Determine for which prices and income the constraint is binding

Question

Under what conditions constraints start to bind and how to find it

I was trying the following optimization problem:

$$ \mathscr{L} = x_1 x_2 + x_2 + \lambda(M-P_1 x_1-P_2 x_2) + \mu x_1$$

The thing is, as you can see, I operate with $U = x_1 x_2 + x_2$, which implies that $x_1$ can be zero for a problem to provide an interesting solution. That is why I have included the "non-negativity" condition for $x_1$ in Lagrangian. I have performed the Lagrangian according to the classic approach and gained the following results:

\begin{align*} &(1) &&& &(2) \\ \mu &= 0 &&& \mu &>0 \\ x_1 &= \frac{M-P_1}{2 P_1} &&& x_1 &= 0 \\ x_2 &= \frac{M-P_1}{2P_2} + \frac{P_1}{P_2} &&& x_2 &=\frac{M}{P_2} \end{align*}

These are the demands for each case of $\mu$... However, what I would like to know is when each case becomes relevant? For which prices and income is relevant $(2)$ instead of $(1)$ etc...

The thing is during computation I came to the following condition (in case of $\mu > 0$):

$$ x_2 + \mu = P_1 \frac{1}{2} $$

However, when solving graphically, there were multiple cases in which $\mu>0$ and they depended on both the prices and the income.

Can I find out for which prices $\mu = 0$ and for which ones $\mu > 0$? How can I systematically find for which prices the constraint binds and for which ones it does not, without relying on graphical solution?

Answer 1

For problems involving non-negativity constraints, you’d solve the usual Lagrangian without them and check if the optimal variable(s) ended up negative or not.

If they end up negative, we set them $= 0$.

The Lagrangian would be

$\mathcal{L} = x_1 x_2 + x_2 + \lambda (M - P_1 x_1 - P_2 x_2)$

Solving this yields the demands

$x_1^\star = \frac{M - P_1}{2 P_1}$

$x_2^\star = \frac{M + P_1}{2 P_2}$

Note the expression for $x_1^\star$ can be negative, this happens when $P_1 > M$.

This implies that if $P_1 \leq M$, the marshallian demands are the $x_1^\star, x_2^\star$ we got above.

If $P_1 > M$, then we set $x_1^\star = 0$.

Then we get $x_2^\star = \frac{M}{P_2}$.

By the way, your method yielded the correct demands for both cases.

In your method, what you would do is check the case where the non-negativity constraint is not binding $(\mu = 0)$.

If the non-negativity constraint is not violated (the expression for $x_1^\star$ is not negative), that is your solution.

If it is violated (the expression for $x_1^\star$ is negative), the other case is your solution.