<p>You already answered your question: the first order conditions resulting from your Lagrangian are necessary conditions for a local optimum of your optimization problem. You will obtain all candidates for a global maximum.</p>

CC BY-SA 4.0
2022-03-13T16:56:08.660Z
Or a local maximum or local minimum
Bertrand
CC BY-SA 4.0
2022-03-14T08:14:45.027Z
That's why I wrote "candidates". You still need to check the second order conditions to see whether you are talking about minima, maxima or saddle points.
jpfeifer
CC BY-SA 4.0
2022-03-13T16:29:39.133Z
2022-03-13T16:29:39.133Z
jpfeifer
<p>Suppose the graph below shows three Indifference Curves such that <span class="math-container">$t &gt; s &gt; r$</span>, and the budget line <span class="math-container">$p_1x + p_2y = I$</span>. I was wondering if we set the Lagrangian as <span class="math-container">$\mathcal{L}= U(x,y) - \lambda (I - p_1x - p_2y)$</span>, do we get all the three tangency points or just the two with greater utility (that is, on <span class="math-container">$U = s$</span>)?</p>
<p>Since the Lagrangian finds both local maxima and local minima (because it essentially finds saddle points rather than maxima/minima), I feel we will get all three points.</p>
<p><a href="https://i.sstatic.net/3Zyvtm.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/3Zyvtm.png" alt="enter image description here" /></a></p>

CC BY-SA 4.0
2022-03-14T15:49:31.460Z
You have to check the second order condition using Bordered Hessian, if you want to know which one is maxima or minima.
Shaikh Ammar
CC BY-SA 4.0
2022-03-14T20:05:22.333Z
@ShaikhAmmar This works only when there's an interior optimum, right? Otherwise, for corner solutions, the maxima may not necessarily correspond to a convex part.
not tdm's twin
CC BY-SA 4.0
2022-03-14T20:18:02.683Z
True, because corner solutions don’t show up in Lagrangian. The only thing Lagrangian does is check if the obj function is tangential to the constraint function.
Shaikh Ammar
<microeconomics><utility><preferences><lagrangian>
Lagrangian when ICs are tangent to the budget line
not tdm's twin
# Lagrangian when ICs are tangent to the budget line

(Asked by: username_0 on 2022-03-13)

Suppose the graph below shows three Indifference Curves such that $t > s > r$, and the budget line $p\_1x + p\_2y = I$. I was wondering if we set the Lagrangian as $\mathcal{L}= U(x,y) - \lambda (I - p\_1x - p\_2y)$, do we get all the three tangency points or just the two with greater utility (that is, on $U = s$)?

Since the Lagrangian finds both local maxima and local minima (because it essentially finds saddle points rather than maxima/minima), I feel we will get all three points.

[![enter image description here](https://i.sstatic.net/3Zyvtm.png)](https://i.sstatic.net/3Zyvtm.png)

**Tags:** microeconomics, utility, preferences, lagrangian
> **Comments:**
> **username_1**: You have to check the second order condition using Bordered Hessian, if you want to know which one is maxima or minima.  
> **username_0**: @ShaikhAmmar This works only when there's an interior optimum, right? Otherwise, for corner solutions, the maxima may not necessarily correspond to a convex part.  
> **username_1**: True, because corner solutions don’t show up in Lagrangian. The only thing Lagrangian does is check if the obj function is tangential to the constraint function.  

---

## 1 answers

### Answer by username_2 on 2022-03-13. Score: 2 | Accepted answer

You already answered your question: the first order conditions resulting from your Lagrangian are necessary conditions for a local optimum of your optimization problem. You will obtain all candidates for a global maximum.

> **Comments:**
> **username_3**: Or a local maximum or local minimum  
> **username_2**: That's why I wrote "candidates". You still need to check the second order conditions to see whether you are talking about minima, maxima or saddle points.  

---

