 Solution

ID: past-exam-of-the-year-1-of-the-computer-science-course-of-the-university-of-oxford/2019/discrete-mathematics-continuous-mathematics-and-probability-resit/4/c/solution

Let

h (x, y) = 2 y^{3} + x y^{2} - y^{2} - 4 x

, so

g = e^{h} > 0

. Its gradient is

\nabla g = g \nabla h

, and

\nabla h = (y^{2} - 4 2 y (3 y + x - 1)) .

y = \pm 2

and

x = 1 - 3 y

, giving $(- 5, 2)$ and $(7, - 2)$ . To classify them, use

H_{g} = g (H_{h} + \nabla h \nabla h^{T}), H_{h} = (0 2 y 2 y 12 y + 2 x - 2) .

\nabla h = 0

and

12 y + 2 x - 2 = 6 y

. Hence

det H_{g} = g^{2} (- 4 y^{2}) = - 16 g^{2} < 0

. The Hessian has one positive and one negative eigenvalue, so both are saddle points. There are no other stationary points.

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