What if there are infinite stationary points?

Question

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?

Answer 1

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.

Answer 2

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 $\inf f(x,y)=-\infty$.

Answer 3

Yes, your calculations were correct

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)

Answer 4

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.