Past exam of the mathematics course of the University of Cambridge / 2015 / iii / Paper 38 / 1 / d

For any feasible $x,y$, increasing $z$ lowers the objective, so the additional upper bound makes $z^{\ast }=0$. Feasibility then requires $2x-y\le 2$. The lowest feasible value of $y$ on the circle occurs at the lower intersection with $2x-y=2$. Solving gives the candidateA global certificate avoids relying on the circle sketch. Introduce Lagrange multipliers for $g\le 0$, $z\le 0$ and $h=0$:ChooseThe coefficient of $z$ vanishes, and $2\mu x^{\ast }+2\lambda =0$, $2\mu y^{\ast }+3-\lambda =0$. Therefore completing the square yieldsThe candidate minimizes $L$ globally and satisfies both active inequality constraints, so complementary slackness and the Lagrangian sufficiency theorem give