Question: Let f (x, y) = x 2 + y 2 4xy. Show that f does not have a relative minimum at its critical point

Let f (x, y) = x² + y² − 4xy. Show that f does not have a relative minimum at its critical point (0, 0), even though it does have a relative minimum at (0, 0) in both the x and y directions.

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So 0 0 is a critical point and 0 0 is a saddle point The above is true but not asked for I... View full answer

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