I'm learning nonlinear programming and got stuck on this problem. Here is the problem:
Find the minimum of $\theta(x,y) = x^2 + y^2 - 2xy + 3y + 5$, with $x, y \in R$ which satislies $g(x,y) = x^2 + y^2 + 3x - 2y - 3 \le 0$
I tried to use Kuhn-Tucker sufficient optimality theorem, which I need to solve the system of equation and in equation:
$$ \left\{ \begin{array}{c} \nabla \theta(x) + u\nabla g(x) = 0 \\ g(x) \le 0 \\ ug(x) = 0 \\ u \ge 0 \end{array} \right. $$
which is equivalent to
$$ \left\{ \begin{array}{c} 2x - 2y + u(2x + 3y) = 0\\ 2y - 2x + 3 + u(2y - 2) = 0 \\ x^2 + y^2 + 3x - 2y - 3 \le 0 \\ u(x^2 + y^2 + 3x - 2y - 3) = 0 \\ u \ge 0 \end{array} \right. $$
then I have this system of equation for $x, y$ (I removed $u$ from the first 2 equations, it's easy to see that $u \neq 0$):
$$ \left\{ \begin{array}{c} x^2 + y^2 + 3x - 2y - 3 = 0 \\ 4x^2 - 4y^2 - 4x - 2y = 9 \\ \end{array} \right. $$
It looks not so difficult, but so bad that I can't solve it. I don't know that I missed something here. Can anyone help me to solve this problem using Kuhn-Tucker sufficient optimality theorem? Thanks so much.
Consider your last equations $$F=x^2 + y^2 + 3x - 2y - 3 = 0 $$ $$G=4x^2 - 4y^2 - 4x - 2y - 9=0$$ Now $$4F-G=16 x+8 y^2-6 y-3=0$$ makes $$x=\frac{1}{16} \left(-8 y^2+6 y+3\right)$$ Plug this expression in $F$ to get $$F=\frac{y^4}{4}-\frac{3 y^3}{8}-\frac{35 y^2}{64}-\frac{47 y}{64}-\frac{615}{256}=0$$ This last equation has an "obvious" root $y_1=-\frac 32$ to which corresponds $x_1=-\frac 32$.
Performing the long division, $$F=\frac{1}{128} \left(y+\frac{3}{2}\right) \left(32 y^3-96 y^2+74 y-205\right)=0$$ Now, using Cardano, you could show that the cubic part has only one real root (which is not rational $y_2 \approx 2.95182$, to which corresponds $x_2 \approx -0.909113$).
Now, computing $$f(x_1,y_1)=\frac 12 \quad g(x_1,y_1)=0$$ $$f(x_2,y_2)\approx 28.7622 \quad g(x_2,y_2)=-2.09126$$