<p>Touch wood that I did not make any mistakes.</p>
<p>Consider the production function <span class="math-container">$X = (K + \alpha)(L + \beta)$</span>.</p>
<p>The elasticity of substitution is given by:
<span class="math-container">$$
\frac{\partial \ln(K/L)}{\partial\ln(MP_L/MP_K)} = \frac{\partial(\ln(K) - \ln(L)}{\partial(\ln(K + \alpha)- \ln(L + \beta))}
$$</span>
Let's take the derivative of both numerator and denominator with respect to <span class="math-container">$L$</span>, writing <span class="math-container">$K$</span> as a function of <span class="math-container">$L$</span> itself:
<span class="math-container">$$
\dfrac{\dfrac{K_L}{K}- \dfrac{1}{L}}{\dfrac{K_L}{K + \alpha} - \dfrac{1}{L + \beta}},\\
$$</span>
Now, rewriting the production function, we have that: <span class="math-container">$K = \dfrac{X}{L + \beta} - \alpha$</span>, so keeping <span class="math-container">$X$</span> fixed we have:
<span class="math-container">$$
K_L = -\frac{X}{(L+\beta)^2}
$$</span>
Substitution gives:
<span class="math-container">$$
\begin{align*}
&amp;\frac{-\dfrac{X}{(L+\beta)^2 K} - \dfrac{1}{L}}{-\dfrac{X}{(L+\beta)^2(K+\alpha)}-\dfrac{1}{L + \beta}},\\
&amp;=\frac{\dfrac{-XL-(L+\beta)^2 K}{(L + \beta)^2 KL}}{-\dfrac{X}{X(L+\beta)}-\dfrac{1}{L+\beta}},\\
&amp;= \frac{\dfrac{-(K + \alpha)(L + \beta)L - (L+\beta)^2 K}{(L + \beta)^2KL}}{-\dfrac{2}{(L+\beta)}},\\
&amp;= \dfrac{(K+\alpha)(L + \beta)L+(L+\beta)^2K}{2(L+\beta)KL},\\
&amp;= \dfrac{(K+\alpha)L + (L + \beta)K}{2KL},\\
&amp;= \dfrac{\alpha L+ \beta K + 2 KL}{2KL},\\
&amp;= \dfrac{\alpha L + \beta K}{2KL} + 1
\end{align*}
$$</span></p>

CC BY-SA 4.0
2021-09-30T17:52:08.710Z
2021-09-30T17:52:08.710Z
tdm
<p>For this function, the marginal rate of technical substitution is given by (K+alpha)/(L+beta). Generally we solve for K/L in terms of MRTS of two factors. Then differentiate to solve for elasticity of substitution. In this case, K/L is becoming a function of both MRTS and L also. How to approach the problem in this case.</p>

<microeconomics><production-function><aggregate-production>
What is the elasticity of Substitution of the function X = (K+alpha)(L+beta)
Anonymous
# What is the elasticity of Substitution of the function X = (K+alpha)(L+beta)

(Asked by: username_0 on 2021-09-30)

For this function, the marginal rate of technical substitution is given by (K+alpha)/(L+beta). Generally we solve for K/L in terms of MRTS of two factors. Then differentiate to solve for elasticity of substitution. In this case, K/L is becoming a function of both MRTS and L also. How to approach the problem in this case.

**Tags:** microeconomics, production-function, aggregate-production

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## 1 answers

### Answer by username_1 on 2021-09-30. Score: 2

Touch wood that I did not make any mistakes.

Consider the production function $X = (K + \alpha)(L + \beta)$.

The elasticity of substitution is given by:
$$
\frac{\partial \ln(K/L)}{\partial\ln(MP\_L/MP\_K)} = \frac{\partial(\ln(K) - \ln(L)}{\partial(\ln(K + \alpha)- \ln(L + \beta))}
$$
Let's take the derivative of both numerator and denominator with respect to $L$, writing $K$ as a function of $L$ itself:
$$
\dfrac{\dfrac{K\_L}{K}- \dfrac{1}{L}}{\dfrac{K\_L}{K + \alpha} - \dfrac{1}{L + \beta}},\\
$$
Now, rewriting the production function, we have that: $K = \dfrac{X}{L + \beta} - \alpha$, so keeping $X$ fixed we have:
$$
K\_L = -\frac{X}{(L+\beta)^2}
$$
Substitution gives:
$$
\begin{align\*}
&\frac{-\dfrac{X}{(L+\beta)^2 K} - \dfrac{1}{L}}{-\dfrac{X}{(L+\beta)^2(K+\alpha)}-\dfrac{1}{L + \beta}},\\
&=\frac{\dfrac{-XL-(L+\beta)^2 K}{(L + \beta)^2 KL}}{-\dfrac{X}{X(L+\beta)}-\dfrac{1}{L+\beta}},\\
&= \frac{\dfrac{-(K + \alpha)(L + \beta)L - (L+\beta)^2 K}{(L + \beta)^2KL}}{-\dfrac{2}{(L+\beta)}},\\
&= \dfrac{(K+\alpha)(L + \beta)L+(L+\beta)^2K}{2(L+\beta)KL},\\
&= \dfrac{(K+\alpha)L + (L + \beta)K}{2KL},\\
&= \dfrac{\alpha L+ \beta K + 2 KL}{2KL},\\
&= \dfrac{\alpha L + \beta K}{2KL} + 1
\end{align\*}
$$


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