Use the Karush-Kuhn-Tucker conditions to find the optimal solution to the following non-linear programming problem.
$$
\begin{gathered}
\min z=\left(x_{1}-1\right)^{2}+\left(x_{2}-2\right)^{2} \\
\text { s.t. }-x_{1}+x_{2}=1 \\
x_{1}+x_{2} \leq 2 \\
x_{1}, x_{2} \geq 0
\end{gathered}
$$
I followed this site to solve this problem,
Note: at the optimum, it is known that the inequality constraint is satisfied but not
binding. Take advantage of this information.
Question: I couldn't understand the note mentioned that site.
Dropping the inequality constraint, we get,
$$
\begin{align}
\pmatrix{2x_1-2\\2x_2-4}-\lambda_1\pmatrix{-1\\2}&=\pmatrix{0\\0}\\
\pmatrix{2x_1\\2x_2}+\pmatrix{\lambda_1\\-2\lambda_1}&=\pmatrix{2\\4}
\end{align}
$$
Use the equality constraint $-x_1+x_2+0.\lambda_1=1$,
$$
A=\pmatrix{-1&1&0\\2&0&1\\0&2&-2}\quad b=\pmatrix{1\\2\\4}
$$
$$
\pmatrix{x_1\\x_2\\\lambda_1}=A^{-1}b=\pmatrix{1\\2\\0}
$$
Hence, the $z_{\min}=(1-1)^2+(2-2)^2=0$.
But the solution provided in my textbook was $z_{\min}=\frac12,x_1=\frac12,x_2=\frac32$. Did their procedure is wrong? I used that because it seems short. But now it wasn't giving the exact solution.
Best Answer
The information given to you is wrong. The inequality is binding as we can see that the circle touches the halfspace.
To make use of your working, you can assume that you were considering the case where it is not binding and you found a contradiction. Hence we know that it is binding, now you can solve the problem by solving $-x_1+x_2=1, x_1+x_2=2$.