# Determine for which prices and income the constraint is binding

## 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?

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.