I need help in the following problem:
Consider the following optimization problem $$ \min_{x_1,x_2}-x_1-x_2\quad\text{s.t.}\quad x_1^2+x_ 2^2-1=0,\quad x_1,x_2\geqslant 0.$$ Show that the linearized in $\mathbf x=(1/2,1/2)^T$ feasible set is empty.
I have $g(x) = x_1^2+x_2^2-1$ and $h(x) = x_i, i={1,2}.$
I use the formula A: {x+$\Delta x$ | g(x) + $∇g(x)^T\Delta x = 0$} and B: {x+$\Delta x | h(x) + ∇h(x)^T\Delta x >= 0$}, where T stands for transpose.
I calculate $∇g(x) = (2x_1, 2x_2)^T$ - in point $(\frac12, \frac12)$ and have $∇ g(x)=(1,1)^T$. Additionally, $∇h(x) = 1$ for both $x_1$ and $x_2$.
Values of the functions are $g(x) = -\frac12$ and $h(x) = \frac12$ for both $x_1$ and $x_2$.
From formula A I obtain $x_2 = - x_1 +\frac32$ and from formula B I get $x_i >= 0$, for i ={1,2}. When I put conditions A and B on one plot I do not obtain empty set...
Can anyone help cause I don't see what is missing?