<p>Since we are interested in finding the unit cost let the output needed to be produced be equal to 1 i.e., <span class="math-container">$y(i)\overset{set}=1$</span>
We need to solve: <span class="math-container">$$\begin{align}
\min_{l(i),x(i),h(i)\geq0} \quad &amp; w_Ll(i)+x(i)+w_{H}h(i)\\
\textrm{s.t.} \quad &amp; [l(i)^{\frac{\epsilon - 1}{\epsilon}} +\alpha(i)(\tilde{\gamma}x(i))^{\frac{\epsilon - 1}{\epsilon}}]^{\frac{\epsilon \beta}{\epsilon-1}}h(i)^{1-\beta}=1\\
\textrm{and} \quad &amp; \alpha(i)=\begin{cases}1 &amp; \text{if } i\text{ is automated}\\ 0 &amp; \text{otherwise}\end{cases}
\end{align}$$</span></p>
<p>I am going to use the indicator function and solve for two different cases and then try to combine them using the indicator.</p>
<p><strong>Case 1:</strong> If <span class="math-container">$i$</span> is not automated i.e., <span class="math-container">$ \alpha(i)=0$</span></p>
<p>If <span class="math-container">$i$</span> is not automated, then the cost minimization problem becomes a standard one with a Cobb-Douglas  production function
<span class="math-container">$$\begin{align}
\min_{l,x,h\geq 0} \quad &amp; w_Ll+x+w_{H}h\\
\textrm{s.t.} \quad &amp; l^{\beta}h^{1-\beta}=1\\
\end{align}$$</span></p>
<p>This is just a standard cost minimization problem with a cobb-douglas production function, so I think you can verify yourself that the above problem  gives: <span class="math-container">$l(i)=\left(\frac{w_H\beta}{w_L(1-\beta)}\right)^{1-\beta}, \quad x(i)=0, \quad h(i)= \left(\frac{w_L(1-\beta)}{w_{H} \beta}\right)^{\beta}$</span></p>
<p>therefore, <span class="math-container">$$\begin{eqnarray}
&amp; c=w_L^\beta\left(\frac{w_H\beta}{1-\beta}\right)^{1-\beta}+w_H^{1-\beta}\left(\frac{w_L(1-\beta)}{\beta}\right)^{\beta}\\\\
\implies &amp; {\boxed{c=w_L^\beta w_H^{1-\beta} \beta^{-\beta}(1-\beta)^{\beta-1}}}\tag{a}
\end{eqnarray}$$</span></p>
<p><strong>Case 2:</strong> If <span class="math-container">$i$</span> is automated
<span class="math-container">$$\begin{align}
\min_{l,x,h\geq 0} \quad &amp; w_Ll+x+w_{H}h\\
\textrm{s.t.} \quad &amp; [l^{\frac{\epsilon - 1}{\epsilon}} +(\tilde{\gamma}x)^{\frac{\epsilon - 1}{\epsilon}}]^{\frac{\epsilon \beta}{\epsilon-1}}h^{1-\beta}=1\\
\end{align}$$</span></p>
<p>let us write the above problem as:
<span class="math-container">$$\begin{align}
{\min_{0\leq h \leq 1}
\underbrace{\begin{pmatrix} \underset{l,x}{\min} \quad &amp; w_Ll+w_Hh+x \\
\textrm{s.t.} \quad &amp; [l^{\frac{\epsilon - 1}{\epsilon}} +(\tilde{\gamma}x)^{\frac{\epsilon - 1}{\epsilon}}]^{\frac{\epsilon}{\epsilon-1}}=h^\frac{\beta -1}{\beta} \end{pmatrix}}_{\text{auxiliary problem}}} \tag{1} 
\end{align}$$</span></p>
<p>We can solve the above problem in two stages:</p>
<p><strong>Stage 1:</strong> first solve the auxiliary problem for <span class="math-container">$l$</span> and <span class="math-container">$x$</span> holding <span class="math-container">$h$</span> fixed</p>
<p><strong>Stage 2:</strong> Substitute the solutions of the auxiliary problem<span class="math-container">$-$</span> <span class="math-container">$l(h)$</span> and <span class="math-container">$x(h)$</span> <span class="math-container">$-$</span> into <span class="math-container">$(1)$</span> and solve the problem to find optimal <span class="math-container">$h$</span></p>
<p><strong>Stage 1:</strong>
we need to solve: <span class="math-container">$$\begin{align} 
\underset{l,x}{\min} \quad &amp; w_Ll+w_Hh+x \\
\textrm{s.t.} \quad &amp; [l^{\frac{\epsilon - 1}{\epsilon}} +(\tilde{\gamma}x)^{\frac{\epsilon - 1}{\epsilon}}]^{\frac{\epsilon}{\epsilon-1}}=h^\frac{\beta -1}{\beta}\\
\text{gives:} \quad &amp; {\left.\begin{matrix} l(h)=\frac{h^\frac{\beta -1}{\beta}}{[1+(w_L\tilde \gamma)^{\epsilon -1}]^\frac{\epsilon}{\epsilon-1}} \\ x(h)=\frac{w_L^\epsilon \tilde{\gamma}^{\epsilon -1} h^\frac{\beta -1}{\beta}}{[1+(w_L\tilde \gamma)^{\epsilon -1}]^\frac{\epsilon}{\epsilon-1}}
\end{matrix}\right\}}\tag{2}
\end{align}$$</span></p>
<p>Because we are given that <span class="math-container">$\epsilon&gt;1$</span>, You can verify that isoquants will be convex. Thus, I was able to solve the above by equating ratios of MP to MC for <span class="math-container">$l$</span> and <span class="math-container">$x$</span> and substituting them into the constraint.</p>
<p><strong>Stage 2:</strong> substituting <span class="math-container">$(2)$</span> in <span class="math-container">$(1)$</span> we get the final problem solving which will give optimal <span class="math-container">$h$</span>
<span class="math-container">$$\begin{align}
\min_{0\leq h\leq 1} \quad &amp; w_Hh+w_L[1+(w_L\tilde \gamma)^{\epsilon -1}]^\frac{1}{1-\epsilon} h^\frac{\beta -1}{\beta}\tag{3}
\end{align}$$</span></p>
<p>the above objective is convex in <span class="math-container">$h \; \because \beta&lt;1$</span>, so assuming the parametric condition gives us a stationary point that lies in the interval <span class="math-container">$(0,1)$</span> we can solve for the optimal h in the above problem and substitute it in the problem itself to get <span class="math-container">$c$</span> for <span class="math-container">$\alpha=1$</span>:</p>
<p><span class="math-container">$$h^*=\left(\frac{(1-\beta)w_L[1+(w_L\tilde \gamma)^{\epsilon -1}]^\frac{1}{1-\epsilon}}{w_H\beta}\right)^\beta$$</span></p>
<p>substituting <span class="math-container">$h$</span> in <span class="math-container">$(3)$</span> give us:
<span class="math-container">$$\boxed{c=w_H^{1-\beta}(1-\beta)^{\beta-1}\beta^{-\beta}\left[w_L^{1-\epsilon}+\tilde \gamma^{\epsilon-1}\right]^\frac{\beta}{1-\epsilon}}\tag{b}$$</span></p>
<p>we can combine the solutions to <em>case 1</em> and <em>case 2</em> , i.e., <span class="math-container">$(a)$</span> and <span class="math-container">$(b)$</span>
Therefore, we have</p>
<p><span class="math-container">$c=\begin{cases} w_H^{1-\beta} \beta^{-\beta}(1-\beta)^{\beta-1}w_L^\beta &amp; \text{if } \alpha(i)=0 \\
w_H^{1-\beta}(1-\beta)^{\beta-1}\beta^{-\beta}\left[w_L^{1-\epsilon}+\tilde \gamma^{\epsilon-1}\right]^\frac{\beta}{1-\epsilon} &amp; \text{if } \alpha(i)=1
\end{cases}$</span></p>
<p>using <span class="math-container">$\alpha(i)$</span>, the above can also be written as:
<span class="math-container">$$\boxed{c=w_H^{1-\beta}\beta^{-\beta}(1-\beta)^{\beta-1}\left(w_L^{1-\epsilon}+\tilde \gamma^{\epsilon-1} \alpha(i) \right)^\frac{\beta}{1-\epsilon}}$$</span></p>

