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In this, at the critical points, I'm getting the following values of Hessian matrix. H(0,0) = 20 H(-5/3,0) = 40/3 H(-1,2) = -16 H(-1,-2) = -16

Based on these, how did they classify the critical points as local max, local min or saddle point?

And how to check for global optimum points. By letting x and y approach infinity and then comparing the function values (which equals infinity) to function values at critical points?

  • $\begingroup$ here are some Infos to the second derivative test mathworld.wolfram.com/SecondDerivativeTest.html $\endgroup$ – Dr. Sonnhard Graubner Sep 16 '17 at 20:00
  • $\begingroup$ Okay, so the principal minors I get are the following - At (0,0) - 10, 20. Since both positive, so positive definite and hence local minimum. Similarly, at (-1,-2), I get -2,-16 as principal minors. So, saddle point. Is this correct? $\endgroup$ – LUCIFER Sep 16 '17 at 20:02
  • $\begingroup$ it Looks good to me $\endgroup$ – Dr. Sonnhard Graubner Sep 16 '17 at 20:04
  • $\begingroup$ Okay. And can you also see if my reasoning for global min and max is correct in the question above (edited version) $\endgroup$ – LUCIFER Sep 16 '17 at 20:10
  • $\begingroup$ en.wikipedia.org/wiki/… $\endgroup$ – Raffaele Sep 16 '17 at 20:15

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