Given the following Lagrangian function:
$$L(x, y, λ_1, λ_2)=x-y^2+2y+11-λ_1(3x+2y-4)+λ_2x$$
And given the first order conditions:
I. $\frac{∂L}{∂x}=1-3λ_1+1λ_2=0$
II. $\frac{∂L}{∂y}=-2y+2-2λ_1=0$
III. $λ_1(3x+2y-4)=0$
IV. $λ_2x=0$
Find the optimal solution.
I've done similar exercises before, but in this case I'm having trouble with the $λ_1$ variable.
So far I have, similar to the solution key:
$λ_2x = 0$
$λ_2 = 0$
Then:
$1-3λ_1+1λ_2=0$
$1-3λ_1=0$
$λ_1=\frac{1}{3}$
Following with the II. equation:
$-2y+2-2λ_1=0$
$-2y+2-2(\frac{1}{3})=0$
$2y = \frac{4}{3}$
$y = \frac{2}{3}$
Now, with the third equation, I thought I had to do the following:
$λ_1(3x+2y-4) = 0$
$\frac{1}{3}(3x+2(\frac{2}{3})-4) = 0$
$3x-\frac{4}{3} = 0$
$x = \frac{4}{9}$
However, the answer key leaves out the $λ_1$ coefficient:
$3x+2y-4=0$
$3x+2(\frac{2}{3})-4 = 0$
$3x = \frac{8}{3}$
$x = \frac{8}{9}$
And substituting the values, $x = \frac{8}{9}$ does check out.
What am I misunderstanding about the conditions? Why shouldn't $λ_1$ be included in the equation?
I know the solutions aren't wrong, but I genuinely can't see what I'm doing wrong despite it probably being glaringly obvious.