<p>The question "how are the first order conditions" seems very unclear to me, and I am providing a set-up for finding and writing them out, while explaining the Kuhn-Tucker conditions that are easy to struggle with.</p>

<p>Though we try to avoid giving away basic study question answers, it's a positively voted question without an answer, and I still think these questions have value for future users. Enough time has also passed where there's no risk of helping someone get through a class, if that was the intention.</p>

<hr>

<p>Usually in cost minimization problems where the production function is a constraint, there is a set quantity to be produced that then leads to minimizing cost with respect to that. Here in this profit maximization problem, you have a minimum amount of production required of both goods, which I find interesting.</p>

<p>We can note the objective function (profit) is linear, and therefore concave and differentiable. The constraints are more interesting because the Cobb-Douglas form is not necessarily concave, namely when the exponents sum to more than 1. So you can ask yourself when the constraints will be convex (they are certainly differentiable). These ideas are related to the <strong>sufficient</strong> conditions for an optimum. A solution <strong>exists</strong> by the Weierstrass Theorem (constraint is compact and the objective function is continuous).</p>

<p>To look at the first order <strong>necessary</strong> conditions, we can omit your last two equalities since they don't add anything. We take:</p>

<p>\begin{equation}
\begin{array}{l}
\max_{q_{x}, q_{y}, l_{x}, l_{y}, k_{x}, l_{x}} \quad p_{x}q_{x}+p_{y}q_{y}-w(l_{x}+l_{y})-v(k_{x}+k_{y}) \\ 
\text{s.t.} \\
q_{x} - k_{x}^{\alpha}l_{x}^{\beta} \leq 0 \\
q_{y} - k_{y}^{\gamma}l_{y}^{\delta} \leq 0 \\
\end{array}
\end{equation}</p>

<p>Form the Lagrangean:</p>

<p>$$\max \quad \mathscr{L} = p_{x}q_{x}+p_{y}q_{y}-w(l_{x}+l_{y})-v(k_{x}+k_{y}) - \mu_x (q_{x} - k_{x}^{\alpha}l_{x}^{\beta}) - \mu_y (q_{y} - k_{y}^{\gamma}l_{y}^{\delta})$$</p>

<p>(note the signs in front of the Lagrangean multipliers)</p>

<p>Then, in place of the derivative with respect to the multipliers, we have our constraints with complementary slackness conditions. We take the other constraints as well.</p>

<p>$$\frac{\partial \mathscr{L}}{\partial \ ( \cdot )} \leq 0 \quad ; \quad ( \cdot ) \geq 0 \quad \forall ( \cdot )$$
$$\text{constraint} \quad ; \quad \mu_{(\cdot)} \geq 0$$</p>

<p>And from there you can move on to solve the system, but you may find yourself in a weird spot because we haven't been able to establish <strong>uniqueness</strong>.</p>

CC BY-SA 3.0
2017-09-02T05:36:51.263Z
2017-09-02T05:36:51.263Z
Kitsune Cavalry
<p>I am currently reading the book "Microeconomics: Principles and Analysis" by Cowell on my own. I'm interested in the section of the multiproduct firm, but i'm confused with the use of the profit function. Specifically, about how to pose the target for the maximization problem. Here an example with two goods $x$ and $y$, two factor $k$ and $l$ and two Cobb-Douglas production functions.</p>

<p>\begin{equation}
\begin{array}{l}
\text{Max}\quad \pi=p_{x}q_{x}+p_{y}q_{y}-w(l_{x}+l_{y})-v(k_{x}+k_{y}) \\ 
\text{subject to:} \\
 q_{x}\leq k_{x}^{\alpha}l_{x}^{\beta} \\
q_{y}\leq k_{y}^{\gamma}l_{y}^{\delta} \\
L=l_{x}+l_{y} \\ 
K=k_{x}+l_{y}
\end{array}
\end{equation}</p>

<p>Am I posing the question well? How are the first order conditions?</p>

CC BY-SA 3.0
2015-07-08T05:49:12.010Z
What exactly is causing you problems?
Do you know the Karush-Kuhn-Tucker theorem? Here is an example:http://www.math.ubc.ca/~israel/m340/kkt2.pdf
Giskard
CC BY-SA 3.0
2015-07-09T00:59:52.253Z
In the first order conditions I must differentiate with respect to all variables ($q_{x}, q_{y}, l_{x}, l_{y}, k_{x}, l_{x}$) and solve the equations system associated?
Héctor Garrido
CC BY-SA 3.0
2015-07-09T01:06:59.573Z
Differentiate the Lagrangian and solve the resulting equation system, yes.
Giskard
<microeconomics><mathematical-economics><producer-theory><competitive-equilibrium>
Maximisation problem in a multiproduct firm
Héctor Garrido
# Maximisation problem in a multiproduct firm

