Past exam of the mathematics course of the University of Oxford / 2015 / Prelims / Mathematics III / 3

The gradient of the given polynomial isIts second component vanishes when $y=0$ or $x=-1$. The first component then gives exactly three critical points:The Hessian matrix and its determinant areAt $(-1/2,0)$ the Hessian matrix is $\mathrm{diag}(2,1)$, whose positive definiteness proves a strict local minimum by the second-derivative test. At $(-1,\pm 1)$ its determinant is $-4$, so both points are saddle points of a scalar function. Their values and classifications areThere are no local maxima.

At a regular point of the optimization constraint, meaning $\nabla g\ne 0$, the necessary Lagrange multiplier condition isIndeed, every allowed tangent vector $t$ to the level set $g=0$ satisfies $\nabla g\cdot t=0$. At a constrained extremum, the directional derivative $\nabla f\cdot t$ must vanish too. Thus both gradients are normal to the same tangent line and are parallel. This includes $\nabla f=0$. At a singular point with $\nabla g=0$, the displayed condition need not be necessary; those points require a separate examination.

For $g=(x-1{)}^2+y^2-29$, the gradient $\nabla g=2(x-1,y)$ never vanishes on the circle. The Lagrange multiplier equations areIf $y=0$, the two possibilities are $(1\pm \sqrt{29},0)$. If $y\ne 0$, then $\lambda =1+x$, and the first equation gives $y^2=2x^2-2x-3$. Combining this with $y^2=28+2x-x^2$ yieldsPutBoth $h_{-}$ and $h_{+}$ are positive, giving the other four points $(r_{-},\pm h_{-})$ and $(r_{+},\pm h_{+})$. To classify all six, restrict $f$ to the circle by eliminating $y^2$:Its derivatives are $H^{\prime }=-3x^2+4x+31$ and $H^{\prime \prime }=4-6x$. In the interior, $H^{\prime \prime }(r_{-})=2s>0$ and $H^{\prime \prime }(r_{+})=-2s<0$. At either endpoint $H^{\prime }<0$; along the circle, $x$ has a strict local minimum at the left endpoint and a strict local maximum at the right endpoint. Consequently the six constrained extrema areThe circle is a compact set, so the extreme value theorem guarantees that its global minimum and global maximum occur among these six points. Their values give the corrected sharp boundThe printed lower bound $-20$ is false. A particularly short counterexample isThe sole local minimum in part (a), $(-1/2,0)$, has value $-1/4$, larger than values on this circle; it is therefore not a global minimum. In fact $f(-2,y)=2-y^2\to -\infty$, so $f$ has no global minimum on the plane.