Past exam of the mathematics course of the University of Oxford / 2023 / Prelims / Mathematics III / 3 / c

Assume $f,F$ are continuously differentiable near the point and $F_z\ne 0$. The implicit function theorem supplies a local function $z(x,y)$ withAt an interior extremum of $g(x,y)=f(x,y,z(x,y))$, the multivariable chain rule givesMultiply these identities by $F_y$ and $F_x$, respectively, and subtract. The terms containing $f_z$ cancel, givingNo assumption $f_z\ne 0$ is needed for this elimination; dividing by $f_z$ would unnecessarily exclude valid cases. If $F_z=0$, this particular local graph argument is unavailable and a different chart or the full Lagrange multiplier condition is needed.

On the constraint $x^2-z^2=1$, substitute $x^2=1+z^2$ to obtainTherefore the two global minima areEvery pair $(y,z)$ is allowed on either smooth branch $x=\pm \sqrt{1+z^2}$, so these are also the only local extrema. There is no local maximum or global maximum: the displayed positive quadratic in $(y,z)$ has no other stationary point and is unbounded above. At the minima $F_z=-2z=1$, consistent with the hypothesis in part i.