CC BY-SA 4.0
2023-05-02T11:33:19.387Z
Thanks so much, I'm going to check the computations in a while and see if I can get useful hints to fully answer my question.
John M.
CC BY-SA 4.0
2023-05-02T12:15:27.943Z
consider rewarding me the bounty if you do
mynameparv
CC BY-SA 4.0
2023-05-02T12:20:37.707Z
For sure, don't you worry about that
John M.
CC BY-SA 4.0
2023-05-02T15:29:27.000Z
How do you get $h$? I mean $h$ just after $(3)$
John M.
CC BY-SA 4.0
2023-05-02T16:21:57.307Z
I think that one problem with your procedure is that you omitted $\alpha(i)$ into the constraint on $(2)$
John M.
CC BY-SA 4.0
2023-05-03T04:57:29.667Z
You get $h$ after solving the optimization problem in $(3)$ and no I haven't omitted $\alpha$ at all. I have used its property as an indicator and substituted its value. Don't worry I'll update in a few days and make it more comprehensive if possible.
mynameparv
CC BY-SA 4.0
2023-05-19T14:40:45.447Z
@JohnM. Hi, sorry was busy before and couldn't find the time to complete the answer, I've edited the previous mistakes and tried to make the procedure more comprehensible for you. According to your question $\gamma=\tilde\gamma^{epsilon}$ but I think it should $\gamma=\tilde\gamma^{epsilon-1}$ instead. So, it could be that I am getting the exponent wrong due to some error as the algebra is quite tedious or that I've done it right and there is an issue in the problem itself. I can also share my workings with you as I did them on a tablet
mynameparv
CC BY-SA 4.0
2023-05-02T11:16:45.427Z
2023-05-19T18:15:16.127Z
2023-05-19T18:15:16.127Z
mynameparv
<p>I'm going to provide the steps to follow for solving this optimization problem:</p>
<p>The Cost Function for <span class="math-container">$y(i)$</span> is given by:</p>
<p><span class="math-container">\begin{equation*}
C(w_H,w_L,y(i))= w_H h(i)^{\ast} + w_L l(i)^{\ast} + x(i)^{\ast}
\label{cost}
\end{equation*}</span></p>
<p>The production function is homogenous of degree 1, that is,it exhibits constant return to scale. This makes average cost equal to the marginal cost.</p>
<p>For convenience, write the production function as</p>
<p><span class="math-container">\begin{equation}
    y(i)=z(i)^{\beta}h^{1-\beta} , \text{$~~~$ with $~~$} z(i) \equiv [l(i)^{\frac{\epsilon -1}{\epsilon}} + (\alpha(i) \tilde{\gamma})^{\frac{\epsilon -1}{\epsilon}}]^{\frac{\epsilon}{\epsilon -1}}
\end{equation}</span>
The cost function reads as</p>
<p><span class="math-container">\begin{equation}
    C(w_H,w_z,y(i))= w_H h(i)^{\ast} + w_z z(i)^{\ast}
    \label{cost2}
\end{equation}</span></p>
<p>In the first stage, you minimize:</p>
<p><span class="math-container">\begin{equation}
\begin{array}{rrclcl}
min ~~~~~{ w_H h(i) + w_z z(i)}\\
\textrm{s.t.} &amp;  z(i)^{\beta}h^{1-\beta} \ge y(i)
\end{array}
\end{equation}</span></p>
<p>Solving this cost minimization problem  you get <span class="math-container">$h(i)^{\ast},z(i)^{\ast}$</span>. Then, plug these conditional factor input demands in the objective function. As a result, you get the cost function (which depends upon <span class="math-container">$w_H,w_Z,y(i))$</span>.</p>
<p>In the second stage, you minimize</p>
<p><span class="math-container">\begin{equation}
\begin{array}{rrclcl}
min ~~~~~{ w_L l(i) + x(i)}\\
\textrm{s.t.} &amp;  \left[ l(i)^{\frac{\epsilon -1}{\epsilon}} +\alpha(i)(\tilde{\gamma}x(i))^{\frac{\epsilon -1}{\epsilon}} \right]^{\frac{\epsilon }{\epsilon -1}} = z(i)
\end{array}
\end{equation}</span></p>
<p>Solving this cost minimization problem, you find <span class="math-container">$x(i)^{\ast},l(i)^{\ast}, \lambda^{\ast}$</span></p>
<p>Now, since the composite good, <span class="math-container">$z(i)$</span>, is priced at its marginal cost, you get <span class="math-container">$w_z=\lambda^{\ast}$</span>. Using this result into the cost function coming from the first stage, you are done, you have just found the total cost function. Divide it by <span class="math-container">$y(i)$</span>, and you get the unit cost of production</p>

