Cobb-Douglas model – Lagrange multipliers

calculuslagrange multipliermultivariable-calculus

I came across this problem and I almost solved it… but not quite, I think.

It's about this model.

Cobb-Douglas Production Function

OK, so let's say we have this function

$f(x,y) = K . x^\alpha . y^{1-\alpha}$

Here $K > 0, \alpha \in (0,1)$ are some constants.

We want to find the maximum of $f$ under the additional condition that

$g(x,y) = mx + ny = p$

Here $m,n,p > 0$ are also constants.

Also, we have that $x \ge 0, y \ge 0$
because $x$ is labor input and $y$ is capital input in the Cobb-Douglas model.

My book tells me to use the Lagrange multipliers method to find the maximum of $f$.

OK, so I applied the Lagrange multipliers method i.e. I created this system of three equations (for $x,y,\lambda$), and I solved it.

$f'_x(x,y) = \lambda g'_x (x,y)$
$f'_y(x,y) = \lambda g'_y (x,y)$
$g(x,y) = p$

So I was able to find that the function $f$ has an extremum at the point $(x_0, y_0) = (\alpha . p / m, (1-\alpha).p / n)$

But then I have some doubts.

  1. How do I know that this extremum is a maximum and not a minimum? How do I justify that it's a maximum? Is it enough to compare $f(x_0, y_0)$ with say $f(0, p /n) = f(p/m, 0) = 0$ in order to conclude that we have a maximum at $(x_0, y_0)$ Somehow this doesn't sound very convincing to me.

  2. It also bothers me that $f$ is defined only when $x \ge 0, y \ge 0$. So because of these additional restrictions on $x$ and $y$, is the Lagrange multipliers method applicable here at all? I guess it is because that's the method my book tells me to use but still… wanted to ask that here. These additional restrictions bother me because of what I learned yesterday in another question that I asked.

Lagrange Multipliers Question – some extremum points are missed (not detected) by the method

Best Answer

Method for solving a two-variable maximization problem with one equality constraint

Let $f$ and $g$ be continuously differentiable functions and let $c$ be a number. If the problem

$$\text{$\max_{x,y}f(x,y)$ subject to $g(x,y)=c$ and $(x,y)\in S$}$$

has a solution, then it may be found as follows:

  1. Find all the values of $(x,y,\lambda)$ for which

$$\begin{align*}&\text{(a) $(x,y)$ is an interior point of $S$}\\ &\text{(b) $(x,y,\lambda)$ satisfies the first-order conditions and the constraint:}\end{align*}$$

$$\begin{align*}L_x(x,y)&=0\\ L_y(x,y)&=0 \\ g(x,y)&=c. \end{align*}$$

  1. Find all the points $(x,y)$ in the interior of $S$ that satisfy

$$\begin{align*}g_x(x,y)&=0\\ g_y(x,y)&=0 \\ g(x,y)&=c. \end{align*}$$

  1. If the set $S$ has any boundary points, find all the boundary points that solve the problem $\max_{x,y}f(x,y)$ subject to the two conditions $g(x,y) = c$ and $(x,y)$ is a boundary point of $S$.

  2. The points $(x,y)$ you have found at which the value $f(x,y)$ is largest are the maximizers of the function $f$.

You have already carried out step 1. In many cases that will give your solution. However, you have to check points that do not satisfy the constraint qualification. A solution $(x^*,y^*)$ satisfies the constraint qualification if $g_x(x^*,y^*)\neq 0$ or $g_y(x^*,y^*)\neq 0$. Step 2 finds all the points in the constraint set that do not satisfy the constraint qualification, as they could potentially be solutions that are not found in step 1. Step 3 addresses your concern related to your previous question by also checking the boundary points of $S$.

In your case, steps 2 and 3 are as follows:

  1. $g_x(x,y)=m\neq 0$ and $g_y(x,y)=n\neq 0$, so there are no points satisfying the conditions in step 2. (The constraint qualification is satisfied at all points.)

  2. $S=\{(x,y)\ |\ x\geq 0, y\geq 0\}$, and the boundary points are where $x=0$ or $y=0$. When $x=0$ or $y=0$ we have $f(x,y)=0$. Since the point you found in step 1 gives $f(x,y)>0$, there is no need to go any further.

To address you first concern, once we have proved a maximum exist, then we know it must be found in step 1, 2, or 3. If a minimum exists, then we know it must satisfy the conditions in step 1, 2, or 3 with $\max$ replaced by $\min$. It is step 4 that then allows you to distinguish between which candidates are minimizers or maximizers.

To prove existence of a maximum and minimum usually involves application of the extreme value theorem. In your case the constraint set is compact (it is the line $g(x,y)=p$ for $x\geq 0$ and $y\geq 0$) and the objective function is continuous, so you know both a maximum and minimum exist.

In your case, the minimum is $0$, and this occurs at boundary points of $S$ satisfying the constraint, i.e. at the points $(0,p/n)$ and $(p/m,0)$.

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