 Solution

ID: past-exam-of-the-year-1-of-the-computer-science-course-of-the-university-of-oxford/2017/discrete-mathematics-continuous-mathematics-and-probability/4/a/solution

f_{x} = 3 x^{2} + 2 x y - 10 x - 9 = 0, f_{y} = x^{2} - 9 = 0.

Thus

x = 3

gives

y = 2

, while

x = - 3

gives

y = 8

H = (6 x + 2 y - 10 2 x 2 x 0), det H = - 4 x^{2} .

- 36 < 0

, so the Hessian is indefinite and the second-derivative test classifies both as saddle points. Direct substitution gives

(3, 2), (- 3, 8) are saddle points, and f (3, 2) = f (- 3, 8) = - 30.

f + 30 = (x^{2} - 9) (x + y - 5)

also shows the crossing level-set branches through each saddle.

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