CC BY-SA 4.0
2023-05-06T07:48:20.693Z
2023-05-06T14:28:23.280Z
2023-05-06T14:28:23.280Z
Tony
<p>I have the following CES nested in a Cobb-Douglas production function:</p>
<p><span class="math-container">$$y(i)=[l(i)^{\frac{\epsilon - 1}{\epsilon}} +\alpha(i)(\tilde{\gamma}x(i))^{\frac{\epsilon - 1}{\epsilon}}]^{\frac{\epsilon \beta}{\epsilon-1}}h(i)^{1-\beta}$$</span></p>
<p><span class="math-container">$\beta \in (0,1)$</span>, <span class="math-container">$\tilde{\gamma}$</span> is a productivity parameter, and <span class="math-container">$\epsilon$</span>&gt;1 is the EOS between low-skilled workers,<span class="math-container">$l(i)$</span>, and machines <span class="math-container">$x(i)$</span>. Also, <span class="math-container">$\alpha(i)$</span> is an indicator function taking value one if the product <span class="math-container">$i$</span> is automated and zero otherwise. Mathematically, <span class="math-container">$\alpha(i) \in \{0,1\}$</span>. Finally, <span class="math-container">$h(i)$</span> are high-skilled workers.</p>
<p>The unit cost function of product <span class="math-container">$y(i)$</span> is:</p>
<p><span class="math-container">$$c= \beta^{-\beta}(1-\beta)^{-(1-\beta)}(w_L^{1-\epsilon}+\gamma \alpha(i))^{\frac{\beta}{1-\epsilon}}w_H^{1-\beta}$$</span></p>
<p>where <span class="math-container">$w_L$</span> is the unit cost of the input <span class="math-container">$l(i)$</span>, one is the unit cost of input <span class="math-container">$x(i)$</span> and <span class="math-container">$w_H$</span> is the unit cost of input <span class="math-container">$h(i)$</span>. Finally <span class="math-container">$\gamma  \equiv \tilde{\gamma^\epsilon}$</span></p>
<p>Could you tell me how to derive this unit cost function?</p>

