Past exam of the year 1 of the computer science course of the University of Oxford / 2019 / Discrete mathematics, continuous mathematics, and probability / 6 / a / Solution

Use the LagrangianThe Lagrange multiplier equations areFeasibility makes $x\ne 0$, so $\lambda =2x^2>0$. The second equation becomes $2y(8y^2+\lambda )=0$, forcing $y=0$ and then $x=\pm 1$. The constraint gradient $(2x,-2y)$ is nonzero on the feasible set, so the multiplier calculation has no missing singular feasible point.

To verify the global minimum rather than just stationarity, eliminate $x^2=1+y^2$:Equality holds exactly at $(x,y)=(1,0)$ or $(-1,0)$, with minimum value $1$.