(Asked by: username_0 on 2015-07-08)

I am currently reading the book "Microeconomics: Principles and Analysis" by Cowell on my own. I'm interested in the section of the multiproduct firm, but i'm confused with the use of the profit function. Specifically, about how to pose the target for the maximization problem. Here an example with two goods $x$ and $y$, two factor $k$ and $l$ and two Cobb-Douglas production functions.

\begin{equation}
\begin{array}{l}
\text{Max}\quad \pi=p\_{x}q\_{x}+p\_{y}q\_{y}-w(l\_{x}+l\_{y})-v(k\_{x}+k\_{y}) \\
\text{subject to:} \\
q\_{x}\leq k\_{x}^{\alpha}l\_{x}^{\beta} \\
q\_{y}\leq k\_{y}^{\gamma}l\_{y}^{\delta} \\
L=l\_{x}+l\_{y} \\
K=k\_{x}+l\_{y}
\end{array}
\end{equation}

Am I posing the question well? How are the first order conditions?

**Tags:** microeconomics, mathematical-economics, producer-theory, competitive-equilibrium
> **Comments:**
> **username_1**: What exactly is causing you problems? Do you know the Karush-Kuhn-Tucker theorem? Here is an example:http://www.math.ubc.ca/~israel/m340/kkt2.pdf  
> **username_0**: In the first order conditions I must differentiate with respect to all variables ($q_{x}, q_{y}, l_{x}, l_{y}, k_{x}, l_{x}$) and solve the equations system associated?  
> **username_1**: Differentiate the Lagrangian and solve the resulting equation system, yes.  

---

## 1 answers

### Answer by username_2 on 2017-09-02. Score: 2 | Accepted answer

The question "how are the first order conditions" seems very unclear to me, and I am providing a set-up for finding and writing them out, while explaining the Kuhn-Tucker conditions that are easy to struggle with.

Though we try to avoid giving away basic study question answers, it's a positively voted question without an answer, and I still think these questions have value for future users. Enough time has also passed where there's no risk of helping someone get through a class, if that was the intention.

---

Usually in cost minimization problems where the production function is a constraint, there is a set quantity to be produced that then leads to minimizing cost with respect to that. Here in this profit maximization problem, you have a minimum amount of production required of both goods, which I find interesting.

We can note the objective function (profit) is linear, and therefore concave and differentiable. The constraints are more interesting because the Cobb-Douglas form is not necessarily concave, namely when the exponents sum to more than 1. So you can ask yourself when the constraints will be convex (they are certainly differentiable). These ideas are related to the **sufficient** conditions for an optimum. A solution **exists** by the Weierstrass Theorem (constraint is compact and the objective function is continuous).

To look at the first order **necessary** conditions, we can omit your last two equalities since they don't add anything. We take:

\begin{equation}
\begin{array}{l}
\max\_{q\_{x}, q\_{y}, l\_{x}, l\_{y}, k\_{x}, l\_{x}} \quad p\_{x}q\_{x}+p\_{y}q\_{y}-w(l\_{x}+l\_{y})-v(k\_{x}+k\_{y}) \\
\text{s.t.} \\
q\_{x} - k\_{x}^{\alpha}l\_{x}^{\beta} \leq 0 \\
q\_{y} - k\_{y}^{\gamma}l\_{y}^{\delta} \leq 0 \\
\end{array}
\end{equation}

Form the Lagrangean:

$$\max \quad \mathscr{L} = p\_{x}q\_{x}+p\_{y}q\_{y}-w(l\_{x}+l\_{y})-v(k\_{x}+k\_{y}) - \mu\_x (q\_{x} - k\_{x}^{\alpha}l\_{x}^{\beta}) - \mu\_y (q\_{y} - k\_{y}^{\gamma}l\_{y}^{\delta})$$

(note the signs in front of the Lagrangean multipliers)

Then, in place of the derivative with respect to the multipliers, we have our constraints with complementary slackness conditions. We take the other constraints as well.

$$\frac{\partial \mathscr{L}}{\partial \ ( \cdot )} \leq 0 \quad ; \quad ( \cdot ) \geq 0 \quad \forall ( \cdot )$$
$$\text{constraint} \quad ; \quad \mu\_{(\cdot)} \geq 0$$

And from there you can move on to solve the system, but you may find yourself in a weird spot because we haven't been able to establish **uniqueness**.


---