CC BY-SA 4.0
2022-09-12T09:30:16.627Z
In this post, you can see the steps that are generally applicable, https://economics.stackexchange.com/a/5273/61
Alecos Papadopoulos
CC BY-SA 4.0
2023-05-01T05:54:47.547Z
@John M. you can try to use the indicator function, first suppose $\alpha=0$ then find $c$ for that case, and then find $c$ for $\alpha=1$, the algebra is hellish for this case as I'm trying to do it myself. Will upload a solution if I am able to do it myself. Also, for the three input case you can try to solve the cost minimization problem in stages
mynameparv
CC BY-SA 4.0
2023-05-01T21:14:49.473Z
@mynameparv can you show me your attempt?
John M.
CC BY-SA 4.0
2023-05-02T04:05:00.993Z
I've written an answer but it's still in complete, the case for $\alpha=1$ is not at all simple. Do you want me to post the incomplete solution and you try the rest yourself?
mynameparv
CC BY-SA 4.0
2023-05-02T04:19:31.630Z
@John M. The solution contains enough information and explanation so that you may attempt the problem on your own and I will get to completing it once I'm a bit more free. Let me know if I should post that attempt or not
mynameparv
CC BY-SA 4.0
2023-05-02T09:52:01.287Z
@mynameparv Yes pls, at least I can upvote your answer
John M.
CC BY-SA 4.0
2023-05-02T11:21:14.777Z
@JohnM. check it and give it a try yourself to see if you can complete the solution yourself. I'll get to it after a few weeks as I'm really busy right now
mynameparv
CC BY-SA 4.0
2023-05-02T16:27:31.423Z
@AlecosPapadopoulos I've seen your answer at the link you posted and it is great, but it seems that does not apply for my case. Even if you don't go into the algebra, do you mind if you can spell out all the steps for this special case, i.e. CES nested in a Cobb-Douglas.  OF course, I'll pick the best answer for the bounty
John M.
CC BY-SA 4.0
2023-05-03T02:46:50.850Z
@JohnM. Up to an including eq. (6) in my post, the steps are applicable to an homogeneous production function- and your production function is homogeneous of degree 1 in your three inputs (my "h" is equal to 1 in your case), and irrespective of whether the alpha indicator function is equal to zero or one. Have you gotten there? After that, you need to differentiate your production function to obtain the three marginal products.
Alecos Papadopoulos
CC BY-SA 4.0
2023-05-04T20:21:14.377Z
@AlecosPapadopoulos I've tried but it didn't work. Is there any source where I can read something that helps?
John M.
<microeconomics><mathematical-economics>
CES nested in a Cobb-Douglas Function
John M.
# CES nested in a Cobb-Douglas Function

