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Suppose we have a function $f(x,y) = x^2 -xy +y^2 +3x -2y +1$

To find critical points of $f$, we compute its gradient:

$\nabla f=(3+2x-y,\ -2 -x +2y)$

The Hessian matrix for function $f$ is:

$$ \nabla^2 f = \begin{pmatrix} 2 & -1 \\ -1 & 2 \\ \end{pmatrix} $$

Since the determinant of this self-adjoint matrix is in the form:

$det(\nabla^2 f) = 2*2 -(-1*-1)= 3 > 0$

Then the determinant of $\nabla^2 f$ is positive

Looking at each of the critical points,

and

$det(\nabla^2 f({-4 \over 3}, {1 \over 3})) >0 $, which is negative, therefore $[{-4 \over 3}, {1 \over 3}]$ is a local mimimum of $f$

Is this result correct?

Paul-Henri GUET
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  • Please don’t keep posting the same question over and over. If you’ve got information to add to a question, edit the question or, if you don’t have enough reputation to do that yet, put the additional information in as a comment. – amd Apr 25 '17 at 21:32
  • @amd I don't think you are referring to this OP. – Roby5 Apr 25 '17 at 21:33
  • @Roby5 It’s the same user with a slightly different user name: “GUET” vs. “Guet.” – amd Apr 25 '17 at 21:39
  • Also, it appears that you’re now using two different logins with a slightly different spelling. Why? – amd Apr 25 '17 at 21:40
  • @amd Sorry for the confusion. – Roby5 Apr 25 '17 at 21:50

1 Answers1

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Yes, your result is correct, just the way you present your whole solution is a bit confusing. Try to always have a clear form of writing and presenting your work.

Anyway, yes. The global minimum of $f$ is the value :

$\min\{x^2−xy+y^2+3x−2y+1\}= -\frac{4}{3} $ at $(x,y)=(-\frac{4}{3},\frac{1}{3}) $

Wolfram Link Here

Rebellos
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