\begin{equation}
\mathcal{L} = x – \lambda_1 ( – \frac{1}{2}x^2 + 2x-y) – \lambda_2 ( \frac{1}{2}x^2 -2x + 4 -y)
\end{equation}
This gives the following FOC's
\begin{equation}
\begin{aligned}
\frac{\partial \mathcal{L}}{\partial x} & = 1 + \lambda_1x -2\lambda_1 – \lambda_2x + 2\lambda_2 = 0 \\
\frac{\partial \mathcal{L}}{\partial y} & = \lambda_1 + \lambda_2 = 0 \\
y &= -\frac{1}{2}x^2 + 2x\\
y &= \frac{1}{2}x^2 -2x + 4
\end{aligned}
\end{equation}
Here we have $-\lambda_1=\lambda_2$ substituting this into the first equation gives us
\begin{equation}
x = 2\lambda_1 – \frac{1}{2}
\end{equation}
Substituting this result into first constraint gives:
\begin{equation}
y = -2\lambda_1^ 2 + \frac{1}{8}
\end{equation}
substituting into the next constraint gives:
\begin{equation}
0 = 4\lambda_1^2 – 7/4 \lambda_1 + 71/16
\end{equation}
the last equation has no real roots so I cant solve the problem yet I know there should be a solution. Also I am not one 100% sure if I understand how to check for the second order conditions in this case. Could someone please help? Also, this Lagrangian should be also solvable using level curve method. Could someone explain me how that would work?
Best Answer
If the problem is
$$ (\min,\max) x \ \ \mbox{s.t.}\ \ \left\{\begin{array}{rcl}g_1(x,y) & = & -\frac 12 x^2 + 2x - y=0\\ g_2(x,y) & = & \frac 12 x^2-2x-y+4=0\end{array}\right. $$
we have that $g_1(x,y)\cap g_2(x,y) = (2,2)$ so the feasible solution is at $x = 2$