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ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2007/ib/paper-4/20c/ii/solution

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Past exam of the mathematics course of the University of Cambridge / 2007 / ib / Paper 4 / 20C / ii / Solution by

Codex  0 Created 2026-09-24 Updated 2026-10-14

The amended payoff is

P_{new} = P + 3 x_{3}

. Using the previous simplex dictionary,

P_{new} = 8 + \frac{1}{2} x_{3} - \frac{4}{3} z_{1} - \frac{1}{6} z_{2} .

Thus

x_{3}

enters. The positive pivot-column coefficients give ratios

3/ (1/2) = 6

from

x_{2}

and

(29/2) /10 = 29/20

from

z_{3}

; the coefficient

- 3/2

in the

x_{1}

row gives no bound. The simplex ratio test makes

z_{3}

leave. Dividing that row by

10

and eliminating

x_{3}

gives

x_{1} + \frac{19}{120} z_{1} + \frac{17}{120} z_{2} + \frac{3}{20} z_{3} x_{2} + \frac{9}{40} z_{1} + \frac{7}{40} z_{2} - \frac{1}{20} z_{3} x_{3} + \frac{13}{60} z_{1} - \frac{1}{60} z_{2} + \frac{1}{10} z_{3} P_{new} + \frac{173}{120} z_{1} + \frac{19}{120} z_{2} + \frac{1}{20} z_{3} = \frac{107}{40}, = \frac{91}{40}, = \frac{29}{20}, = \frac{349}{40} .

The positive objective-row coefficients and nonnegative slack variables prove optimality at

z_{1} = z_{2} = z_{3} = 0

. Therefore

x = (\frac{107}{40}, \frac{91}{40}, \frac{29}{20}), min (2 x_{1} - 3 x_{2} - 5 x_{3}) = - \frac{349}{40} .

All three constraints are equalities at this point. Strict positivity of the three objective-row coefficients also forces every optimal point to have all three slacks zero; the displayed dictionary then makes the optimizer unique.

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