Past exam of the year 1 of the computer science course of the University of Oxford / 2018 / Discrete mathematics, continuous mathematics, and probability / 5

The coordinate sum is $2\sin t$, while the sum of their cubes is $6\sin t-4{\sin }^{3}t$. Along the parametrized curve, the composite function is therefore $-3\sin t/2-{\sin }^{3}t/3$. Differentiate, or apply the multivariable chain rule, to obtain

Apply Newton's method to $F(t)=1-3\sin t/2-{\sin }^{3}t/3$. The derivative in part (a) givesChoose a starting point near a desired root, such as $t_0=0.6$ for the root in $(0,\pi /2)$; do not divide when $\cos t_k=0$. On $(0,\pi /2)$, $F$ is strictly decreasing and changes sign between its endpoints, so the intermediate value theorem gives exactly one root there. Its derivative is nonzero, giving the usual local convergence of Newton's method. The periodic function has other roots; the iteration does not impose global uniqueness or convergence from every starting value.

The gradient vanishes exactly when $x^2/4-1=0$ and $y^2/4-1=0$, giving the four points $(\pm 2,\pm 2)$. The Hessian matrix is $\mathrm{diag}(x/2,y/2)$. It is positive definite at $(2,2)$, negative definite at $(-2,-2)$, and indefinite at the two mixed-sign points. Thus $(2,2)$ is a strict local minimum, $(-2,-2)$ a strict local maximum, and $(2,-2),(-2,2)$ are saddle points by the second-derivative test. Their respective function values are $-8/3$, $8/3$, and zero at both saddle points.

Substituting $y=-x$ cancels both the cubic and linear terms, leaving $f(x,-x)=0$ for every real $x$. Thus the constrained maximum and minimum are both zero, attained everywhere on the constraint line. The feasible set is unbounded, but the restricted function is constant; compactness is not needed here.

Write $f(x,y)=g(x)+g(y)$ with $g(s)=s^3/12-s$. On $[0,4]$, its derivative is $g^{\prime }(s)=s^2/4-1$, negative below two and positive above two. Compare the critical and endpoint values:Each coordinate can be chosen independently. Therefore the global minimum is $-8/3$ at $(2,2)$, and the global maximum is $8/3$ at $(4,4)$. This comparison includes the boundary of the rectangle, rather than classifying only interior stationary points.

The continuous function attains a global maximum and minimum on the closed disc by the extreme value theorem. All four unconstrained stationary points have squared distance eight, so none lies in its interior. The extrema must therefore lie on $x^2+y^2=6$.

The constraint gradient is nonzero on that circle, so Lagrange multipliers apply:Subtracting gives $(x-y)((x+y)/4-2\lambda )=0$. If $x=y$, the constraint gives $x=y=\pm \sqrt{3}$. Otherwise $2\lambda =(x+y)/4$; substitution into either equation gives $xy=-4$. This is impossible, since $|xy|\le (x^2+y^2)/2=3$ on the circle. Thus the diagonal points are the only candidates. Comparing their values gives