How to Find Critical Points of function f(x,y,z)

[Gh. Soleimani]

TL;DR

undefined critical points for first partial derivatives

Hi everybody

If we have not any answers for critical points after first partial derivatives equal to zero, how can we continue to find local MAX, local MIN and Saddle point?. For example: Suppose we have below equations for first partial derivatives:


∂ƒ/∂x = y + 5 , ∂ƒ/∂y = 2z , ∂ƒ/∂z = y
As you can see, for ∇ƒ= 0 , there are not any answers (undefined)

 

[Mark44]

The three first partials are zero at (x, -5, 0) and (x, 0, 0), with x being arbitrary.

 

[pasmith]

∂ƒ/∂x = y + 5 , ∂ƒ/∂y = 2z , ∂ƒ/∂z = y
As you can see, for ∇ƒ= 0 , there are not any answers (undefined)

There is a fundamental problem here: \nabla \times (y + 5, 2z, y) = (-1,0,-1) is not zero, so (y + 5, 2z, y) cannot be the gradient of a function.

The three first partials are zero at (x, -5, 0) and (x, 0, 0), with x being arbitrary.

y + 5 and y do not vanish simultaneously.

 

[Mark44]

y + 5 and y do not vanish simultaneously.

Yes, I understand that.

 

[Gh. Soleimani]

Sorry. I think that I took a mistake to introduce the function. First of all, let me write true function as follows:


ƒ(x, y, z) = x +yz - xy


∇ƒ = (1-y) i +(z-x) j + y k

where:

∂ƒ/∂x = 1 - y =0

∂ƒ/∂y = z - x =0

∂ƒ/∂z = y =0

As you can see, there are not any answer for critical point because we have: 1- y =0 , also y =0

My question is: Can we make below H - matrix and evaluate eigenvalues?

 

[Mark44]

I agree with your values for the first partials; namely, $f_x = 1 - y, f_y = z - x; f_z = y$, but I disagree that there are no critical points.

The three first partials are zero at on the sets of points {(x, 0, x)} and {(x, 1, x)}, where x is arbitrary. The first set is a line in the x-z plane and the second set is a line in the plane y = 1.

Setting the first partials to zero results in the following matrix equation, but I'm not sure that doing this is helpful or why you need to find eigenvalues.

$\begin{bmatrix} 0&1&0 \\1&0&-1 \\0 & 1& 0\end{bmatrix} \begin{bmatrix} x\\y \\ z\end{bmatrix} = \begin{bmatrix} 1\\ 0 \\ 0\end{bmatrix}$

 

[Gh. Soleimani]

I agree with your values for the first partials; namely, $f_x = 1 - y, f_y = z - x; f_z = y$, but I disagree that there are no critical points.

The three first partials are zero at on the sets of points {(x, 0, x)} and {(x, 1, x)}, where x is arbitrary. The first set is a line in the x-z plane and the second set is a line in the plane y = 1.

Setting the first partials to zero results in the following matrix equation, but I'm not sure that doing this is helpful or why you need to find eigenvalues.

$\begin{bmatrix} 0&1&0 \\1&0&-1 \\0 & 1& 0\end{bmatrix} \begin{bmatrix} x\\y \\ z\end{bmatrix} = \begin{bmatrix} 1\\ 0 \\ 0\end{bmatrix}$

Thank you so much for your answer.

Finally, I need to obtain local minimum, local maximum and saddle points. As you know, we have to use Hessian matrix which will be made by second derivative. In fact, to examine eigenvalues of H - matrix in critical point says to us where is local minimum and local maximum.

 

[Mark44]

$|H| = \begin{vmatrix} 0&1&0 \\1&0&-1 \\0 & 1& 0\end{vmatrix}$ does not depend on the values of x, y, or z. What do you get for |H|?

According to this wiki article, https://en.wikipedia.org/wiki/Hessian_matrix, if det(H) = 0 at a critical point, the critical point is degenerate. If det(H) is not 0, the critical point is non-degenerate.

 

[Gh. Soleimani]

Yes. You are right. Here, the eigenvalue of the “H” matrix is equal to zero; therefore, the test is inclusive, and I have to use a contour map to obtain the local minimum/maximum and saddle point.
But perhaps there is another way, which is to utilize some data analysis models by machine for a selected domain.

 

[pasmith]

$|H| = \begin{vmatrix} 0&1&0 \\1&0&-1 \\0 & 1& 0\end{vmatrix}$ does not depend on the values of x, y, or z. What do you get for |H|?

According to this wiki article, https://en.wikipedia.org/wiki/Hessian_matrix, if det(H) = 0 at a critical point, the critical point is degenerate. If det(H) is not 0, the critical point is non-degenerate.

A Hessian matrix must be symmetric; this matrix is not. This is a consequence of the problem I noted in my earlier post: the vector field specified by the OP is not the gradient of a scalar function.