Critical points and second derivative test for two optimization problems

8 replies · 3K views · last reply
tcuay
Messages
8
Reaction score
0
These 2 questions,I have attempted them for hours but still no outcome:confused:
For the first question:
part a) i take the ∇f(x) and set it to be zero;
then find out(x1,x2)=(0,-1) or((-2a-1)/3,(-2a-4)/3);
but then for part b), after using second-order necessary condition,
i have no idea to continue;
part c): i know if i can prove the function is convex or concave, i can surely conclude the local minimum/maximum is a global minimum/maximum; but after taking the gradient for 2 times, i don't know to work for it in further;

For Question 2,
part a) i got 5 possible points (x1,x2)=(0.5,0),(1,-1),(1,1),(-1,-sqrt(3)),(-1,sqrt(3));
but in part b) i can only prove (0.5,0) is a local maximizer while other 4 points don't satisfy the 2nd conditions(because their determinant of Hessian Matrix is smaller than zero;)
part c) same, no idea how to do it;
part d) i find the directional vectors are both [0,0]'; does it mean f(1,-1) is a local minimum?

Thanks for help:frown:
 
Attachments
  • Handwritten homework problems: f(x) = x1 + a x1^2 + x1^3 - x2 - (1/2) x2^2 + x1 x2 with parts (a) to (c) on critical points, local extrema and global extrema, and a second problem f(x) = x1^2 x2^2 - 2x2^2 - x2^2 + 2x1 with parts (a) to (d).
    optimization.png
    17.8 KB · Views: 508
  • Handwritten working on lined paper: second-order condition for a function f(x) with gradient-squared and Hessian matrix, giving second derivatives 2a+bx and a conclusion that (0, -1) is a local maximizer.
    Q1.png
    41 KB · Views: 522
  • Handwritten derivation of the second-order condition: Hessian matrix of f with entries 2x1^2-4, 4x1x2, 4x1x2, 2x2^2-2; evaluated at (1/2, 0) giving diagonal -4 and -3/2 with eigenvalues -4 and -3/2; and a 2x2 table of Hessian determinants at points (1,1), (1,-1), (-1,-1) and (-1,1).
    Q2.png
    47 KB · Views: 498
tcuay said:
no body can help?:cry::cry:

Looking for stationary points and examining Hessians only works if the function has a (local) max or min at the point of interest. You can only apply these methods to global optima if you know that the functions have global optima, so first you need to examine that issue: does the function have a finite global max? Global min?
 
Ray Vickson said:
Looking for stationary points and examining Hessians only works if the function has a (local) max or min at the point of interest. You can only apply these methods to global optima if you know that the functions have global optima, so first you need to examine that issue: does the function have a finite global max? Global min?
First of all, really thanks for your reply;
These 2 questions both are required to find their global optima, it implies they should have a finite global optima; but what i am confused is how to determine whether those local optima are also global optima? And Even though i know the methods to find local optima.( e,g first order necessary condition, second order condition necessary/sufficient condition...), but i still cannot solve these 2 problems
Very difficult to me. Could you spend some more time to help me solve them :frown:
 
tcuay said:
First of all, really thanks for your reply;
These 2 questions both are required to find their global optima, it implies they should have a finite global optima; but what i am confused is how to determine whether those local optima are also global optima? And Even though i know the methods to find local optima.( e,g first order necessary condition, second order condition necessary/sufficient condition...), but i still cannot solve these 2 problems
Very difficult to me. Could you spend some more time to help me solve them :frown:

No, I cannot do any more, because that would involve my solving the problem for you. I will just say that your reasoning for the existence of global max or min is faulty; just because somebody asks you to find it does not mean it exists--it may, or may not!
 
tcuay said:
These 2 questions both are required to find their global optima, it implies they should have a finite global optima;
As Ray says, don't deduce that.
They are both smooth functions, defined everywhere. If a global extremum is not also a local extremum, where is it?
 
Okay, you are right. Then i also want to ask, for Q2 part d, if i find a steepest descent directional vector at a point, i.e x*, is zero; can i say that the point is a local minimizer? can the directional vector be zero when the point is a saddle point or maximum point?
 
tcuay said:
Okay, you are right.
Umm.. about what exactly?
When I wrote
If a global extremum is not also a local extremum, where is it?
there is an answer to that question.
can the directional vector be zero when the point is a saddle point or maximum point?
Yes, it will be. The grad is zero if and only if the tangent plane is horizontal. To know whether it's a min, a max, a saddle, or even a horizontal valley or ridge, you need to look at higher derivatives.
 
  • Like
Likes   Reactions: 1 person
haruspex said:
Umm.. about what exactly?
When I wrote

there is an answer to that question.

Yes, it will be. The grad is zero if and only if the tangent plane is horizontal. To know whether it's a min, a max, a saddle, or even a horizontal valley or ridge, you need to look at higher derivatives.
Okay, thanks very much;
And all answers are solved, thanks for your reminder;