Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2016 / Paper 1 / 12 / b

The first partial derivatives factor asIf either coordinate vanishes, the other equation forces the other coordinate to vanish as well. If both are nonzero, subtracting the two bracket equations gives $x^2=y^2$, and then $x^2=y^2=1$. Thus the complete list of stationary points is $(0,0)$ and the four points $(\pm 1,\pm 1)$.

The Hessian matrix isAt the origin it is $8I$, which is a positive-definite matrix. At the four other points its eigenvalues are $32$ and $-16$, independently of the signs. The second-derivative test therefore givesThere are no local maxima. The minimum at the origin is not global, since $g(t,t)=8t^2-4t^4\to -\infty$ as $|t|\to \infty$.

Within the requested first-quadrant square, the stationary points visible in the contour plot are the corner minimum $(0,0)$ and the saddle $(1,1)$. Near the origin, $g\simeq 4(x^2+y^2)$, so the low positive contours look like quarter-circles. Along either coordinate axis, $g=4t^2+t^4$ rises monotonically. Along the diagonal, $g(t,t)=8t^2-4t^4$ rises to $4$ at $t=1$, returns to zero at $t=\sqrt{2}$, then becomes negative.

The saddle contour $g=4$ crosses at $(1,1)$ and separates the central positive contours from the outer diagonal region. Negative contours occur toward $(2,2)$, where $g=-32$, while the axis corners $(2,0)$ and $(0,2)$ have $g=32$. These checks determine both the contour ordering and the change of connectivity at the saddle; the figure shows these level sets.