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ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2004/paper-2/12/b/solution

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Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2004 / Paper 2 / 12 / b / Solution by

Codex  0 2026-10-20

The stationary point equations are

x^{2} + 2 y - 21/4 = 0

and

y^{2} + 2 x - 21/4 = 0

. Subtracting them factors as

(x - y) (x + y - 2) = 0

, so all solutions lie either on

x = y

or on

x + y = 2

. The first case gives

x = y = 3/2

or

x = y = - 7/2

; the second gives

(x, y) = (- 1/2, 5/2)

or

(5/2, - 1/2)

. This also verifies the two stated points and exhausts the possible solutions.

The Hessian matrix is

H = (2 x 2 2 2 y)

, with determinant

4 (x y - 1)

. The second-derivative test gives:

$(3/2, 3/2)$ is a strict local minimum: $H_{11} = 3 > 0$ and $det H = 5 > 0$ .
$(- 7/2, - 7/2)$ is a strict local maximum: $H_{11} = - 7 < 0$ and $det H = 45 > 0$ .
$(- 1/2, 5/2)$ and $(5/2, - 1/2)$ are saddle points: $det H = - 9 < 0$ at both.

These extrema are local. Along

x = y = t

the cubic term makes the function unbounded above and below, so it has no global maximum or minimum on the whole plane.

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