Past exam of the year 1 of the computer science course of the University of Oxford / 2022 / Discrete mathematics, continuous mathematics and probability resit / 4 / c

Use $F=x^6+\frac{3}{2}{y}^4$ and $G=x^4-2y^2-3$. Every feasible point has $x\ne 0$, so $\nabla G=(4x^3,-4y)$ cannot vanish. The Lagrange multiplier equations areThe first gives $\lambda =\frac{3}{2}{x}^2>0$. If $y\ne 0$, the second would give $\lambda =-\frac{3}{2}{y}^2<0$, a contradiction. Therefore $y=0$, and the constraint gives $x^4=3$. The only constrained stationary points are $(\pm 3^{1/4},0)$.

For a global certificate, feasibility implies $x^4\ge 3$, and henceEquality requires $y=0$ and $x^4=3$. ThusThe case split is essential: dividing by $y$ would discard precisely the minimizers.