<blockquote>
<p><span class="math-container">$$H=\begin{bmatrix} -\frac{5\sqrt{5}}{8y^{3/2}} &amp; 0 \\ 0 &amp; -\frac{5\sqrt{5}}{2y^{3/2}}  \end{bmatrix}$$</span>
Now this is where things get messy for me. I know that if the Hessian is semi positive definite, then we have a minimum. If we have that the hessian is semi negative definite, then we have a maximum. This Hessian doesn't appear to be either. I believe this indicates to us that we have a saddle point.</p>
</blockquote>
<p>The Hessian above has two eigenvalues (denoted by <span class="math-container">$\lambda$</span>):
<span class="math-container">$$
\lambda_1 = -\frac{5\sqrt{5}}{8y^{3/2}}
\ \text{ and } \
\lambda_2 = -\frac{5\sqrt{5}}{2y^{3/2}}
$$</span>
The expression <span class="math-container">$y^{3/2}$</span> is defined only when <span class="math-container">$y \geq 0$</span>. This implies that both eigenvalues are negative or undefined. Thus the Hessian does not appear to be indefinite.</p>

CC BY-SA 4.0
2021-10-20T16:08:45.257Z
Thank you, this was a rookie mistake. This next question is more of a math question,  why am I finding that |H|^(1)<0 and |H|^(2)=|H|>0 (ie. since the upper left entry is negative and the lower right entry is negative). Wouldn't the determinant be positive?
Mistah White
CC BY-SA 4.0
2021-10-20T16:10:39.783Z
What exactly are you denoting by $|H|^{(1)}<0$?
Giskard
CC BY-SA 4.0
2021-10-20T16:11:35.340Z
@MistahWhite Perhaps you should edit your calculations into the body of the question, as otherwise it may be nigh-on-impossible to determine why you are reaching erroneus results.
Giskard
CC BY-SA 4.0
2021-10-20T16:18:23.230Z
Thank you for your suggestion. I just edited the calculations into the question.
Mistah White
CC BY-SA 4.0
2021-10-20T16:21:42.290Z
@MistahWhite I think you may have misread [Sylvester's criterion](https://en.wikipedia.org/wiki/Sylvester%27s_criterion)? Or why is it that you think this contradicts negative definiteness?
Giskard
CC BY-SA 4.0
2021-10-20T16:25:53.557Z
For semi negative definiteness, don't we need all upper left submatrices to have a negative or zero valued determinant?
Mistah White
CC BY-SA 4.0
2021-10-20T16:42:40.403Z
@MistahWhite Can you support the statement (that we need this) with a reasoning or a reference?
Giskard
CC BY-SA 4.0
2021-10-20T15:56:34.387Z
2021-10-20T15:56:34.387Z
Giskard
<p>I am currently getting my Masters in Economics. I did not get any exposure to optimization with inequality constraints in my undergrad. I would like to ensure that I am doing this problem correctly. Note: I understand that Khun-Tucker is not necessary for solving this following problem. That being said, it is important that I master this technique, so I want all the practice I can get:</p>
<p>Problem: A firm has two plants with cost functions <span class="math-container">$c_{1}(y_{1})=4\sqrt{y_{1}}$</span> and <span class="math-container">$c_{2}(y_{2})=2\sqrt{y_{2}}$</span>. What is the cost of producing <span class="math-container">$y=y_{1}+y_{2}$</span>? Assume that <span class="math-container">$y_{1},y_{2} \geq 0$</span></p>
<p>I set up the Lagrange in the following way:</p>
<p><span class="math-container">$$\mathcal{L}=4\sqrt{y_{1}}+2\sqrt{y_{2}}-\lambda(y_{1}+y_{2}-y)-\mu_{1}y_{1}-\mu_{2}y_{2}$$</span></p>
<p>The FOC:
<span class="math-container">$$\mathcal{L}_{y_{1}}=\frac{2}{\sqrt{y_{1}}}-\lambda-\mu_{1}=0$$</span>
<span class="math-container">$$\mathcal{L}_{y_{2}}=\frac{1}{\sqrt{y_{2}}}-\lambda-\mu_{2}=0$$</span>
<span class="math-container">$$\mathcal{L}_{\lambda}=0 \Longrightarrow y_{1}+y_{2}=y$$</span>
<span class="math-container">$$\mu_{1}y_{1}=0$$</span>
<span class="math-container">$$\mu_{2}y_{2}=0$$</span></p>
<p>My complementary slackness conditions are:
<span class="math-container">$$\mu_{1} \geq 0; \ \mu_{1}=0 \ if \ y_{1}&gt;0$$</span>
<span class="math-container">$$\mu_{2} \geq 0; \ \mu_{2}=0 \ if \ y_{2}&gt;0$$</span></p>
<p>I first check for an interior solution. That is, I look to see if there is a solution where <span class="math-container">$y_{1},y_{2}&gt;0$</span>. This means that <span class="math-container">$\mu_{1}=\mu_{2}=0$</span>. This means that
<span class="math-container">$$\frac{2}{\sqrt{y_{1}}}=\lambda \ \ and \ \ \frac{1}{\sqrt{y_{2}}}=\lambda$$</span>
I can take the ratio to find that
<span class="math-container">$$\frac{2 \sqrt{y_{2}}}{\sqrt{y_{1}}}=1$$</span>
Using the fact that <span class="math-container">$y_{1}+y_{2}=y$</span>, we can see that we have the following solution
<span class="math-container">$$y_{1}=\frac{4}{5}y \ \ and \ \ y_{2}=\frac{1}{5}y$$</span></p>
<p>I now check the second order conditions. We have <span class="math-container">$C(y)=4\sqrt{y_{1}}+2\sqrt{y_{2}}$</span> which means
<span class="math-container">$$C_{1}=\frac{2}{\sqrt{y_{1}}}, \ \ C_{2}=\frac{1}{\sqrt{y_{2}}}$$</span>
<span class="math-container">$$\Longrightarrow C_{11}=-\frac{1}{y_{1}^{3/2}}, \ \ C_{22}=-\frac{1}{2y_{2}^{3/2}}, \ \ C_{12}=C_{21}=0$$</span>
Plugging in the optimal values into the hessian give us:</p>
<p><span class="math-container">$$H=\begin{bmatrix} -\frac{5\sqrt{5}}{8y^{3/2}} &amp; 0 \\ 0 &amp; -\frac{5\sqrt{5}}{2y^{3/2}}  \end{bmatrix}$$</span></p>
<p>Now this is where things get messy for me. I know that if the Hessian is semi positive definite, then we have a minimum. If we have that the hessian is semi negative definite, then we have a maximum. This Hessian doesn't appear to be either. I believe this indicates to us that we have a saddle point.</p>
<p>That is, using principal minors, we have:</p>
<p><span class="math-container">$$\mid H \mid ^{(1)}&lt;0, \ \ and \ \ \mid H \mid ^{(2)}=|H|&gt;0$$</span></p>
<p>That is,
<span class="math-container">$$|H|^{(1)}=det(-\frac{5\sqrt{5}}{8y^{3/2}})=-\frac{5\sqrt{5}}{8y^{3/2}}&lt;0$$</span></p>
<p><span class="math-container">$$|H|^{(2)}=|H|=det(\begin{bmatrix} -\frac{5\sqrt{5}}{8y^{3/2}} &amp; 0 \\ 0 &amp; -\frac{5\sqrt{5}}{2y^{3/2}}  \end{bmatrix})=(-\frac{5\sqrt{5}}{8y^{3/2}})(-\frac{5\sqrt{5}}{2y^{3/2}})-0=\frac{125}{16y^{3}}&gt;0$$</span></p>
<p>That being said, I'm pretty sure we are supposed to find that the interior solution is a maximum. I believe I have done something wrong, but I am not sure what.</p>

