KKT condition - minimization problem

Question

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

FEASIBILITY:

$-x^2-y^2+9\geq0$

$y\geq0$

STATIONARITY:

$8/(x+4) + 2x γ_1=0$

$2y+2yγ_1-γ_2=0$

NON NEGATIVE MULTIPLIERS:

$γ_1 \geq 0$

$γ_2 \geq 0$

COMPLEMENTARY:

$ γ_1(x^2+y^2-9)=0$

$γ_2(-y) = 0$

So I have to consider 4 cases.

CASE 1)
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.

CASE 2)
The first costraint is non-active, the second constraint is active:

$y=0 , γ_1=0$

So with stationarity conditions I have :

$1/(x+4)=0$

that is not possibile.

CASE 3)
First constraint active, second constrait non-active: $$\left\{\begin{array}{l} x^2+y^2-9=0\\ γ_2=0 \end{array}\right.$$

CASE 4)
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.

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 $y^2+8\ln(x+4)$.