= Solution
The <stationary point> equations are $x^2+2y-21/4=0$ and $y^2+2x-21/4=0$. Subtracting them factors as $(x-y)(x+y-2)=0$, so all solutions lie either on $x=y$ or on $x+y=2$. The first case gives $x=y=3/2$ or $x=y=-7/2$; the second gives $(x,y)=(-1/2,5/2)$ or $(5/2,-1/2)$. This also verifies the two stated points and exhausts the possible solutions.
The <Hessian matrix> is $H=\begin{pmatrix}2x&2\\2&2y\end{pmatrix}$, with determinant $4(xy-1)$. The <second-derivative test> gives:
* \b[$(3/2,3/2)$ is a strict local minimum:] $H_{11}=3>0$ and $\det H=5>0$.
* \b[$(-7/2,-7/2)$ is a strict local maximum:] $H_{11}=-7<0$ and $\det H=45>0$.
* \b[$(-1/2,5/2)$ and $(5/2,-1/2)$ are saddle points:] $\det H=-9<0$ at both.
These extrema are local. Along $x=y=t$ the cubic term makes the function unbounded above and below, so it has no global maximum or minimum on the whole plane.
Back to article page
