<p>The only thing that changes is the <span class="math-container">$k$</span> first order condition.</p>
<p>The one you got with the non-negativity constraint Lagrange multiplier is</p>
<p><span class="math-container">$-r + \lambda \alpha L^{1-\alpha} + \mu_k \leq 0$</span></p>
<p>Substracting <span class="math-container">$\mu_k$</span>, and since <span class="math-container">$\mu_k \geq 0$</span>,</p>
<p><span class="math-container">$-r + \lambda \alpha L^{1-\alpha} \leq - \mu_k \leq 0$</span></p>
<p>Therefore,</p>
<p><span class="math-container">$-r + \lambda \alpha L^{1-\alpha} \leq 0$</span>.</p>
<p>This last inequality is the one used in the screenshot’s solution.</p>
<p>This is consistent with <span class="math-container">$k = 0$</span> because as long as this last inequality holds, there exists values of <span class="math-container">$\mu_k \geq 0$</span> such that your inequality holds.</p>
<p>You already found those values of <span class="math-container">$\mu_k$</span>.</p>

CC BY-SA 4.0
2023-05-06T17:11:10.250Z
Nicolas this looks like what i might be looking for! Which would be amazing, i'll comment again when i've properly reviewed, thanks!
CormJack
CC BY-SA 4.0
2023-05-08T11:00:04.510Z
Hi Nicholas, i've thought about this some more. It's so helpful. I wish i could upvote it twice! It's so obvious now, but this little trick is really helpful. Would you agree implicitly this is what they are doing with the screenshot? As surely when you test the case for k = 0, they should be using a non-negativity constraint. So this is how they have got rid of it. Thanks so much! Any other clever tricks like this to do with KKT, or constraints you think i should know?
CormJack
CC BY-SA 4.0
2023-05-08T11:25:03.897Z
One thing i'm uncertain about is why don't they state the specific condition on $u_k$ that i found? And don't we need to show that $ u_k \ge 0$ i'm not sure how to show this here?
CormJack
CC BY-SA 4.0
2023-05-08T15:53:08.913Z
@CormJack Probably in the context of the problem, the value of $\mu_k$ isn’t relevant. I think when the non-negativity constraint is binding, $\mu_k$ is the marginal utility of $k$, as this would be analogous to $\lambda$ being the marginal utility of the budget/production constraint in consumer/firm theory aka shadow price.
Nicolas Torres
CC BY-SA 4.0
2024-04-22T19:26:46.523Z
Hi @NicolasTorres, hope you're well! I hadn't noticed there's the comment from you before it's actually very helpful. Thank you! I I have a quick question regarding a minimisation problem and the non-negativity constraint. If you get a chance to look, that would be amazing: https://economics.stackexchange.com/questions/55496/minimisation-problem-turned-into-maximisation
CormJack
CC BY-SA 4.0
2023-05-06T16:42:48.203Z
2023-05-06T16:42:48.203Z
Nicolas Torres
<p>I'm expanding on my question here as it's still causing trouble. It's a real pain so any additional thoughts are very appreciated!!</p>
<ul>
<li><p>I believe that given their is a possibility <span class="math-container">$k = 0$</span> then the Lagrangian should have a non-negativity constraint on <span class="math-container">$k$</span> i.e. <span class="math-container">$k \ge 0 \implies -k \le 0$</span></p>
</li>
<li><p>This means that their Lagrangian is missing this constraint and the corresponding multiplier <span class="math-container">$λ_2$</span> i.e. <span class="math-container">$ + λ_2(k)$</span></p>
</li>
<li><p>They clearly think this is needed as they consider the case for <span class="math-container">$k = 0$</span> i.e. this constraint is binding.</p>
</li>
<li><p>In which case i do not understand why they never use <span class="math-container">$λ_2$</span>? I also don't see why it's trivial, as including <span class="math-container">$λ_2$</span> seems to ruin the analysis at the end. See in red in the screenshot.</p>
</li>
</ul>
<p><a href="https://i.sstatic.net/PhBt9.jpg" rel="nofollow noreferrer"><img src="https://i.sstatic.net/PhBt9.jpg" alt="enter image description here" /></a></p>

