 Solution

ID: past-exam-of-the-year-1-of-the-computer-science-course-of-the-university-of-oxford/2019/discrete-mathematics-continuous-mathematics-and-probability/6/a/solution

L = x^{4} + 4 y^{4} - λ (x^{2} - y^{2} - 1) .

4 x^{3} - 2 λ x = 0, 16 y^{3} + 2 λ y = 0, x^{2} - y^{2} = 1.

Feasibility makes

x \neq = 0

, so

λ = 2 x^{2} > 0

. The second equation becomes

2 y (8 y^{2} + λ) = 0

y = 0

and then

x = \pm 1

. The constraint gradient

(2 x, - 2 y)

is nonzero on the feasible set, so the multiplier calculation has no missing singular feasible point.

To verify the global minimum rather than just stationarity, eliminate

x^{2} = 1 + y^{2}

x^{4} + 4 y^{4} = 1 + 2 y^{2} + 5 y^{4} \geq 1.

Equality holds exactly at $(x, y) = (1, 0)$ or $(- 1, 0)$ , with minimum value $1$ .

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