Cobb-Douglas model - Lagrange multipliers

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

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

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 $$(p/m,0)$$.

• Thank you. This procedure seems very detailed and clear. Commented Aug 16, 2023 at 12:08
• The set $S$ captures the constraints other than the equality constraint. The constraint set for the maximization problem is the intersection of $S$ with the set defined by the equality constraint. In your case the constraint set is the intersection of $S=\{(x,y) \ | \ x\geq 0, y\geq 0\}$ and the line defined by $g(x,y)=p$. So the set $S$ does have interior points; any points where $x>0$ and $y>0$ are interior.
– smcc
Commented Aug 16, 2023 at 12:13
• Yeah, got it, that's why I deleted my comment. But you can keep yours. Thank you. I will reread all this and think it over. Commented Aug 16, 2023 at 12:17
• Is there any book you can recommend, where this and similar optimization problems are described so well as in your answer? Commented Aug 16, 2023 at 13:52
• These online notes by the economist Martin Osborne are very useful (it is where I got the method): mjo.osborne.economics.utoronto.ca/index.php/tutorial/index/1/… The Sydsaeter and Hammond book Mathematics for Economic Analysis (referred to in Osborne's notes) is quite similar in level to his notes.
– smcc
Commented Aug 16, 2023 at 13:56