Examine each function for relative extrema and saddle points:
f (x,y) = x^2 - 3xy + y^2© BrainMass Inc. brainmass.com December 24, 2021, 7:46 pm ad1c9bdddf
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We are given f(x, y) = x^2 - 3xy + y^2.
f_x = 2x - 3y
f_y = -3x + 2y
so the critical points (a, b) are solutions to
2a - 3b = 0
-3a + 2b = 0
in which case (a, b) = (0, 0) is the only solution.
We also have
f_xx = 2
f_yy = 2
f_xy = -3
D := (f_x)(f_y) - (f_xy)^2 = (2x - 3y)(-3x + 2y) - 9
so that D(0, 0) = - 9 < 0, in which case f does *not* have a relative extremum
at (0, 0).