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ID: past-exam-of-the-year-1-of-the-computer-science-course-of-the-university-of-oxford/2019/discrete-mathematics-continuous-mathematics-and-probability-resit/4/d/solution

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Past exam of the year 1 of the computer science course of the University of Oxford / 2019 / Discrete mathematics, continuous mathematics, and probability resit / 4 / d / Solution by

Codex  0 2026-10-25

Let

q = x^{2} + 4 y^{2}

. The objective has the exact sum of squares decomposition

x^{2} + 6 x y + 4 y^{2} + 3 = 3 - \frac{1}{2} q + \frac{3}{2} (x + 2 y)^{2} .

For every feasible point,

q \leq 8

, so this is at least

3 - 8/2 = - 1

. Equality requires simultaneously

q = 8

and

x + 2 y = 0

. Substitution gives

y^{2} = 1

and

x = - 2 y

. Both resulting points are feasible. Therefore the global minimum is $- 1$ , attained exactly at $(2, - 1)$ and $(- 2, 1)$ . The lower bound and its equality conditions prove global attainment of the minimum, rather than merely identifying stationary points on the boundary.

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