(Asked by: username_0 on 2022-09-11)

I have the following CES nested in a Cobb-Douglas production function:

$$y(i)=[l(i)^{\frac{\epsilon - 1}{\epsilon}} +\alpha(i)(\tilde{\gamma}x(i))^{\frac{\epsilon - 1}{\epsilon}}]^{\frac{\epsilon \beta}{\epsilon-1}}h(i)^{1-\beta}$$

$\beta \in (0,1)$, $\tilde{\gamma}$ is a productivity parameter, and $\epsilon$>1 is the EOS between low-skilled workers,$l(i)$, and machines $x(i)$. Also, $\alpha(i)$ is an indicator function taking value one if the product $i$ is automated and zero otherwise. Mathematically, $\alpha(i) \in \{0,1\}$. Finally, $h(i)$ are high-skilled workers.

The unit cost function of product $y(i)$ is:

$$c= \beta^{-\beta}(1-\beta)^{-(1-\beta)}(w\_L^{1-\epsilon}+\gamma \alpha(i))^{\frac{\beta}{1-\epsilon}}w\_H^{1-\beta}$$

where $w\_L$ is the unit cost of the input $l(i)$, one is the unit cost of input $x(i)$ and $w\_H$ is the unit cost of input $h(i)$. Finally $\gamma \equiv \tilde{\gamma^\epsilon}$

Could you tell me how to derive this unit cost function?

