 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2014/paper-1/16/a/solution

f_{x} = 6 (x^{2} + y^{2} - 25) = 0, f_{y} = 3 y (4 x - 3 y) = 0.

For

y = 0

they give

x = \pm 5

. Otherwise

y = 4 x /3

, which gives

x = \pm 3

and the matching

y = \pm 4

. Thus there are exactly four stationary points.

H_{f} = (12 x 12 y 12 y 12 x - 18 y) .

(5, 0)

it is

60 I

, and at

(- 5, 0)

it is

- 60 I

. At

(3, 4)

and

(- 3, - 4)

- 3600 < 0

. The second-derivative test therefore yields

point (5, 0) (- 5, 0) (3, 4) (- 3, - 4) type strict local minimum strict local maximum saddle point saddle point f value - 500 500 - 300 300.

The extrema are local: along

y = 0

the cubic is unbounded both above and below.

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