In order for the parallelogram to be a
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Looks like .
- If
, then
- critical point at (0, 0).
- If
, then
- two critical points at
and
The latter two critical points occur outside of since
so we ignore those points.
The Hessian matrix for this function is
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
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
and we get the same extrema as in the previous case: 8 at (-1, 1), and 10 at (-1, -1).
- If
which doesn't tell us about anything we don't already know (namely that 8 is an extreme value).
- If
, then
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).