 Solution

ID: past-exam-of-the-year-1-of-the-computer-science-course-of-the-university-of-oxford/2022/discrete-mathematics-continuous-mathematics-and-probability/continuous-mathematics/6/c/solution

For the equality-constrained minimization problem,

y = 8/ x

and

x \neq = 0

. Put

t = x^{2} > 0

. The restricted objective is

f (x, 8/ x) = t^{2} + \frac{128}{t} - 28 =: F (t), F^{'} (t) = 2 t - \frac{128}{t ^{2}} .

The only zero has

t^{3} = 64

, so

t = 4

. Since

F^{''} (t) = 2 + 256/ t^{3} > 0

t

. Consequently the equality constraint has minimizers $(2, 4)$ and $(- 2, - 4)$ , with minimum $20$ .

For the inequality-constrained minimization problem, the identity

f = (x^{2} - 1)^{2} + 2 (y - x)^{2} + 3

supplies the universal lower bound

3

. Both equality points for that bound,

(1, 1)

and

(- 1, - 1)

, satisfy

x y = 1 \leq 8

. Hence the inequality constraint has exactly these two minimizers, with minimum $3$ .

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