Past exam of the year 1 of the computer science course of the University of Oxford / 2017 / Discrete mathematics, continuous mathematics and probability / 4

The gradient equations areThus $x=3$ gives $y=2$, while $x=-3$ gives $y=8$. The Hessian matrix isAt both critical points the determinant is $-36<0$, so the Hessian is indefinite and the second-derivative test classifies both as saddle points. Direct substitution givesThe factorization $f+30=(x^2-9)(x+y-5)$ also shows the crossing level-set branches through each saddle.

Write $\dot{x}=dx/du$, $\ddot{x}=d^{2}x/d{u}^2$, $A_v=dA/dv$, $A_{vv}=d^{2}A/d{v}^2$, and $y_w=dy/dw$, $y_{ww}=d^{2}y/d{w}^2$. Put $S=A+A^T$. The product rule and chain rule giveHere $w^{\prime },w^{\prime \prime }$ are derivatives with respect to $v$. In the first derivative, $x^{T}A\dot{x}={\dot{x}}^T{A}^{T}x$ because the expression is scalar. Likewise ${\dot{x}}^{T}S\dot{x}=2{\dot{x}}^{T}A\dot{x}$ if desired. This differentiation of a parameter-dependent quadratic form does not assume that $A$ is symmetric. The positive linear term $y^{T}b$ contributes the positive chain-rule terms in the $v$ derivatives; it is independent of $u$.

Complete the square:The minimum locus of a squared function occurs where its residual vanishes. Hence the minimum is $-6$ on $y=-x^2/2$, provided these points lie in the ellipse. Setting $t=x^2$ in the constraint gives $t+t^2/2\le 9$, or $0\le t\le \sqrt{19}-1$. ThusFor the maximum, let $q=x^2+2y$. The ellipse givesAlso $q\ge 2y\ge -3\sqrt{2}$, whose absolute value is less than $19/2$. Therefore $q^2\le (19/2{)}^2$. Equality is attained at $y=1/2$, $x^2=17/2$, givingThese attained bounds are global, not merely stationary-point tests. The minimizing locus is a parabola segment.