<p><strong>The Statement of the Problem:</strong></p>

<p>Let $f,g$ be two functions on $\mathbb R^n$ and assume $\nabla f $and$ \nabla g$ are nonzero everywhere. Write the first and second order Langrangian-multiplier conditions for the following two problems:</p>

<p>$$(1) \text{ minimize } f(\mathbf x) \text{ subject to constraint } g(\mathbf x)=b $$</p>

<p>$$(2) \text{ maximize } g(\mathbf x) \text{ subject to constraint } f(\mathbf x)=c  $$</p>

<p>Given a solution $\mathbf y$ to first and second order conditions for the minimization problem $(1)$, for some value of $b$, set $c=f(\mathbf y)$ and consider the maximization problem $(2)$; when is $\mathbf y$ a solution to this problem as well?</p>

<p><strong>My Questions:</strong></p>

<p>I believe I have the first-order conditions for $(1)$:</p>

<p>$$ \frac{\partial \mathcal{L}(\mathbf x)}{\partial x_i} = \frac{\partial f(\mathbf x)}{\partial x_i}-\lambda \frac{\partial g(\mathbf x)}{\partial x_i}=0$$</p>

<p>But what are the second-order conditions? I know that, in order to minimize $f$, $\mathcal{L}$ must be convex and that the constraint set must be convex, but are those the "second-order conditions"? If so, how do I determine them? By way of bordered Hessians? That doesn't seem right. Anyway, any help here would be appreciated. Thanks!</p>

CC BY-SA 3.0
2015-04-20T15:46:23.827Z
Why that doesn't seem right by way of bordered Hessians?
Metta World Peace
CC BY-SA 3.0
2015-04-20T17:02:42.953Z
What you write as a "first-order condition", is not -it is just the first derivative of the Lagrangean. What is _the condition_ on it? Also, convexity is a convenient property to guarantee that the second-order conditions hold, and so that the first-order conditions are sufficient and necessary for a minimum. As with univariate optimization, where we look into second derivatives, here we have to look into Hessians/border Hessians. These _are_ the 2nd-order conditions, not a "way to verify them".
Alecos Papadopoulos
CC BY-SA 3.0
2015-04-20T17:38:44.957Z
Thanks. So, yeah, I forgot to put the "$=0$" in there for the "first-order conditions." And, right, so by checking the definite-ness of the bordered Hessians, I can determine whether or not $\mathcal{L}$ is concave/convex... which is the second-order condition. However, I don't really have a way of determining the definiteness of the Hessian because I don't actually have any values to work with. Furthermore, don't I have to consider "regularity"? Sorry, I'm really confused about all of this...
thisisourconcerndude
CC BY-SA 3.0
2015-04-20T18:35:15.553Z
Since you are given no other information, you might as well assume convexity, or just "for $y$ to be a solution, second order conditions _must_ be satisfied". After all, this is obviously an exercise in duality.
Alecos Papadopoulos
<mathematical-economics><optimization>
First and Second Order Lagrangian-Multiplier Conditions for Optimization
thisisourconcerndude
# First and Second Order Lagrangian-Multiplier Conditions for Optimization

(Asked by: username_0 on 2015-04-20)

**The Statement of the Problem:**

Let $f,g$ be two functions on $\mathbb R^n$ and assume $\nabla f $and$ \nabla g$ are nonzero everywhere. Write the first and second order Langrangian-multiplier conditions for the following two problems:

$$(1) \text{ minimize } f(\mathbf x) \text{ subject to constraint } g(\mathbf x)=b $$

$$(2) \text{ maximize } g(\mathbf x) \text{ subject to constraint } f(\mathbf x)=c $$

Given a solution $\mathbf y$ to first and second order conditions for the minimization problem $(1)$, for some value of $b$, set $c=f(\mathbf y)$ and consider the maximization problem $(2)$; when is $\mathbf y$ a solution to this problem as well?

**My Questions:**

I believe I have the first-order conditions for $(1)$:

$$ \frac{\partial \mathcal{L}(\mathbf x)}{\partial x\_i} = \frac{\partial f(\mathbf x)}{\partial x\_i}-\lambda \frac{\partial g(\mathbf x)}{\partial x\_i}=0$$

But what are the second-order conditions? I know that, in order to minimize $f$, $\mathcal{L}$ must be convex and that the constraint set must be convex, but are those the "second-order conditions"? If so, how do I determine them? By way of bordered Hessians? That doesn't seem right. Anyway, any help here would be appreciated. Thanks!

**Tags:** mathematical-economics, optimization
> **Comments:**
> **username_1**: Why that doesn't seem right by way of bordered Hessians?  
> **username_2**: What you write as a "first-order condition", is not -it is just the first derivative of the Lagrangean. What is _the condition_ on it? Also, convexity is a convenient property to guarantee that the second-order conditions hold, and so that the first-order conditions are sufficient and necessary for a minimum. As with univariate optimization, where we look into second derivatives, here we have to look into Hessians/border Hessians. These _are_ the 2nd-order conditions, not a "way to verify them".  
> **username_0**: Thanks. So, yeah, I forgot to put the "$=0$" in there for the "first-order conditions." And, right, so by checking the definite-ness of the bordered Hessians, I can determine whether or not $\mathcal{L}$ is concave/convex... which is the second-order condition. However, I don't really have a way of determining the definiteness of the Hessian because I don't actually have any values to work with. Furthermore, don't I have to consider "regularity"? Sorry, I'm really confused about all of this...  
> **username_2**: Since you are given no other information, you might as well assume convexity, or just "for $y$ to be a solution, second order conditions _must_ be satisfied". After all, this is obviously an exercise in duality.  

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## 0 answers

