$y^2-8 \ln(x+4)\rightarrow$ min,

such that $-x^2 -y^2+9 \geq 0, y \geq 0$

*I have to find all possible optimal points.
* Lagragian function is:

$L(x,y,γ_1,γ_2) = y^2 - 8\ln(x+4)+γ_1(x^2+y^2-9) + γ_2(-y)$

KKT conditions are the following:





$8/(x+4) + 2x γ_1=0$



$γ_1 \geq 0$

$γ_2 \geq 0$


$ γ_1(x^2+y^2-9)=0$

$γ_2(-y) = 0$

So I have to consider 4 cases.

Both constraints are active:

$$\left\{\begin{array}{l} x^2+y^2-9=0\\ y=0 \end{array}\right.$$ There are two real solutions, but just one point respects non negative multipliers condition: $$\left\{\begin{array}{l} x=-3\\ y=0 \end{array}\right.$$ (-3,0) is a possible optimal point.

I have some difficulties with the other cases.

The first costraint is non-active, the second constraint is active:

$y=0 , γ_1=0$

So with stationarity conditions I have :


that is not possibile.

First constraint active, second constrait non-active: $$\left\{\begin{array}{l} x^2+y^2-9=0\\ γ_2=0 \end{array}\right.$$

Both constraint are non-active: $$\left\{\begin{array}{l} γ_1=0\\ γ_2=0 \end{array}\right.$$

I dont't understand how to use the KKT conditions to find possible(or not) optimal points in case 2, case 3 and case 4.


1 Answer 1


Case 1: You need also to check whether $(-3,0)$ satisfies all other conditions: from stationarity $8-6\gamma_1=0$ $\Leftrightarrow$ $\gamma_1=\frac43\ge 0$ OK and $\gamma_2=0\ge 0$ OK.

Case 2: No solution. (It does not have to have a solution in all cases.)

Case 3: From stationarity $2y+2y\gamma_1=0$ $\Leftrightarrow$ $2y(1+\gamma_1)=0$ $\Rightarrow$ $y=0$ since $\gamma_1\ge 0$. It means that the second constraint is active which contradicts the assumption for Case 3. Thus no solution.

Case 4: Similar to Case 2, no solution.

Conclusion: the only point that satisfies the KKT conditions is $(-3,0)$.

P.S. The solution is not reasonable since the function $y^2-8\ln(x+4)$ is clearly smaller when the logarithm is larger, that is the correct solution would be $(3,0)$. The error is in your stationarity condition: the partial derivative wrt $x$ is minus $8/(x+4)$. So you actually minimized the function $y^2+8\ln(x+4)$.

  • $\begingroup$ Thank you. Excellent answer. You are right, I made a mistake with the partial derivative. Just one last question about complementary conditions: let us assume to be in the case where a constraint, with a γa multiplier associated, is active; besides, using stationarity condition, γa is 0. Can I say there are not any solutions for this case? $\endgroup$ Oct 10, 2015 at 11:52
  • $\begingroup$ @DistribuzioneGaussiana It may happen that for an active constraint the corresponding multiplier is zero, it does not contradict anything. If all other conditions are satisfied it gives a legitimate KKT point. $\endgroup$
    – A.Γ.
    Oct 10, 2015 at 12:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.