<optimization><cost-functions>
Solving the following cost minimization problem using Kuhn-Tucker conditions
Mistah White
# Solving the following cost minimization problem using Kuhn-Tucker conditions

(Asked by: username_0 on 2021-10-20)

I am currently getting my Masters in Economics. I did not get any exposure to optimization with inequality constraints in my undergrad. I would like to ensure that I am doing this problem correctly. Note: I understand that Khun-Tucker is not necessary for solving this following problem. That being said, it is important that I master this technique, so I want all the practice I can get:

Problem: A firm has two plants with cost functions $c\_{1}(y\_{1})=4\sqrt{y\_{1}}$ and $c\_{2}(y\_{2})=2\sqrt{y\_{2}}$. What is the cost of producing $y=y\_{1}+y\_{2}$? Assume that $y\_{1},y\_{2} \geq 0$

I set up the Lagrange in the following way:

$$\mathcal{L}=4\sqrt{y\_{1}}+2\sqrt{y\_{2}}-\lambda(y\_{1}+y\_{2}-y)-\mu\_{1}y\_{1}-\mu\_{2}y\_{2}$$

The FOC:
$$\mathcal{L}\_{y\_{1}}=\frac{2}{\sqrt{y\_{1}}}-\lambda-\mu\_{1}=0$$
$$\mathcal{L}\_{y\_{2}}=\frac{1}{\sqrt{y\_{2}}}-\lambda-\mu\_{2}=0$$
$$\mathcal{L}\_{\lambda}=0 \Longrightarrow y\_{1}+y\_{2}=y$$
$$\mu\_{1}y\_{1}=0$$
$$\mu\_{2}y\_{2}=0$$

My complementary slackness conditions are:
$$\mu\_{1} \geq 0; \ \mu\_{1}=0 \ if \ y\_{1}>0$$
$$\mu\_{2} \geq 0; \ \mu\_{2}=0 \ if \ y\_{2}>0$$