**Tags:** microeconomics, mathematical-economics
> **Comments:**
> **username_1**: In this post, you can see the steps that are generally applicable, https://economics.stackexchange.com/a/5273/61  
> **username_2**: @John M. you can try to use the indicator function, first suppose $\alpha=0$ then find $c$ for that case, and then find $c$ for $\alpha=1$, the algebra is hellish for this case as I'm trying to do it myself. Will upload a solution if I am able to do it myself. Also, for the three input case you can try to solve the cost minimization problem in stages  
> **username_0**: @mynameparv can you show me your attempt?  
> **username_2**: I've written an answer but it's still in complete, the case for $\alpha=1$ is not at all simple. Do you want me to post the incomplete solution and you try the rest yourself?  
> **username_2**: @John M. The solution contains enough information and explanation so that you may attempt the problem on your own and I will get to completing it once I'm a bit more free. Let me know if I should post that attempt or not  
> **username_0**: @mynameparv Yes pls, at least I can upvote your answer  
> **username_2**: @JohnM. check it and give it a try yourself to see if you can complete the solution yourself. I'll get to it after a few weeks as I'm really busy right now  
> **username_0**: @AlecosPapadopoulos I've seen your answer at the link you posted and it is great, but it seems that does not apply for my case. Even if you don't go into the algebra, do you mind if you can spell out all the steps for this special case, i.e. CES nested in a Cobb-Douglas.  OF course, I'll pick the best answer for the bounty  
> **username_1**: @JohnM. Up to an including eq. (6) in my post, the steps are applicable to an homogeneous production function- and your production function is homogeneous of degree 1 in your three inputs (my "h" is equal to 1 in your case), and irrespective of whether the alpha indicator function is equal to zero or one. Have you gotten there? After that, you need to differentiate your production function to obtain the three marginal products.  
> **username_0**: @AlecosPapadopoulos I've tried but it didn't work. Is there any source where I can read something that helps?  

---

## 2 answers

### Answer by username_3 on 2023-05-06. Score: 4 | Accepted answer

I'm going to provide the steps to follow for solving this optimization problem:

The Cost Function for $y(i)$ is given by:

\begin{equation\*}
C(w\_H,w\_L,y(i))= w\_H h(i)^{\ast} + w\_L l(i)^{\ast} + x(i)^{\ast}
\label{cost}
\end{equation\*}

The production function is homogenous of degree 1, that is,it exhibits constant return to scale. This makes average cost equal to the marginal cost.

For convenience, write the production function as

\begin{equation}
y(i)=z(i)^{\beta}h^{1-\beta} , \text{$~~~$ with $~~$} z(i) \equiv [l(i)^{\frac{\epsilon -1}{\epsilon}} + (\alpha(i) \tilde{\gamma})^{\frac{\epsilon -1}{\epsilon}}]^{\frac{\epsilon}{\epsilon -1}}
\end{equation}
The cost function reads as

\begin{equation}
C(w\_H,w\_z,y(i))= w\_H h(i)^{\ast} + w\_z z(i)^{\ast}
\label{cost2}
\end{equation}

In the first stage, you minimize:

\begin{equation}
\begin{array}{rrclcl}
min ~~~~~{ w\_H h(i) + w\_z z(i)}\\
\textrm{s.t.} & z(i)^{\beta}h^{1-\beta} \ge y(i)
\end{array}
\end{equation}

Solving this cost minimization problem you get $h(i)^{\ast},z(i)^{\ast}$. Then, plug these conditional factor input demands in the objective function. As a result, you get the cost function (which depends upon $w\_H,w\_Z,y(i))$.

In the second stage, you minimize

\begin{equation}
\begin{array}{rrclcl}
min ~~~~~{ w\_L l(i) + x(i)}\\
\textrm{s.t.} & \left[ l(i)^{\frac{\epsilon -1}{\epsilon}} +\alpha(i)(\tilde{\gamma}x(i))^{\frac{\epsilon -1}{\epsilon}} \right]^{\frac{\epsilon }{\epsilon -1}} = z(i)
\end{array}
\end{equation}

Solving this cost minimization problem, you find $x(i)^{\ast},l(i)^{\ast}, \lambda^{\ast}$

Now, since the composite good, $z(i)$, is priced at its marginal cost, you get $w\_z=\lambda^{\ast}$. Using this result into the cost function coming from the first stage, you are done, you have just found the total cost function. Divide it by $y(i)$, and you get the unit cost of production


