find the KKT point of the following problem:
$$\min\quad f(x_1,x_2)=(x_1-3)^{2}+(x_2-2)^{2}\\
subject\quad to\qquad \qquad \qquad \qquad \qquad\qquad\\
x_1^{2}+x_2^{2}\le5\\
x_1+2x_2=4\\
x_1\ge0,x_2\ge0$$
what I tried:
for the general problem:
$$min\quad f(\textbf x)\\
s.t.\quad h_i(\textbf x)=0\quad (i=1,2,\dots ,m)\\
g_j(\textbf x)\ge0\quad (j=1,2,\dots,l)$$
where $\textbf x=(x_1,x_2,\dots,x_n)^{T}$
the KKT condition is
$$\nabla f(x^{*})-\sum_{j=1}^{l}\gamma_j^*\nabla g_j(x^*)-\sum_{i=1}^m\lambda_i^*\nabla h_i(x^*)=0\\
\gamma_j^*g_j(x^*)=0\quad (j=1,2,\dots,l)\\
\gamma_j^*\ge0\quad (j=1,2,\dots,l)$$
where $x^*=(x_1,x_2,\dots,x_n)^{T}$,$\gamma^*=(\gamma_1^*,\gamma_2^*,\dots,\gamma_l^*)^{T}$,$\lambda^*=(\lambda_1^*,\lambda_2^*,\dots,\lambda_m^*)^{T}$
$x^*$ is the KKT point if $x^*$ satisfies the KKT condition.
with this knowledge, for the specific problem above,
Let
$$f(\textbf x)=(x_1-3)^{2}+(x_2-2)^{2}\\
g_1(\textbf x)=-x_1^2-x_2^2+5\\
g_2(\textbf x)=x_1\\
g_3(\textbf x)=x_2
h_1(x)=x_1+2x_2-4$$
(change $x_1^2+x_2^2\le5$ to the general form $-x_1^2-x_2^2+5\ge0$)
and $x^*=(x_1,x_2)^{T}$ is the KKT point,then we have:
$$\nabla f(x^*)=
\begin{bmatrix}
2(x_1-3)\\
2(x_2-2)\\
\end{bmatrix}$$
$$\nabla g_1(x^*)=
\begin{bmatrix}
-2x_1\\
-2x_2\\
\end{bmatrix}$$
$$\nabla g_2(x^*)=
\begin{bmatrix}
1\\
0\\
\end{bmatrix}$$
$$\nabla g_2(x^*)=
\begin{bmatrix}
0\\
1\\
\end{bmatrix}$$
$$\nabla h_1(x^*)=
\begin{bmatrix}
1\\
2\\
\end{bmatrix}$$
the KKT condition is:
$$\begin{bmatrix}
2(x_1-3)\\
2(x_2-2)\\
\end{bmatrix}-\gamma_1\begin{bmatrix}
-2x_1\\
-2x_2\\
\end{bmatrix}-\gamma_2\begin{bmatrix}
1\\
0\\
\end{bmatrix}-\gamma_3\begin{bmatrix}
0\\
1\\
\end{bmatrix}-\lambda_1\begin{bmatrix}
1\\
2\\
\end{bmatrix}=0$$
$$\gamma_1(-x_1^2-x_2^2+5)=0$$
$$\gamma_2 x_1=0$$
$$\gamma_3 x_2=0$$
$$\gamma_1,\gamma_2,\gamma_3\ge0$$
we should get the KKT point $x^*$ by solving the equation set above.
For me, the problem here is that the equation set seems not solvable, so how to solve the equation, or I did it completely the wrong way?
Best Answer
Note that the general form of KKT you wrote is $$ \nabla f(x^{*})-\sum_{j=1}^{l}\gamma_j^*\nabla g_j(x^*)-\color{red}{\sum_{i=1}^m\lambda_i^*\nabla h_i(x^*)}=0 $$ while for this problem you have got $$ \begin{bmatrix} 2(x_1-3)\\ 2(x_2-2)\\ \end{bmatrix}-\gamma_1\begin{bmatrix} -2x_1\\ -2x_2\\ \end{bmatrix}-\gamma_2\begin{bmatrix} 1\\ 0\\ \end{bmatrix}-\gamma_3\begin{bmatrix} 0\\ 1\\ \end{bmatrix}=0 $$ that is without the equlity part. The right equation is $$ \begin{bmatrix} 2(x_1-3)\\ 2(x_2-2)\\ \end{bmatrix}-\gamma_1\begin{bmatrix} -2x_1\\ -2x_2\\ \end{bmatrix}-\gamma_2\begin{bmatrix} 1\\ 0\\ \end{bmatrix}-\gamma_3\begin{bmatrix} 0\\ 1\\ \end{bmatrix}-\color{red}{\lambda\begin{bmatrix} 1\\ 2\\ \end{bmatrix}}=0 $$ together with $x_1+2x_2=4$.
Now $x_2=0$ gives $x_1=4$ which is infeasible point ($x_1^2+x_2^2>5$), hence $x_2>0$ and $\gamma_3=0$.
Similarly, $x_1=0$ gives $x_2=2$ (and $\gamma_3=0$), $x_1^2+x_2^2<5$ (and $\gamma_1=0$), which leads to $\lambda=0$ and $\gamma_2=-6$. It is impossible as $\gamma_2\ge 0$, hence, $x_1>0$ and $\gamma_2=0$.
Next step is to try $\gamma_1=0$, it will give after some algebra $\lambda=-6/5$ and an infeasible point ($x_1^2+x_2^2>5$), then $\gamma_1>0$ and $x_1^2+x_2^2=5$.
Solving two equations \begin{align} x_1+2x_2&=4,\\ x_1^2+x_2^2&=5 \end{align} gives only one feasible point $(2,1)$. Check that all KKT conditions are satisfied ($\gamma_1=1/3$, $\lambda=-2/3$), then it is a KKT point.