CC BY-SA 4.0
2023-05-06T14:33:53.550Z
2023-05-06T14:33:53.550Z
CormJack
<p>We have the constrained maximisation problem:</p>
<blockquote>
<p>A perfectly competitive firm produces one output with two inputs, capital <span class="math-container">$(k)$</span> and labour <span class="math-container">$(l)$</span>. The rental cost of capital is equal to <span class="math-container">$r &gt;0$</span> and the wage rate is equal to <span class="math-container">$w&gt;0$</span>. The production function is <span class="math-container">$f(k, l)=(k + 1)^α l^{1-α}$</span> with <span class="math-container">$0 &lt;α&lt;1$</span>. The firm
wishes to minimise total costs while achieving a given level of production <span class="math-container">$y&gt;0$</span>.</p>
</blockquote>
<p>Now for the sake of this course we construct this as a maximisation, problem I am particularly interested in the case where <span class="math-container">$k=0$</span> and <span class="math-container">$l&gt;0$</span>. While they do specifically consider this case they <strong>don't write the non-negativity constraint into the Lagrangian, and i believe this effects the analysis at the end.</strong> I have attached a screenshot of the &quot;answer.&quot;</p>
<blockquote>
<p>UPDATE: The specific equation they derive in the screenshot actually goes on to be very important in the following question. So thoughts are very much appreciated, thank!</p>
</blockquote>
<p><span class="math-container">$\mathcal{L}(k, l, \lambda, \mu_k, \mu_l) = -rk - wl + \lambda [(k + 1)^\alpha l^{1 - \alpha} - y] + \mu_k k + \mu_l l$</span>`</p>
<ol>
<li>With respect to k:</li>
</ol>
<p><span class="math-container">$\frac{\partial \mathcal{L}}{\partial k} = -r + \lambda \alpha (k + 1)^{\alpha - 1} l^{1 - \alpha} + \mu_k$</span></p>
<ol start="2">
<li>With respect to l:</li>
</ol>
<p><span class="math-container">$\frac{\partial \mathcal{L}}{\partial l} = -w + \lambda (1 - \alpha) (k + 1)^\alpha l^{-\alpha} + \mu_l$</span></p>
<ol start="3">
<li>With respect to λ:</li>
</ol>
<p><span class="math-container">$\frac{\partial \mathcal{L}}{\partial \lambda} = (k + 1)^\alpha l^{1 - \alpha} - y$</span></p>
<p>Now, consider the specific case where <span class="math-container">$k = 0$</span> and <span class="math-container">$l &gt; 0$</span>:`</p>
<ol>
<li><p><span class="math-container">$\frac{\partial \mathcal{L}}{\partial k}\bigg|_{k=0} = -r + \lambda \alpha l^{1 - \alpha} + \mu_k \le 0$</span></p>
</li>
<li><p><span class="math-container">$\frac{\partial \mathcal{L}}{\partial l}\bigg|_{k=0} = -w + \lambda (1 - \alpha) l^{-\alpha} = 0$</span></p>
</li>
<li><p><span class="math-container">$\frac{\partial \mathcal{L}}{\partial \lambda}\bigg|_{k=0} = l^{1 - \alpha} - y = 0$</span></p>
</li>
</ol>
<ul>
<li><span class="math-container">$k^* = 0$</span></li>
<li><span class="math-container">$l^* = y^{\frac{1}{1-\alpha}}$</span></li>
<li><span class="math-container">$λ = \frac{w}{1-\alpha}y^{\frac{\alpha}{1-\alpha}}$</span></li>
</ul>
<ol>
<li><strong>They solve for for a condition on <span class="math-container">$y$</span> (see screenshot)</strong> consistent with <span class="math-container">$k = 0$</span> but their lack of an explicit use of a <span class="math-container">$\mu_k$</span>  for our non-negativity on <span class="math-container">$k$</span> results in a different condition to what i would get, which would include <span class="math-container">$\mu_k$</span> . <strong>Why have they not explicitly used <span class="math-container">$\mu_k$</span> and <span class="math-container">$\mu_l$</span> as i have??</strong></li>
<li>I also get a weird condition on <span class="math-container">$\mu_k$</span> by just substituting in <span class="math-container">$l^*$</span> and <span class="math-container">$λ^*$</span> does it looks right?</li>
</ol>
<ul>
<li><span class="math-container">$\mu_k \le r - \frac{\alpha wy^{\frac{1}{1-\alpha}}}{1-\alpha}$</span></li>
</ul>
<p><strong>Screen Shot from Answer showing condition on y</strong></p>
<p><a href="https://i.sstatic.net/WUyJE.jpg" rel="nofollow noreferrer"><img src="https://i.sstatic.net/WUyJE.jpg" alt="enter image description here" /></a></p>

