 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-oxford/2017/prelims/mathematics-iii/3/b/solution

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

Substitute

x = - y /2

to obtain

y (3 y - 1/2) = 0

. Thus the only critical points are

(0, 0)

and

(- 1/12, 1/6)

H = (21 1 6 y), D = 12 y - 1.

(0, 0)

D = - 1

, so this is a saddle point. At

(- 1/12, 1/6)

D = 1

and

f_{xx} = 2 > 0

, so this is a strict local minimum. Therefore

(0, 0) : saddle; (- 1/12, 1/6) : strict local minimum .

The value at the local minimum is

- 1/432

; there is no local maximum.

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