 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2018/paper-1/14/c/solution

Set both components of the gradient to zero. After dividing out the nonzero exponential, they are equivalent to

x (1 - 4 x y) + 2 y = 0, y (1 - 4 x y) + 2 x = 0.

Adding and subtracting gives

(x + y) (3 - 4 x y) = 0

and

(x - y) (1 + 4 x y) = 0

. If

x = y

, the first equation reduces to

x (3 - 4 x^{2}) = 0

, giving the origin and the two diagonal points. If

x = - y

, it reduces to

x (4 x^{2} - 1) = 0

, giving the origin and the two opposite-diagonal points. The alternative simultaneous values

x y = 3/4

and

x y = - 1/4

are incompatible, so these cases exhaust the possibilities. Thus the five stationary points are

(0, 0), (\frac{3}{2}, \frac{3}{2}), (- \frac{3}{2}, - \frac{3}{2}), (\frac{1}{2}, - \frac{1}{2}), (- \frac{1}{2}, \frac{1}{2}) .

They are marked in both panels of the figure.

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