I am trying to understand the interpretation of an equality constraint minimization problem which I posted in a previous question here.
The function being constrained is $C(x,y)=tx+wy$ where the constants $t, w > 0$. The constraint function is $P=ay^2 + ax$ where the constants $P, a > 0$.
In the accepted solution to the question linked to above, the Lagrangian multiplier technique is used to get the following solution:
$P_c = (\frac{P}{a} - \frac{w^2}{4t^2}, \frac{w}{2t}, \frac{t}{a})$
I understand how the process to reach this result was done, but I don't understand why it is that the second term, ie. $y = \frac{w}{2t}$, does not have a dependence on $P$ as the first term ($x$) does. Intuitively, I would expect changing the value of $P$ would result in a some change in $y$, but it seems not to be the case as both $w$ and $t$ are constants.
Why is this so?
This is a result of the special nature of your equations. By imposing your constraint you are restricting the set of points you are considering to a parabola $x=P/a-y^2$ (this parabola is "horizontal"). Minimizing $C$ involves finding the leftmost (or rightmost, depending on the signs of some of the coefficients) of a family of parallel lines that intersects this parabola. If you change $P$ you will move the parabola left or right, and the leftmost curve intersecting it will also move up by the same amount. The tangency point (which is where your extremal point is) will correspondingly move left or right (but not up or down), meaning that the $y$ coordinate of the optimal point is independent of $P$. Another way to say it, is that changing $P$ in the constraint function happens to be the same as replacing $y$ by $y+const$, and this happens to also move lines $tx+wy=const$ to each other. The reason $\lambda$ is unchanged is a bit trickier, but it can be explained in similar way -- both the constraint function and the family of level sets of the cost function are simply shifted, so the proportionality between the constraint and level set gradients (which is what $\lambda$ is) is preserved.
BTW, your original problem does not need Lagrange multipliers. Just solve the constraint equation to get $x=P/a-y^2$ and plug into the cost to get $C=t(P/a-y^2)+wy=-ty^2+wy+tP/a$, which is a parabola. A parabola $az^2+bz+c$ has extremal value at $z=-b/2a$, so in your case the extremum is attained indeed at $y=w/2t$ (if $t>0$ this is a maximum, if $t<0$ a minimum). Correspondingly $x=P/a-y^2=P/a-w^2/4t^2$, as before.