<p>The optimization problem you solve is</p>
<p><span class="math-container">$\min w_1 x_1 + w_2 x_2$</span></p>
<p>s.t. <span class="math-container">$x_1^{0.5} x_2^{0.5} = q $</span></p>
<p>I assume you’re used to solving optimization problems with the Lagrangian.</p>
<p>The Lagrangian is:</p>
<p><span class="math-container">$\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) $</span></p>
<p>Now we find the first order conditions on <span class="math-container">$x_1, x_2$</span></p>
<p><span class="math-container">$\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}} $</span></p>
<p><span class="math-container">$\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}}$</span></p>
<p>Equating both lambdas,</p>
<p><span class="math-container">$-\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}$</span></p>
<p>From the first order conditions on the two control variables <span class="math-container">$x_1,x_2$</span>, we got your equation in your statement about the technical rate of substitution.</p>
<p>In fact, the statement “<span class="math-container">$TRS = \frac{w_2}{w_1}$</span>” can be shown in general, yielding a shortcut to using the Lagrangian:</p>
<p>Let’s solve the general cost minimization problem</p>
<p><span class="math-container">$\min w_1 x_1 + w_2 x_2$</span></p>
<p>s.t. <span class="math-container">$f(x_1,x_2) = q$</span></p>
<p>Its Lagrangian is</p>
<p><span class="math-container">$\mathcal{L}(x_1,x_2,\lambda) = w_1 x_1 + w_2 x_2 + \lambda (f(x_1,x_2) - q)$</span></p>
<p>Its first order conditions are</p>
<p><span class="math-container">$\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}} $</span></p>
<p><span class="math-container">$\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}} $</span></p>
<p>Equating both lambdas,</p>
<p><span class="math-container">$- \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}}$</span></p>
<p>i.e. technical rate of substitution <span class="math-container">$=$</span> relative prices.</p>
<p>The rate of substitution <span class="math-container">$=$</span> 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.</p>
<p>This equation serves as a shortcut to the Lagrangian method. Once I learned it, I haven’t used a Lagrangian in my economics courses.</p>

CC BY-SA 4.0
2023-04-06T18:10:26.353Z
2023-04-06T18:17:07.007Z
2023-04-06T18:17:07.007Z
Nicolas Torres
<p>I don't understand the solution to a question which deals with the long run production function of a firm.</p>
<p><strong>The question is:</strong></p>
<p>Suppose a firm has a production function <span class="math-container">$f(x_1, x_1) = x_1^{0.5}x_2^{0.5}$</span>, and factor prices are <span class="math-container">$w_1 = w_2 = 1$</span>. Derive the long run total cost function.</p>
<p><strong>The solution is:</strong></p>
<p>From the first order condition, i.e. the technical rate of substitution equals the ratio of factor prices, we have <span class="math-container">$ -\frac{x_2}{x_1} = -\frac{1}{1} $</span>, i.e. <span class="math-container">$x_1 = x_2$</span>. Thus the conditional factor demand is <span class="math-container">$x_1 = x_2 = q$</span>. So the long run cost function is <span class="math-container">$TC(q) = w_1 x_1 + w_2 x_2 = 2q$</span>.</p>
<p>__</p>
<p>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.</p>
<p>I could not find an explanation in my textbook.</p>

CC BY-SA 4.0
2023-04-06T20:19:30.440Z
This video might help: https://youtu.be/yPlNthLQUEw
Amit
<microeconomics><econometrics><supply-and-demand><production-function><producer-theory>
Solve long run production function of a firm using technical rate of substitution
juliusphysics
# Solve long run production function of a firm using technical rate of substitution

(Asked by: username_0 on 2023-04-06)

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.

**Tags:** microeconomics, econometrics, supply-and-demand, production-function, producer-theory

> **Comments:**
> **username_1**: This video might help: https://youtu.be/yPlNthLQUEw  

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## 1 answers

### Answer by username_2 on 2023-04-06. Score: 3 | Accepted answer

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.


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