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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?

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    $\begingroup$ Nice question, just a remark: It is not too hard to give an argument that one can find a smooth function with this property, but I am not sure how to make it into a polynomial. $\endgroup$ – Lukas Geyer May 29 '15 at 17:11
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    $\begingroup$ I am not sure, hence not an answer, but a comment. However, intuitively I would think the answer is no. At least, in a 1-variable domain it cannot be done. $\endgroup$ – FundThmCalculus May 29 '15 at 17:13
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    $\begingroup$ Consider the values of the polynomial along the line joining the two extreme points ... $\endgroup$ – Mark Bennet May 29 '15 at 17:38
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    $\begingroup$ One-dimensional max/min problems are very different from two-dimensional ones, and looking at values on a line is not going to help very much. $\endgroup$ – Lukas Geyer May 29 '15 at 17:41
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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, $F(x,y) = x^5 - 2x^3 + 1.1x + xy^2$:enter image description here

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  • $\begingroup$ A nice example! Thank you very much! $\endgroup$ – gžd15 May 30 '15 at 2:31

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