 #1
 Homework Statement:

Find the max/min/saddle points of ##f(x,y) = x^4  y^4## subject to the constraint ##g(x,y) = x^22y^2 1 =0##
Use Lagrange multipliers method
Classify the stationnary points (max/min/saddle) using the definiteness of the Hessian
 Relevant Equations:
 Positive/Negative definite matrix
Hello,
I am using the Lagrange multipliers method to find the extremums of ##f(x,y)## subjected to the constraint ##g(x,y)##, an ellipse.
So far, I have successfully identified several triplets ##(x^∗,y^∗,λ^∗)## such that each triplet is a stationary point for the Lagrangian: ##\nabla \mathscr{L} (x^∗,y^∗,λ^∗) = 0##
Now, I want to classify my triplets as max/min/saddle points, using the positive/negative definiteness of the Hessian like I have been doing for unconstrained optimization, so I compute what I think is the Hessian of the Lagrangian:
$$H_{\mathscr{L}}(x,y,λ)= \begin{pmatrix} 12x^2  2\lambda & 0 \\ 0 & 12y^2  4\lambda \end{pmatrix}$$
Evaluating the Hessian for my first triplet ##(0,\pm \frac{\sqrt{2}}{2},−\frac{1}{2})## gives me:
$$H_{\mathscr{L}}(0,\pm \frac{\sqrt{2}}{2},−\frac{1}{2}) = \begin{pmatrix} 1 & 0 \\ 0 &  4\end{pmatrix}$$
This matrix is diagonal, meaning that we immediately read its eigenvalues on the diagonal: ##\lambda_1 = 1 > 0## and ##\lambda_2 = 4 < 0##. A positive/negative definite matrix has only positive/negative eigenvalues, thus I conclude that this matrix is neither, due to its eigenvalues' opposite signs.
When I was studying unconstrained optimization, I learned that we have in this case a saddle point, so I would like to think that the points ##(0,\pm \frac{\sqrt{2}}{2})## are both saddle points for my function f, however, the solution to this problem affirms these points are minimums, using the following argument:
Using the fact that ##\nabla g(x,y) = (0,\pm \frac{\sqrt{2}}{2})## and that ##w^T \nabla g(x,y) = 0## if and only if ##w = (\alpha, 0), \alpha \in \mathbb{R}^{\ast}##
I thought that it was enough to check for the definiteness of the Hessian, and now I am really confused...
Here are my questions:
Edit: PF destroyed my LaTeX formatting.
I am using the Lagrange multipliers method to find the extremums of ##f(x,y)## subjected to the constraint ##g(x,y)##, an ellipse.
So far, I have successfully identified several triplets ##(x^∗,y^∗,λ^∗)## such that each triplet is a stationary point for the Lagrangian: ##\nabla \mathscr{L} (x^∗,y^∗,λ^∗) = 0##
Now, I want to classify my triplets as max/min/saddle points, using the positive/negative definiteness of the Hessian like I have been doing for unconstrained optimization, so I compute what I think is the Hessian of the Lagrangian:
$$H_{\mathscr{L}}(x,y,λ)= \begin{pmatrix} 12x^2  2\lambda & 0 \\ 0 & 12y^2  4\lambda \end{pmatrix}$$
Evaluating the Hessian for my first triplet ##(0,\pm \frac{\sqrt{2}}{2},−\frac{1}{2})## gives me:
$$H_{\mathscr{L}}(0,\pm \frac{\sqrt{2}}{2},−\frac{1}{2}) = \begin{pmatrix} 1 & 0 \\ 0 &  4\end{pmatrix}$$
This matrix is diagonal, meaning that we immediately read its eigenvalues on the diagonal: ##\lambda_1 = 1 > 0## and ##\lambda_2 = 4 < 0##. A positive/negative definite matrix has only positive/negative eigenvalues, thus I conclude that this matrix is neither, due to its eigenvalues' opposite signs.
When I was studying unconstrained optimization, I learned that we have in this case a saddle point, so I would like to think that the points ##(0,\pm \frac{\sqrt{2}}{2})## are both saddle points for my function f, however, the solution to this problem affirms these points are minimums, using the following argument:
Using the fact that ##\nabla g(x,y) = (0,\pm \frac{\sqrt{2}}{2})## and that ##w^T \nabla g(x,y) = 0## if and only if ##w = (\alpha, 0), \alpha \in \mathbb{R}^{\ast}##
I thought that it was enough to check for the definiteness of the Hessian, and now I am really confused...
Here are my questions:
 When is it enough to check the definiteness of the Hessian to classify stationnary points?
 Why is there this additional step in constrained optimization?
 What am I missing?
Edit: PF destroyed my LaTeX formatting.
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