Please check if my answers are correct specially for part 3 and help me to do the fourth part in this problem
Let f(x,y)=x$^4$-8x$^2$+y$^4$-18y$^2$
- Find the critical points of f
- Determine the nature of the critical points of f
- Find the set of global minimizers of f
- Does f have a global maximizer?Justify?
My answer is:
I first calculated the gradient of f and then let partial derivative of x and y to be equal to 0.Thereby the critical points I found are (0,0)(0,3)(0,-3)(6,0)(6,3)(6,-3).
Then I computed the Hessian matrix :
H(x,y)= $$ \begin{pmatrix} 12x^2-48x & & 0 \\ 0& & 12y^2-36 \\ \\ \end{pmatrix} $$ To find the nature of critical points I computed the Hassian matrix at each critical point.
These are my answers for part 2:
H(0,0)
This can be either positive semi definite or negative semi definite.Hence the critical point can either be a minimizer , maximizer or a saddle point.Therefore it can't be concluded the type of the critical point.
H(0,3)
This is positive semi definite. Therefore as earlier nothing can be said about the critical point.
Same with H(0,-3).
H(6,0)is also positive semi definite.
H(6,3) is a positive definite matrix.Hence there's a local minimum at (6,3)
H(6,-3) is a positive definite matrix.Hence there's a local minimum at (6,-3).
Answers for part 3:
I used the theory
H(z) is negative semidefinite for all z ∈ S ⇒ [x is a global maximizer of f in S if and only if x is a stationary point of f ]
H(z) is positive semidefinite for all z ∈ S ⇒ [x is a global minimizer of f in S if and only if x is a stationary point of f ],
When H(x,y) is positive semidefinite for all x in a particular region, in that region the function is convex and any local minimizer is a global minimizer in that region. Is this a correct statement?
To be positive semi definite det H(1)>0 and det H(2)=0:
Thus 12x$^2$ - 48x > 0.
x$\in $ (-infinity,0)$\cup$(4,infinity)
det H(2)=0 when (12x$^2$-48x)(12y$^2$-36)=0
Since (12x$^2$-48x) >0,
(12y$^2$-36)=0
Thus y=$\pm$$\sqrt3$
Thus hessian matrix is positive definite when x$\in $ (-infinity,0)$\cup$(4,infinity) and y=$\pm$$\sqrt3$ .
But can I say that the matrix is positive definite even if det H(2)>0.
Therefore I selected the critical points where x$\in $ (-infinity,0)$\cup$(4,infinity) and y>$\sqrt3$ and y<-$\sqrt3$.
Thus the global minimizers of f are (6,3)are (6,-3).
This is a very long process and I do not understand how to find the answer for part 4.
Is there a short method to do this kind of a problem and I would prefer if anybody can help me to do this problem using theories based on positive/negative definite/semidefiniteness and using hessian matrix.