I first check for an interior solution. That is, I look to see if there is a solution where $y\_{1},y\_{2}>0$. This means that $\mu\_{1}=\mu\_{2}=0$. This means that
$$\frac{2}{\sqrt{y\_{1}}}=\lambda \ \ and \ \ \frac{1}{\sqrt{y\_{2}}}=\lambda$$
I can take the ratio to find that
$$\frac{2 \sqrt{y\_{2}}}{\sqrt{y\_{1}}}=1$$
Using the fact that $y\_{1}+y\_{2}=y$, we can see that we have the following solution
$$y\_{1}=\frac{4}{5}y \ \ and \ \ y\_{2}=\frac{1}{5}y$$

I now check the second order conditions. We have $C(y)=4\sqrt{y\_{1}}+2\sqrt{y\_{2}}$ which means
$$C\_{1}=\frac{2}{\sqrt{y\_{1}}}, \ \ C\_{2}=\frac{1}{\sqrt{y\_{2}}}$$
$$\Longrightarrow C\_{11}=-\frac{1}{y\_{1}^{3/2}}, \ \ C\_{22}=-\frac{1}{2y\_{2}^{3/2}}, \ \ C\_{12}=C\_{21}=0$$
Plugging in the optimal values into the hessian give us:

$$H=\begin{bmatrix} -\frac{5\sqrt{5}}{8y^{3/2}} & 0 \\ 0 & -\frac{5\sqrt{5}}{2y^{3/2}} \end{bmatrix}$$

Now this is where things get messy for me. I know that if the Hessian is semi positive definite, then we have a minimum. If we have that the hessian is semi negative definite, then we have a maximum. This Hessian doesn't appear to be either. I believe this indicates to us that we have a saddle point.

That is, using principal minors, we have:

$$\mid H \mid ^{(1)}<0, \ \ and \ \ \mid H \mid ^{(2)}=|H|>0$$

That is,
$$|H|^{(1)}=det(-\frac{5\sqrt{5}}{8y^{3/2}})=-\frac{5\sqrt{5}}{8y^{3/2}}<0$$

$$|H|^{(2)}=|H|=det(\begin{bmatrix} -\frac{5\sqrt{5}}{8y^{3/2}} & 0 \\ 0 & -\frac{5\sqrt{5}}{2y^{3/2}} \end{bmatrix})=(-\frac{5\sqrt{5}}{8y^{3/2}})(-\frac{5\sqrt{5}}{2y^{3/2}})-0=\frac{125}{16y^{3}}>0$$

That being said, I'm pretty sure we are supposed to find that the interior solution is a maximum. I believe I have done something wrong, but I am not sure what.

**Tags:** optimization, cost-functions

---

## 1 answers

### Answer by username_1 on 2021-10-20. Score: 2

> $$H=\begin{bmatrix} -\frac{5\sqrt{5}}{8y^{3/2}} & 0 \\ 0 & -\frac{5\sqrt{5}}{2y^{3/2}} \end{bmatrix}$$
> Now this is where things get messy for me. I know that if the Hessian is semi positive definite, then we have a minimum. If we have that the hessian is semi negative definite, then we have a maximum. This Hessian doesn't appear to be either. I believe this indicates to us that we have a saddle point.

The Hessian above has two eigenvalues (denoted by $\lambda$):
$$
\lambda\_1 = -\frac{5\sqrt{5}}{8y^{3/2}}
\ \text{ and } \
\lambda\_2 = -\frac{5\sqrt{5}}{2y^{3/2}}
$$
The expression $y^{3/2}$ is defined only when $y \geq 0$. This implies that both eigenvalues are negative or undefined. Thus the Hessian does not appear to be indefinite.

> **Comments:**
> **username_0**: Thank you, this was a rookie mistake. This next question is more of a math question,  why am I finding that |H|^(1)<0 and |H|^(2)=|H|>0 (ie. since the upper left entry is negative and the lower right entry is negative). Wouldn't the determinant be positive?  
> **username_1**: What exactly are you denoting by $|H|^{(1)}<0$?  
> **username_1**: @MistahWhite Perhaps you should edit your calculations into the body of the question, as otherwise it may be nigh-on-impossible to determine why you are reaching erroneus results.  
> **username_0**: Thank you for your suggestion. I just edited the calculations into the question.  
> **username_1**: @MistahWhite I think you may have misread [Sylvester's criterion](https://en.wikipedia.org/wiki/Sylvester%27s_criterion)? Or why is it that you think this contradicts negative definiteness?  
> **username_0**: For semi negative definiteness, don't we need all upper left submatrices to have a negative or zero valued determinant?  
> **username_1**: @MistahWhite Can you support the statement (that we need this) with a reasoning or a reference?  

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