Past exam of the mathematics course of the University of Cambridge / 2009 / ib / Paper 4 / 5B / Solution

At a regular constrained stationary point, $\nabla g\ne 0$ and the directional derivative of $f$ vanishes along the tangent to $g=0$. Therefore $\nabla f=\lambda \nabla g$ for a Lagrange multiplier $\lambda$. Solve this equation together with the constraint, then compare candidate values and any exceptional nonregular points. Proportionality of the two gradients givesIf $\nabla g=0$ at an exceptional point, this determinant vanishes trivially, although the regular multiplier method need not identify every such point.

For the given ellipse, maximize $r^2=x^2+y^2$. The constraint is $r^2+xy=4$, and $xy\ge -r^2/2$, so $4\ge r^2/2$. Equality requires $x=-y$ and the constraint then gives $(x,y)=(2,-2)$ or $(-2,2)$. ThusThe constraint gradient vanishes only at the origin, which is not on this ellipse, so the regularity condition holds everywhere relevant.