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

The amended payoff is $P_{\mathrm{new}}=P+3x_3$. Using the previous simplex dictionary,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 $x_3$ givesThe positive objective-row coefficients and nonnegative slack variables prove optimality at $z_1=z_2=z_3=0$. ThereforeAll 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.