 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/ib/paper-3/19h/c/solution

Codex  0 Created 2026-09-24 Updated 2026-09-25

Put

y_{i} = 1 - x_{i}

. Feasibility gives

y_{i} \geq 0

and

y_{1} + y_{2} + y_{3} \geq \frac{1}{2},

while the objective is

x_{1} + 2 x_{2} + 4 x_{3} = 7 - (y_{1} + 2 y_{2} + 4 y_{3}) .

To maximize it, the required total deficit

1/2

must be assigned entirely to the cheapest variable

y_{1}

. Thus

y_{1} = \frac{1}{2}, y_{2} = y_{3} = 0,

and the unique optimizer is

x^{*} = (\frac{1}{2}, 1, 1), x_{1}^{*} + 2 x_{2}^{*} + 4 x_{3}^{*} = \frac{13}{2} .

This is the vertex

A

from part (b).

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