CC BY-SA 4.0
2023-05-04T13:05:30.570Z
The production function is not well-defined for $l<0$ which would imply a fractional (in the sense of non-integer) power of a negative quantity - see [here](https://math.stackexchange.com/questions/317528/how-do-you-compute-negative-numbers-to-fractional-powers).
Adam Bailey
CC BY-SA 4.0
2023-05-04T13:24:33.990Z
Your FOC's 2 and 3 are equivalent to those in the screen shot, so should lead to the same formula for $\lambda$.  Possibly it's just a typo, but your formula should have $y^{\frac{\alpha}{1-\alpha}}$ not $y^{\frac{1}{1-\alpha}}$.  The "weird" condition on $\mu_k$ will then become $\mu_k \leq r-\frac{\alpha wy^{\frac{1}{1-\alpha}}}{1-\alpha}$.
Adam Bailey
CC BY-SA 4.0
2023-05-06T14:34:48.563Z
Hi Adam, thanks again for your help. I understand why we don't worry about $l < 0$, but regarding $k = 0$ and the negativity constraint i have tried to elaborate why it's bothering me below. Thanks!
CormJack
CC BY-SA 4.0
2023-05-06T16:46:47.503Z
I just edited out the typos I spotted
Nicolas Torres
<mathematical-economics><optimization><producer-theory>
Missing Non-Negativity Constraint?
CormJack
# Missing Non-Negativity Constraint?

(Asked by: username_0 on 2023-04-21)

We have the constrained maximisation problem:

> A perfectly competitive firm produces one output with two inputs, capital $(k)$ and labour $(l)$. The rental cost of capital is equal to $r >0$ and the wage rate is equal to $w>0$. The production function is $f(k, l)=(k + 1)^α l^{1-α}$ with $0 <α<1$. The firm
> wishes to minimise total costs while achieving a given level of production $y>0$.

Now for the sake of this course we construct this as a maximisation, problem I am particularly interested in the case where $k=0$ and $l>0$. While they do specifically consider this case they **don't write the non-negativity constraint into the Lagrangian, and i believe this effects the analysis at the end.** I have attached a screenshot of the "answer."

> UPDATE: The specific equation they derive in the screenshot actually goes on to be very important in the following question. So thoughts are very much appreciated, thank!

$\mathcal{L}(k, l, \lambda, \mu\_k, \mu\_l) = -rk - wl + \lambda [(k + 1)^\alpha l^{1 - \alpha} - y] + \mu\_k k + \mu\_l l$`

1. With respect to k:

$\frac{\partial \mathcal{L}}{\partial k} = -r + \lambda \alpha (k + 1)^{\alpha - 1} l^{1 - \alpha} + \mu\_k$

2. With respect to l:

$\frac{\partial \mathcal{L}}{\partial l} = -w + \lambda (1 - \alpha) (k + 1)^\alpha l^{-\alpha} + \mu\_l$

3. With respect to λ:

$\frac{\partial \mathcal{L}}{\partial \lambda} = (k + 1)^\alpha l^{1 - \alpha} - y$

Now, consider the specific case where $k = 0$ and $l > 0$:`

1. $\frac{\partial \mathcal{L}}{\partial k}\bigg|\_{k=0} = -r + \lambda \alpha l^{1 - \alpha} + \mu\_k \le 0$
2. $\frac{\partial \mathcal{L}}{\partial l}\bigg|\_{k=0} = -w + \lambda (1 - \alpha) l^{-\alpha} = 0$
3. $\frac{\partial \mathcal{L}}{\partial \lambda}\bigg|\_{k=0} = l^{1 - \alpha} - y = 0$

* $k^\* = 0$
* $l^\* = y^{\frac{1}{1-\alpha}}$
* $λ = \frac{w}{1-\alpha}y^{\frac{\alpha}{1-\alpha}}$

1. **They solve for for a condition on $y$ (see screenshot)** consistent with $k = 0$ but their lack of an explicit use of a $\mu\_k$ for our non-negativity on $k$ results in a different condition to what i would get, which would include $\mu\_k$ . **Why have they not explicitly used $\mu\_k$ and $\mu\_l$ as i have??**
2. I also get a weird condition on $\mu\_k$ by just substituting in $l^\*$ and $λ^\*$ does it looks right?

