# What if there are infinite stationary points?

I want to calculate extremes of certain multivariable function $$f(x,y)=(6−x−y)x^2y^3$$. After solving system of derivatives $$f_x=0$$ and $$f_y=0$$ I got something like this:

$$P_1=(x,0),x\in \mathbb R$$

$$P_2=(0,y),y\in \mathbb R$$

$$P_3=(2,3)$$

First two conditions are satisfied with infinite number of $$x$$ and $$y$$. How am I supposed to act in such situation? Do I have to check the first two points in some way? If so, how should I do this?

• Can you provide the multivariable function ? Sep 5, 2019 at 7:21
• Because we can't be sure that you did it in right way Sep 5, 2019 at 7:21
• Of course, this is it: $f(x,y)=(6−x−y)x^2y^3$. I have updated $P_3$ value in my question. Sep 5, 2019 at 7:24

Based on your comments I think that you have a problem with the second derivative test. Note that the Hessian of $$f$$ is

$$\begin{pmatrix} 12 y^3 - 6xy^3 - 2y^4 & 36 xy^2 - 9x^2 y^2 - 8xy^3 \\ 36 xy^2 - 9x^2y^2 - 8xy^3 & 36x^2 y - 6x^3 y - 12 x^2 y^2 \end{pmatrix}.$$

Now we want to check what happens at $$P_1 = (x, 0)$$ with $$x \in \mathbb{R}$$. Plugging this in yields the zero matrix

$$\begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix}.$$

Since this matrix has determinant $$0$$, the second derivative test fails. Similarly, for $$P_2 = (0, y)$$ with $$y \in \mathbb{R}$$, we get

$$\begin{pmatrix} 12y^3 - 2y^4 & 0 \\ 0 & 0 \end{pmatrix}$$

for every $$y \in \mathbb{R}$$. Note that this matrix also has determinant $$0$$ (for every $$y$$). So you have to use other methods for determining what happens there. Some ideas were already given in the answers, have a look also e. g. here.

• Thank you, that helped! Sep 5, 2019 at 8:10

If you see graph of this equation you would see that (0,y) are points of inflection where (X,0)are minimum points. Like sine and cosine have multiple stationary points this graph too have infinite minimum points (X,0)   • But this function does not have minimum. Sep 5, 2019 at 7:45
• Can you elaborate on that? Because I am stuck on hessian matrix. I have calculated $f_{xx}$, $f_{xy}$, $f_{yx}$ and $f_{yy}$ and only stationary point that I know how to handle is $P_3$. Sep 5, 2019 at 7:46
• See, the graph does have minimum points Sep 5, 2019 at 7:50
• Graph is perfectly fine, but how to handle such situation arithmetically? In the case of casual points, with known $x$ and $y$, the only thing I have to do is to substitute variables with coordinates in hessian matrix. How should I act with something like $(x, 0)$? Sep 5, 2019 at 7:52
• Input 0 in place of y and X in place of X. Tell me the result u get Sep 5, 2019 at 7:56

Hint

For $$(x_0,0) \in P_1$$ and $$(x,y) \in \mathbb R^2$$ you have

$$f(x,y) -f(x_0,0) = f(x,y)=(6−x−y)x^2y^3$$

The RHS quantity can take both positive and negative values around $$(x_0,0)$$ for $$x_0 \neq 6$$ as $$y^3$$ takes both positive and negative values around $$0$$. Therefore elements of $$P_1$$ are not minimum nor maximum.

You can prove the same for elements of $$P_2$$.

Now you have to consider the point $$(2,3)$$. For that, compute the hessian at that point. This is a symmetric matrix of dimension $$2$$. Look at the eigenvalues to conclude if that point is a minimum, a maximum or not.

I am assuming that $$f$$ is defined on $$\mathbb R^2$$.

On the curve $$(-t,t)$$ For $$t\in \mathbb R$$ $$f(-t,t)=6t^2t^3$$ so if you let tend $$t\to +\infty , f \to +\infty$$. On the other hand if $$t\to -\infty, f\to -\infty$$.

So we can coclude that $$\sup f(x,y)=+\infty$$ and $$\inf f(x,y)=-\infty$$.

• This is my function: $f(x,y)=(6−x−y)x^2y^3$. I am stuck on hessian matrix. I know how to play with $P_3$ but I don't know how to perform substitution of points like $(x, 0)$ or $(0, y)$. Sep 5, 2019 at 7:50
• But this function is unbounded why do you think that have min and max? Sep 5, 2019 at 7:54
• @MarcoLecci OP looks for local extrema, which can, a priori, exist.
– Jan
Sep 5, 2019 at 7:56