---

### Answer by username_2 on 2023-05-02. Score: 2

Since we are interested in finding the unit cost let the output needed to be produced be equal to 1 i.e., $y(i)\overset{set}=1$
We need to solve: $$\begin{align}
\min\_{l(i),x(i),h(i)\geq0} \quad & w\_Ll(i)+x(i)+w\_{H}h(i)\\
\textrm{s.t.} \quad & [l(i)^{\frac{\epsilon - 1}{\epsilon}} +\alpha(i)(\tilde{\gamma}x(i))^{\frac{\epsilon - 1}{\epsilon}}]^{\frac{\epsilon \beta}{\epsilon-1}}h(i)^{1-\beta}=1\\
\textrm{and} \quad & \alpha(i)=\begin{cases}1 & \text{if } i\text{ is automated}\\ 0 & \text{otherwise}\end{cases}
\end{align}$$

I am going to use the indicator function and solve for two different cases and then try to combine them using the indicator.

**Case 1:** If $i$ is not automated i.e., $ \alpha(i)=0$

If $i$ is not automated, then the cost minimization problem becomes a standard one with a Cobb-Douglas  production function
$$\begin{align}
\min\_{l,x,h\geq 0} \quad & w\_Ll+x+w\_{H}h\\
\textrm{s.t.} \quad & l^{\beta}h^{1-\beta}=1\\
\end{align}$$

This is just a standard cost minimization problem with a cobb-douglas production function, so I think you can verify yourself that the above problem  gives: $l(i)=\left(\frac{w\_H\beta}{w\_L(1-\beta)}\right)^{1-\beta}, \quad x(i)=0, \quad h(i)= \left(\frac{w\_L(1-\beta)}{w\_{H} \beta}\right)^{\beta}$

therefore, $$\begin{eqnarray}
& c=w\_L^\beta\left(\frac{w\_H\beta}{1-\beta}\right)^{1-\beta}+w\_H^{1-\beta}\left(\frac{w\_L(1-\beta)}{\beta}\right)^{\beta}\\\\
\implies & {\boxed{c=w\_L^\beta w\_H^{1-\beta} \beta^{-\beta}(1-\beta)^{\beta-1}}}\tag{a}
\end{eqnarray}$$

**Case 2:** If $i$ is automated
$$\begin{align}
\min\_{l,x,h\geq 0} \quad & w\_Ll+x+w\_{H}h\\
\textrm{s.t.} \quad & [l^{\frac{\epsilon - 1}{\epsilon}} +(\tilde{\gamma}x)^{\frac{\epsilon - 1}{\epsilon}}]^{\frac{\epsilon \beta}{\epsilon-1}}h^{1-\beta}=1\\
\end{align}$$

let us write the above problem as:
$$\begin{align}
{\min\_{0\leq h \leq 1}
\underbrace{\begin{pmatrix} \underset{l,x}{\min} \quad & w\_Ll+w\_Hh+x \\
\textrm{s.t.} \quad & [l^{\frac{\epsilon - 1}{\epsilon}} +(\tilde{\gamma}x)^{\frac{\epsilon - 1}{\epsilon}}]^{\frac{\epsilon}{\epsilon-1}}=h^\frac{\beta -1}{\beta} \end{pmatrix}}\_{\text{auxiliary problem}}} \tag{1}
\end{align}$$

We can solve the above problem in two stages:

**Stage 1:** first solve the auxiliary problem for $l$ and $x$ holding $h$ fixed

**Stage 2:** Substitute the solutions of the auxiliary problem$-$ $l(h)$ and $x(h)$ $-$ into $(1)$ and solve the problem to find optimal $h$

**Stage 1:**
we need to solve: $$\begin{align}
\underset{l,x}{\min} \quad & w\_Ll+w\_Hh+x \\
\textrm{s.t.} \quad & [l^{\frac{\epsilon - 1}{\epsilon}} +(\tilde{\gamma}x)^{\frac{\epsilon - 1}{\epsilon}}]^{\frac{\epsilon}{\epsilon-1}}=h^\frac{\beta -1}{\beta}\\
\text{gives:} \quad & {\left.\begin{matrix} l(h)=\frac{h^\frac{\beta -1}{\beta}}{[1+(w\_L\tilde \gamma)^{\epsilon -1}]^\frac{\epsilon}{\epsilon-1}} \\ x(h)=\frac{w\_L^\epsilon \tilde{\gamma}^{\epsilon -1} h^\frac{\beta -1}{\beta}}{[1+(w\_L\tilde \gamma)^{\epsilon -1}]^\frac{\epsilon}{\epsilon-1}}
\end{matrix}\right\}}\tag{2}
\end{align}$$