* $\mu\_k \le r - \frac{\alpha wy^{\frac{1}{1-\alpha}}}{1-\alpha}$

**Screen Shot from Answer showing condition on y**

[![enter image description here](https://i.sstatic.net/WUyJE.jpg)](https://i.sstatic.net/WUyJE.jpg)

**Tags:** mathematical-economics, optimization, producer-theory
> **Comments:**
> **username_1**: The production function is not well-defined for $l<0$ which would imply a fractional (in the sense of non-integer) power of a negative quantity - see [here](https://math.stackexchange.com/questions/317528/how-do-you-compute-negative-numbers-to-fractional-powers).  
> **username_1**: Your FOC's 2 and 3 are equivalent to those in the screen shot, so should lead to the same formula for $\lambda$.  Possibly it's just a typo, but your formula should have $y^{\frac{\alpha}{1-\alpha}}$ not $y^{\frac{1}{1-\alpha}}$.  The "weird" condition on $\mu_k$ will then become $\mu_k \leq r-\frac{\alpha wy^{\frac{1}{1-\alpha}}}{1-\alpha}$.  
> **username_0**: Hi Adam, thanks again for your help. I understand why we don't worry about $l < 0$, but regarding $k = 0$ and the negativity constraint i have tried to elaborate why it's bothering me below. Thanks!  
> **username_2**: I just edited out the typos I spotted  

---

## 2 answers

### Answer by username_2 on 2023-05-06. Score: 1 | Accepted answer

The only thing that changes is the $k$ first order condition.

The one you got with the non-negativity constraint Lagrange multiplier is

$-r + \lambda \alpha L^{1-\alpha} + \mu\_k \leq 0$

Substracting $\mu\_k$, and since $\mu\_k \geq 0$,

$-r + \lambda \alpha L^{1-\alpha} \leq - \mu\_k \leq 0$

Therefore,

$-r + \lambda \alpha L^{1-\alpha} \leq 0$.

This last inequality is the one used in the screenshot’s solution.

This is consistent with $k = 0$ because as long as this last inequality holds, there exists values of $\mu\_k \geq 0$ such that your inequality holds.

You already found those values of $\mu\_k$.

> **Comments:**
> **username_0**: Nicolas this looks like what i might be looking for! Which would be amazing, i'll comment again when i've properly reviewed, thanks!  
> **username_0**: Hi Nicholas, i've thought about this some more. It's so helpful. I wish i could upvote it twice! It's so obvious now, but this little trick is really helpful. Would you agree implicitly this is what they are doing with the screenshot? As surely when you test the case for k = 0, they should be using a non-negativity constraint. So this is how they have got rid of it. Thanks so much! Any other clever tricks like this to do with KKT, or constraints you think i should know?  
> **username_0**: One thing i'm uncertain about is why don't they state the specific condition on $u_k$ that i found? And don't we need to show that $ u_k \ge 0$ i'm not sure how to show this here?  
> **username_2**: @CormJack Probably in the context of the problem, the value of $\mu_k$ isn’t relevant. I think when the non-negativity constraint is binding, $\mu_k$ is the marginal utility of $k$, as this would be analogous to $\lambda$ being the marginal utility of the budget/production constraint in consumer/firm theory aka shadow price.  
> **username_0**: Hi @NicolasTorres, hope you're well! I hadn't noticed there's the comment from you before it's actually very helpful. Thank you! I I have a quick question regarding a minimisation problem and the non-negativity constraint. If you get a chance to look, that would be amazing: https://economics.stackexchange.com/questions/55496/minimisation-problem-turned-into-maximisation  

---

### Answer by username_0 on 2023-05-06. Score: 0

I'm expanding on my question here as it's still causing trouble. It's a real pain so any additional thoughts are very appreciated!!

* I believe that given their is a possibility $k = 0$ then the Lagrangian should have a non-negativity constraint on $k$ i.e. $k \ge 0 \implies -k \le 0$
* This means that their Lagrangian is missing this constraint and the corresponding multiplier $λ\_2$ i.e. $ + λ\_2(k)$
* They clearly think this is needed as they consider the case for $k = 0$ i.e. this constraint is binding.
* In which case i do not understand why they never use $λ\_2$? I also don't see why it's trivial, as including $λ\_2$ seems to ruin the analysis at the end. See in red in the screenshot.

[![enter image description here](https://i.sstatic.net/PhBt9.jpg)](https://i.sstatic.net/PhBt9.jpg)


---

