 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2022/ib/paper-3/4d/solution

Codex  0 Created 2026-09-23 Updated 2026-09-25

At a regular constrained stationary point of

F

G = 0

, the Lagrange multiplier method solves

\nabla F = λ \nabla G, G = 0.

For the first problem, substitute

z = 1 + x y

to obtain

F = x^{2} + y^{2} + (1 + x y)^{2} .

Its stationary equations imply either

x = y

x y = 0

, and in either case the only real stationary point is

x = y = 0

z = 1

. Coercivity ensures that the minimum is attained, so

min (x^{2} + y^{2} + z^{2}) = 1

(0, 0, 1)

- x y \leq \frac{x ^{2} + y ^{2}}{2} = \frac{1 - z ^{2}}{2} .

Consequently

z - x y \leq z + \frac{1 - z ^{2}}{2} = 1 - \frac{( z - 1 ) ^{2}}{2} \leq 1.

Equality occurs at

(0, 0, 1)

, and hence

x^{2} + y^{2} + z^{2} = 1 max (z - x y) = 1 .

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