Past exam of the mathematics course of the University of Cambridge / 2026 / ib / Paper 1 / 7H (Optimisation)

Replace the quadratic constraint by the two linear inequalities $-2\le x_1-x_2+x_3\le 2$. Simplex pivots lead to the active constraints $x_1=0$, $-x_2+x_3=2$, and $3x_2+x_3=4$. Thus $x^{\ast }=(0,1/2,5/2)$ and the value is $3$. With right-hand side $4+\epsilon$, the same basis remains optimal for small nonzero $\epsilon$, giving $x^{\ast }=(0,1/2+\epsilon /4,5/2+\epsilon /4)$.