Because we are given that $\epsilon>1$, You can verify that isoquants will be convex. Thus, I was able to solve the above by equating ratios of MP to MC for $l$ and $x$ and substituting them into the constraint.

**Stage 2:** substituting $(2)$ in $(1)$ we get the final problem solving which will give optimal $h$
$$\begin{align}
\min\_{0\leq h\leq 1} \quad & w\_Hh+w\_L[1+(w\_L\tilde \gamma)^{\epsilon -1}]^\frac{1}{1-\epsilon} h^\frac{\beta -1}{\beta}\tag{3}
\end{align}$$

the above objective is convex in $h \; \because \beta<1$, so assuming the parametric condition gives us a stationary point that lies in the interval $(0,1)$ we can solve for the optimal h in the above problem and substitute it in the problem itself to get $c$ for $\alpha=1$:

$$h^\*=\left(\frac{(1-\beta)w\_L[1+(w\_L\tilde \gamma)^{\epsilon -1}]^\frac{1}{1-\epsilon}}{w\_H\beta}\right)^\beta$$

substituting $h$ in $(3)$ give us:
$$\boxed{c=w\_H^{1-\beta}(1-\beta)^{\beta-1}\beta^{-\beta}\left[w\_L^{1-\epsilon}+\tilde \gamma^{\epsilon-1}\right]^\frac{\beta}{1-\epsilon}}\tag{b}$$

we can combine the solutions to *case 1* and *case 2* , i.e., $(a)$ and $(b)$
Therefore, we have

$c=\begin{cases} w\_H^{1-\beta} \beta^{-\beta}(1-\beta)^{\beta-1}w\_L^\beta & \text{if } \alpha(i)=0 \\
w\_H^{1-\beta}(1-\beta)^{\beta-1}\beta^{-\beta}\left[w\_L^{1-\epsilon}+\tilde \gamma^{\epsilon-1}\right]^\frac{\beta}{1-\epsilon} & \text{if } \alpha(i)=1
\end{cases}$

using $\alpha(i)$, the above can also be written as:
$$\boxed{c=w\_H^{1-\beta}\beta^{-\beta}(1-\beta)^{\beta-1}\left(w\_L^{1-\epsilon}+\tilde \gamma^{\epsilon-1} \alpha(i) \right)^\frac{\beta}{1-\epsilon}}$$

> **Comments:**
> **username_0**: Thanks so much, I'm going to check the computations in a while and see if I can get useful hints to fully answer my question.  
> **username_2**: consider rewarding me the bounty if you do  
> **username_0**: For sure, don't you worry about that  
> **username_0**: How do you get $h$? I mean $h$ just after $(3)$  
> **username_0**: I think that one problem with your procedure is that you omitted $\alpha(i)$ into the constraint on $(2)$  
> **username_2**: You get $h$ after solving the optimization problem in $(3)$ and no I haven't omitted $\alpha$ at all. I have used its property as an indicator and substituted its value. Don't worry I'll update in a few days and make it more comprehensive if possible.  
> **username_2**: @JohnM. Hi, sorry was busy before and couldn't find the time to complete the answer, I've edited the previous mistakes and tried to make the procedure more comprehensible for you. According to your question $\gamma=\tilde\gamma^{epsilon}$ but I think it should $\gamma=\tilde\gamma^{epsilon-1}$ instead. So, it could be that I am getting the exponent wrong due to some error as the algebra is quite tedious or that I've done it right and there is an issue in the problem itself. I can also share my workings with you as I did them on a tablet  

---

