 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2021/paper-2/12/c/solution

f = c

can be rearranged, without changing its domain, as

(c + 2) x^{2} + c y^{2} - 2 y + c = 0.

The requested special contours are therefore

f = 0 : y = x^{2}; f = 1 : (x, y) = (0, 1); f = - 1 : y + 1 = \pm x .

The first is a parabola, the second a single point, and the last two intersecting straight lines. Moreover

1 - f = \frac{3 x ^{2} + ( y - 1 ) ^{2}}{D} \geq 0,

with equality only at

(0, 1)

. Hence

(0, 1)

(0, - 1)

the crossing critical contour separates sectors with values above and below

- 1

H_{f} (0, 1) = diag (- 3, - 1)

is negative definite, while

H_{f} (0, - 1) = diag (- 1, 1)

is indefinite. There is no stationary minimum.

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