I have the following Lagrangian:
$\frac{xp\eta}{R(x)}[v(x)-\bar\theta R(x)]+(1-\frac{xp\eta}{R(x)})xp\eta-(1+\lambda)tpx-\lambda R(x) + \delta_1x + \delta_2p + \delta_3(1-p) (1)$
Leading to first order conditions:
With respect to $p$:
$\frac{x\eta}{R(x)}[v(x)-x\bar\theta R(x)]+x\eta-\frac{2x^2p\eta^2}{R(x)}-(1+\lambda)tx + \delta_2 - \delta_3 =0 (2)$
And with respect to $\eta$:
$\frac{xp}{R(x)}[v(x)-\bar\theta R(x)]+xp-\frac{2p^2\eta x^2}{R(x)}=0 (3)$
In addition $\bar\theta$ is a coefficient between zero and one. I don't see what I should be doing next here.
When I isolate $p\eta$ in $(3)$ and plug into $(2)$, I obtain this result:
$\delta_2 - \delta_3 = (1+\lambda)tx (4)$
So I'm having trouble interpreting this. For example, we can see that \delta_3 can be zero, when \delta_2 is not, so the lower bound can be binding. We can tell that the upper bound is never binding.
What I'm having trouble understanding is - what if I arbitrarily change the upper bound to be 0.1? Changing the condition to $\delta_3(0.1-p)$ would result in a similar outcome, we would see that the upper bound would never be binding. We can keep doing this arbitrarily to find that $p$ must be the lower bound.
However, I feel like that should be obvious from $(4)$, that p cannot be between $0$ and $1$ and should simply be zero. Is my understanding correct?
This is related to a question I asked earlier in that I made some progress on my own and am hoping to check that my understanding/intuition here is correct.
If I summarize what you explain here and in your other question, you are trying to maximize an objective of the form: $$ F(x,p\eta) -(1+\lambda)tpx $$ under the constraints, $x, \eta \geq 0$ and $0 \leq p \leq 1$. The observation you make that the optimal value is necessarily $p = 0$ can indeed be derived from this very special form $F(x,p\eta)$ which only involves the product $p \eta$. Indeed, you could recast your problem as the maximization of $$ F(x,\sigma) -(1+\lambda)t px $$ with the constraints $x, \sigma \geq 0$ and $0 \leq p \leq 1$. Now it becomes clear that (provided $1+\lambda \geq 0$), since the second term is negative, the best choice is always $p = 0$.