 Solution

ID: past-exam-of-the-mathematical-methods-ii-course-of-the-university-of-oxford/2014-resit/6/b/solution

(2 x + 2, - 2 y - 4)

, so the only stationary point is

(x, y) = (- 1, - 2)

diag (2, - 2)

and its determinant is

- 4 < 0

. Thus

(x, y) = (- 1, - 2), f = 13, saddle point .

Completing squares confirms the classification directly:

f = (x + 1)^{2} - (y + 2)^{2} + 13

. Arbitrarily close to the point, moving in the

x

direction increases the value and moving in the

y

direction decreases it.

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