26f_%7Byy%7D%5Cend%7Bbmatrix%7D%3D%5Cbegin%7Bbmatrix%7D2-2y%26-2x%5C%5C-2x%262%5Cend%7Bbmatrix%7D)
The value of its determinant at (0, 0) is
, which means a minimum occurs at the point, and we have
.
Now consider each boundary:
- If
, then
![f(1,y)=8-y+y^2=\left(y-\dfrac12\right)^2+\dfrac{31}4](https://tex.z-dn.net/?f=f%281%2Cy%29%3D8-y%2By%5E2%3D%5Cleft%28y-%5Cdfrac12%5Cright%29%5E2%2B%5Cdfrac%7B31%7D4)
which has 3 extreme values over the interval
of 31/4 = 7.75 at the point (1, 1/2); 8 at (1, 1); and 10 at (1, -1).
- If
, then
![f(-1,y)=8-y+y^2](https://tex.z-dn.net/?f=f%28-1%2Cy%29%3D8-y%2By%5E2)
and we get the same extrema as in the previous case: 8 at (-1, 1), and 10 at (-1, -1).
- If
, then
![f(x,1)=8](https://tex.z-dn.net/?f=f%28x%2C1%29%3D8)
which doesn't tell us about anything we don't already know (namely that 8 is an extreme value).
- If
, then
![f(x,-1)=2x^2+8](https://tex.z-dn.net/?f=f%28x%2C-1%29%3D2x%5E2%2B8)
which has 3 extreme values, but the previous cases already include them.
Hence
has absolute maxima of 10 at the points (1, -1) and (-1, -1) and an absolute minimum of 0 at (0, 0).