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ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2022/ib/paper-2/7h/solution

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Past exam of the mathematics course of the University of Cambridge / 2022 / ib / Paper 2 / 7H / Solution by

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

The Lagrange sufficiency theorem says that for a convex differentiable objective and convex differentiable inequality constraints, any feasible point satisfying the Karush-Kuhn-Tucker conditions with nonnegative multipliers is a global minimizer.

Write

f = - x_{1} - 3 x_{2}, g_{1} = x_{1}^{2} + x_{2}^{2} - 25, g_{2} = - x_{1} + 2 x_{2} - 5.

(1)

The unconstrained maximizer of

x_{1} + 3 x_{2}

on the disc violates

g_{2} \leq 0

, so both boundaries are active at the optimum. Their intersections are

(- 5, 0) and (3, 4),

(2)

and

(3, 4)

gives the smaller objective.

To certify it, at

x_{*} = (3, 4)

choose

λ = \frac{1}{4}, μ = \frac{1}{2} .

(3)

Then

\nabla f (x_{*}) + λ \nabla g_{1} (x_{*}) + μ \nabla g_{2} (x_{*}) = 0,

(4)

both multipliers are nonnegative, and complementary slackness holds because both constraints are active. The sufficiency theorem therefore gives

x_{1} = 3, x_{2} = 4, f_{m i n} = - 15 .

(5)

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