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The value of the function is here f (0, 0) = 0, and it is obvious that f (x, y) can be both positive and negative in any neighbourhood of (0, 0), so there exists no point in which a maximum or a minimum can be attained. com 33 Calculus 2c-4 Maximum and Minimum 4) It follows from the rules of magnitudes that f (x, y) = exp(x2 + y 2 ) − 4xy → +∞ for x2 + y 2 → +∞, and the function does not have a maximum. The possible stationary points are the solutions of the equations ∂f ∂x ∂f ∂y = 2x exp(x2 + y 2 ) − 4y = 0, = 2y exp(x2 + y 2 ) − 4x = 0.

1) Stationary points. The equations of the stationary points are ∂f = 9x2 + 6y 2 − 18x = 0, ∂x ∂f = 12xy + 12y 2 = 12(x + y)y = 0. ∂y The line x + y = 0 does not intersect the interior of A, so we only get the possibility y = 0. If we put this into the former equation we get 0 = 9x2 + 0 − 18x = 9x(x − 2), hence x = 0 or x = 2, corresponding to the stationary points (0, 0) and (2, 0). Only (2, 0) lies in A. Then compute the value of the function at this point, f (2, 0) = 3 · 23 + 0 + 0 − 9 · 22 = 24 − 36 = −12.

X2 + y 2 − 2y − 3)2 (x2 y2 Inside the domain, these equations are reduced to x{(y − 1)2 − 5} = 0 and y − 1 = 0, hence y = 1 and x = 0. The only stationary point is the centre of the circle (0, 1). This must necessarily be a maximum, because the maximum exists, and it can only be attained at a stationary point (because the function is of class C ∞ ). Hence the maximum is 1 S = f (0, 1) = − . 4 Remark 1. One can also ﬁnd the maximum in the following way without any calculation. Note that the numerator is positive, and the denominator is negative everywhere in the domain.