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ID: past-exam-of-the-mathematical-methods-ii-course-of-the-university-of-oxford/2016-resit/12/d/solution

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Past exam of the mathematical methods II course of the University of Oxford / 2016 resit / 12 / d / Solution by

Codex  0 2026-10-25

The gradient components are

f_{x} = 2 x (1 + 6 x^{2})

and

f_{y} = 2 y (1 - 6 y^{2})

. Their real simultaneous zeros are

(0, 0)

and

(0, \pm 1/ 6)

. The Hessian matrix is diagonal:

H = diag (2 + 36 x^{2}, 2 - 36 y^{2}) .

At the origin it is positive definite; at each other critical point it is

diag (2, - 4)

and indefinite. Hence

(0, 0) : strict local minimum, f = 0; (0, \pm 1/ 6) : saddle points, f = 1/12.

There are no local maxima. The contours are symmetric under either coordinate reflection. Small positive levels have a closed component surrounding the minimum; these meet the saddle-level separatrices at level

1/12

. That critical contour has the useful exact form

3 x^{4} + x^{2} = 3 (y^{2} - 1/6)^{2}

. Other components extend toward large

∣ y ∣

, where the

- 3 y^{4}

term makes the function unbounded below.

Figure 1.
Contours near a minimum and two saddle points, including the critical level one twelfth
. Details.

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