<p>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.</p>
<p>If they end up negative, we set them <span class="math-container">$= 0$</span>.</p>
<p>The Lagrangian would be</p>
<p><span class="math-container">$\mathcal{L} = x_1 x_2 + x_2 + \lambda (M - P_1 x_1 - P_2 x_2)$</span></p>
<p>Solving this yields the demands</p>
<p><span class="math-container">$x_1^\star = \frac{M - P_1}{2 P_1}$</span></p>
<p><span class="math-container">$x_2^\star = \frac{M + P_1}{2 P_2}$</span></p>
<p>Note the expression for <span class="math-container">$x_1^\star$</span> can be negative, this happens when <span class="math-container">$P_1 &gt; M$</span>.</p>
<p>This implies that if <span class="math-container">$P_1 \leq M$</span>, the marshallian demands are the <span class="math-container">$x_1^\star, x_2^\star$</span> we got above.</p>
<p>If <span class="math-container">$P_1 &gt; M$</span>, then we set <span class="math-container">$x_1^\star = 0$</span>.</p>
<p>Then we get <span class="math-container">$x_2^\star = \frac{M}{P_2}$</span>.</p>
<p>By the way, your method yielded the correct demands for both cases.</p>
<p>In your method, what you would do is check the case where the non-negativity constraint is not binding <span class="math-container">$(\mu = 0)$</span>.</p>
<p>If the non-negativity constraint is not violated (the expression for <span class="math-container">$x_1^\star$</span> is not negative), that is your solution.</p>
<p>If it is violated (the expression for <span class="math-container">$x_1^\star$</span> is negative), the other case is your solution.</p>

CC BY-SA 4.0
2023-03-09T12:49:30.133Z
2023-03-09T12:58:38.227Z
2023-03-09T12:58:38.227Z
Nicolas Torres
<h2>Under what conditions constraints start to bind and how to find it</h2>
<p>I was trying the following optimization problem:</p>
<p><span class="math-container">$$ \mathscr{L} = x_1 x_2 + x_2 + \lambda(M-P_1 x_1-P_2 x_2) + \mu x_1$$</span></p>
<p>The thing is, as you can see, I operate with <span class="math-container">$U = x_1 x_2 + x_2$</span>, which implies that <span class="math-container">$x_1$</span> can be zero for a problem to provide an interesting solution. That is why I have included the <em>&quot;non-negativity&quot;</em> condition for <span class="math-container">$x_1$</span> in Lagrangian. I have performed the Lagrangian according to the classic approach and gained the following results:</p>
<p><span class="math-container">\begin{align*}
&amp;(1) &amp;&amp;&amp; &amp;(2)  \\
\mu &amp;= 0 &amp;&amp;&amp; \mu &amp;&gt;0 \\ 
x_1 &amp;= \frac{M-P_1}{2 P_1} &amp;&amp;&amp; x_1 &amp;= 0 \\
x_2 &amp;= \frac{M-P_1}{2P_2} + \frac{P_1}{P_2} &amp;&amp;&amp; x_2 &amp;=\frac{M}{P_2}
\end{align*}</span></p>
<p>These are the demands for each case of <span class="math-container">$\mu$</span>... However, what I would like to know is <strong>when each case becomes relevant?</strong> For which prices and income is relevant <span class="math-container">$(2)$</span> instead of <span class="math-container">$(1)$</span> etc...</p>
<p>The thing is during computation I came to the following condition (in case of <span class="math-container">$\mu &gt; 0$</span>):</p>
<p><span class="math-container">$$ x_2 + \mu = P_1 \frac{1}{2} $$</span></p>
<p>However, when solving graphically, there were multiple cases in which <span class="math-container">$\mu&gt;0$</span> and they depended on both the prices and the income.</p>
<p>Can I find out for which prices <span class="math-container">$\mu = 0$</span> and for which ones <span class="math-container">$\mu &gt; 0$</span>? <strong>How can I systematically find for which prices the constraint binds and for which ones it does not, without relying on graphical solution?</strong></p>

<microeconomics><mathematical-economics><lagrangian>
Determine for which prices and income the constraint is binding
Athaeneus
# Determine for which prices and income the constraint is binding

(Asked by: username_0 on 2023-03-08)

## 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?**

**Tags:** microeconomics, mathematical-economics, lagrangian


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## 1 answers

### Answer by username_1 on 2023-03-09. Score: 3 | Accepted answer

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.


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