Solve long run production function of a firm using technical rate of substitution

Question

I don't understand the solution to a question which deals with the long run production function of a firm.

The question is:

Suppose a firm has a production function $f(x_1, x_1) = x_1^{0.5}x_2^{0.5}$, and factor prices are $w_1 = w_2 = 1$. Derive the long run total cost function.

The solution is:

From the first order condition, i.e. the technical rate of substitution equals the ratio of factor prices, we have $ -\frac{x_2}{x_1} = -\frac{1}{1} $, i.e. $x_1 = x_2$. Thus the conditional factor demand is $x_1 = x_2 = q$. So the long run cost function is $TC(q) = w_1 x_1 + w_2 x_2 = 2q$.

__

While the second sentence of the solution makes sense, I do not understand what the first sentence means, particularly the link to a first order condition and the technical rate of substitution.

I could not find an explanation in my textbook.

Answer 1

The optimization problem you solve is

$\min w_1 x_1 + w_2 x_2$

s.t. $x_1^{0.5} x_2^{0.5} = q $

I assume you’re used to solving optimization problems with the Lagrangian.

The Lagrangian is:

$\mathcal{L}(x_1,x_2,\lambda) = w_1 x_1 + w_2 x_2 + \lambda (x_1^{0.5} x_2^{0.5} - q) $

Now we find the first order conditions on $x_1, x_2$

$\frac{\partial \mathcal{L}}{\partial x_1} = w_1 + \lambda x_1^{-0.5} x_2^{0.5} = 0 \implies \lambda = - \frac{w_1 x_1^{0.5}}{x_2^{0.5}} $

$\frac{\partial \mathcal{L}}{\partial x_2} = w_2 + \lambda x_1^{0.5} x_2^{-0.5} = 0 \implies \lambda = - \frac{w_2 x_2^{0.5}}{x_1^{0.5}}$

Equating both lambdas,

$-\frac{w_1 x_1^{0.5}}{x_2^{0.5}} = -\frac{w_2 x_2^{0.5}}{x_1^{0.5}} \implies \frac{x_1}{x_2} = \frac{w_2}{w_1}$

From the first order conditions on the two control variables $x_1,x_2$, we got your equation in your statement about the technical rate of substitution.

In fact, the statement “$TRS = \frac{w_2}{w_1}$” can be shown in general, yielding a shortcut to using the Lagrangian:

Let’s solve the general cost minimization problem

$\min w_1 x_1 + w_2 x_2$

s.t. $f(x_1,x_2) = q$

Its Lagrangian is

$\mathcal{L}(x_1,x_2,\lambda) = w_1 x_1 + w_2 x_2 + \lambda (f(x_1,x_2) - q)$

Its first order conditions are

$\frac{\partial \mathcal{L}}{\partial x_1} = w_1 + \lambda \frac{\partial f}{\partial x_1} \implies \lambda = - \frac{w_1}{\frac{\partial f}{\partial x_1}} $

$\frac{\partial \mathcal{L}}{\partial x_2} = w_2 + \lambda \frac{\partial f}{\partial x_2} \implies \lambda = - \frac{w_2}{\frac{\partial f}{\partial x_2}} $

Equating both lambdas,

$- \frac{w_1}{\frac{\partial f}{\partial x_1}} = - \frac{w_2}{\frac{\partial f}{\partial x_2}} \implies \frac{w_1}{w_2} = \frac{\frac{\partial f}{\partial x_1}}{\frac{\partial f}{\partial x_2}}$

i.e. technical rate of substitution $=$ relative prices.

The rate of substitution $=$ relative prices equation also holds in the primal profit maximization problem, as well as in consumer theory in both the primal (utility maximization) and dual (expenditure minimization) problems.

This equation serves as a shortcut to the Lagrangian method. Once I learned it, I haven’t used a Lagrangian in my economics courses.