This is unconstrained optimization. First Order Necessary Condition (FONC) gives me two equations that are contradicting each other. x1 + x2 = 5 and x1 + x2 = -5. What does this imply?

  • $\begingroup$ Too little information. Describe the whole problem in more detail. $\endgroup$ May 3, 2014 at 22:50
  • $\begingroup$ Find maximum or minimum of a given function. I have \nabla f. Hessian is [-2, -2; -2, -2] $\endgroup$
    – user100503
    May 4, 2014 at 0:17

1 Answer 1


It's very difficult to answer this question without more information. However, the function


has the Hessian you specified, and if you set $\nabla f(x)=0$, you get the two equations that you've provided in your posting. This particular function has no local min/max points.

It's also possible that you've made an error in computing $\nabla f(x)$ or the Hessian.

  • $\begingroup$ This is exactly the function that I have. So looks like my \nabla and Hessian are correct. How do you find if this has a minimizer of maximizer? I tried the usual First Order Necessary Condition but that does not give me any points. How do you conclude that there are no maximizers or minimizers? Thank you very much. $\endgroup$
    – user100503
    May 4, 2014 at 3:30
  • $\begingroup$ If you fix $x_{1}$, then consider $f$ as a function of $x_{2}$ only, you'll see that the maximum of $f$ keeping $x_{1}$ constant occurs at $x_{2}=-5-x_{1}$ and that the maximum value is $20x_{1}+25$. Thus we can increase $x_{1}$ towards $+\infty$, move $x_{2}$ to $-5-x_{1}$ and keep getting larger and larger values of $f$ without bound. This is not unusual behavior- it's quite common for indefinite quadratic forms to have no maximum or minimum. $\endgroup$ May 4, 2014 at 3:39
  • $\begingroup$ At a different level, you should understand that because there are no solutions to the FONC and $f$ is smooth, then there are no minimizers or maximizers. $\endgroup$ May 4, 2014 at 3:40

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