Extreme values of a two-variable polynomial Is it possible to find a two-variable polynomial which has only two extreme values on the whole plane, one is a local maximum, another is a local minimum, and the local maximum is less than the local minimum?
 A: Yes, it is possible, here is an example. Pick any odd degree $5$ polynomial $f$ with critical points $\pm x_1$ and $\pm x_2$, where $0<x_1<x_2$, such that $f(x)>0$ for $x>0$. This automatically implies that $f$ has local maxima at $-x_2$ and $x_1$, and local minima at $-x_1$ and $x_2$, with $f(-x_2) < 0 < f(x_2)$. (As an example, $f(x) = x^5 - 2x^3 + 1.1 x$ works, but there are lots of such polynomials.)
Now define a polynomial of two variables by $F(x,y) = f(x) + xy^2$. The partial derivatives are
$$
F_x(x,y) = f'(x) + y^2 \quad \text{ and } \quad F_y(x,y) = 2xy,
$$
so the critical points satisfy $x=0$ or $y=0$. Since $f'(0)>0$, there are no critical points with $y=0$, so the only critical points are solutions to $y=0$ and $f'(x) = 0$, i.e., the critical points of $f$ on the $x$-axis. The Hessian matrix (of second-order partials) at those points is
$$
HF(x,y) = \begin{bmatrix} f''(x) & 0 \\ 0 & 2x \end{bmatrix},
$$
so the second derivative test shows that there is a local maximum at $(-x_2, 0)$, a local minimum at $(x_2,0)$, and saddle points at $(\pm x_1,0)$. Lastly, 
$$F(-x_2,0) = f(-x_2) < 0 < f(x_2) = F(x_2, 0),$$
as desired.
Here is a plot of the graph of one example, $F(x,y) = x^5 - 2x^3 + 1.1x + xy^2$:
